STATS

WITHOUT

TEARS

Copyright © 2001–2023 by Stan Brown, https://BrownMath.com/
Updated 12 May 2022

Contents

Help » About This Book

This book is an alternative to the usual textbooks for a one-semester course in statistics. Whether you’re teaching in a classroom or learning on your own, you’ve come to the right place.

Being on the Web, this book will get updated frequently, based on your feedback. You can see the revision dates in the chapter list above, and a revision history is shown at the end of every chapter. at https://BrownMath.com/swt/pfswt.htm .

1A.  Statistics? What’s That?

Summary: We live in an uncertain world. You never have complete information when you make a decision, but you have to make decisions anyway. Statistics helps you make sense of confusing and incomplete information, to decide whether a pattern is meaningful or just coincidence. It’s a way of taming uncertainty, a tool to help you assess the risks and make the best possible decisions.

1A1.  What Should You Expect?

Statistics is different from any other math course.

Yes, you’ll solve problems. But most will be real-world practical problems: Does aspirin make heart attacks less likely? Was there racial bias in the selection of that jury? What’s your best strategy in a casino? (Most examples will be from business, public policy, and medicine, but we’ll hit other fields too.)

There will be very little use of formulas. Real statisticians don’t do things by hand. They use calculators or software, and so will you. Your TI-83 or TI-84 may seem intimidating at first, but you’ll quickly get to know it and be amazed at how it relieves you of drudgery.

With little grunt work to do, you will focus on what your numbers mean. You’re not just a button-pushing calculator monkey; you have to think about what you’re doing and understand it well enough to explain it. Most of the time your answers will be non-technical English, not numbers or statistical jargon. That may seem scary and unfamiliar at first, but if you stick with it you’ll love stretching your brain instead of just following a book’s examples by rote.

1A2.  What Do You Get From the Course?

It may be a required course, so you get that much closer to graduation. ☺ But you can get more than that.

If you do it right, statistics teaches you to think. You become skeptical, unwilling to take everything at face value. Instead, when somebody makes a statement you question how they know that and what they’re not telling you. You can’t be fooled so easily. You become a more thoughtful citizen, a more savvy consumer.

Who knows? You might even have some fun along the way.  So— Let’s get started!

1A3.  Sample and Population

Suppose you want to know about the health of athletes who use steroids versus those who don’t. Or you want to know whether people are likely to buy your new type of chips. Or you want to know whether a new type of glue makes boxes less likely to come apart in shipping. How do you answer questions like that?

With most things you want to know, it’s impossible or impractical to examine every member of the group you want to know about, so you examine part of that group and then generalize to the whole group.

Example 1: You want to know what proportion of likely voters will vote for your candidate, so you poll 850 people. The people you actually ask are your sample, and the likely voters are the population.

Caution!: Your sample is the 850 people you took data from, not just the subgroup that said they would vote for your candidate. The population is all likely voters, regardless of which candidate they prefer. Yes, you want to know who will vote for your candidate, but everybody’s vote counts, so the group you want to know something about — the population — is all likely voters.

Example 2: You’re monitoring quality in a factory that turns out 2400 units an hour, so you test 30 units from each hour’s production.

The units you tested are your sample, and your sample size is 30. All production in that hour is the population, and the population size is 2400.

Isn’t the population the factory’s total production, since you want to know about the overall quality? No! Your sample was all drawn from one hour’s production. A sample from one production run can tell you about that production run, not about overall operations. This is why quality testing never ends.

Example 3: You’re testing a new herpes vaccine. 800 people agree to participate in your study. You divide them randomly into two groups and administer the vaccine to one group and a placebo (something that looks and feels like a vaccine but is medically inactive) to another group. Over the course of the study, a few people drop out, and at the end you have 397 vaccinated individuals and 396 who received the placebo.

You have two samples, individuals who were vaccinated (n₁ = 397) and the control group (n₂ = 396). The corresponding populations are all people who will take this vaccine in the future, and all people who won’t. Both of those populations are uncountable or infinite because more people are being born all the time.

1A4.  Descriptive and Inferential Statistics

Sometimes you want to summarize the data from your sample, and other times you want to use the sample to tell you something about the larger population. Those two situations are the two grand branches of statistics.

Definition: Descriptive statistics is summarizing and presenting data that were actually measured. Inferential statistics is making statements about a population based on measurements of a smaller sample.

Example 4: “52.9% of 1000 voters surveyed said they will vote for Candidate A.” That is descriptive statistics because someone actually measured (took responses from) those 1000 people.

Compare: “I’m 95% confident that 49.8% to 56.0% of voters plan to vote for Candidate A.” That is inferential statistics because no one has asked all voters. Instead, a sample of voters was asked, and from that an estimate was made of the feelings of all voters.

1A5.  Statistic and Parameter

Definitions: A statistic is a numerical summary of a sample. A parameter is a numerical summary of a population.

Mnemonic: sample and statistic begin with s; population and parameter both begin with p.

Continuing with Example 4: “52.9% of 1000 voters surveyed plan to vote for Candidate A.” — 52.9% is a statistic because it summarizes the sample.

“I’m 95% confident that 49.8% to 56.0% of voters plan to vote for Candidate A.” — 49.8% to 56.0% is an estimate of a parameter. (The actual parameter is the exact proportion presently planning to vote for A, which you don’t know exactly.)

A statistic is always a statement of descriptive statistics and is always known exactly, because a statistic is a number that summarizes a sample of actual measured data.

A parameter is usually estimated, not known exactly, and therefore is usually a matter of inferential statistics. The exception is a census, in which data are taken from the whole population. In that case, the parameter is known exactly because you have complete data for the population, so the parameter is then descriptive statistics.

1B.  Good Samples, Bad Samples

Definition: A probability sample is a sample where the members are chosen by a predetermined process that uses the workings of chance and removes discretion from the investigators. Some of the types of probability samples are discussed below.

See also: For lots of examples of good sampling and (usually) clear presentation of data about the American people, you might want to visit the Pew Research Center and its tech-oriented spinoff, Pew Internet. The venerable Gallup Poll also makes available its snapshots of the American public.

1B1.  The Gold Standard: Random Samples

Definition: A random sample (also called a simple random sample) is a sample constructed through a process that gives every member of the population an equal chance of being chosen for the sample.

You always want a random sample, if you can get one. But to create a random sample you need a frame, and in many situations it’s impossible or unreasonably difficult to list all members of the population. The sections below explain alternative types of samples that can lead to statistically valid results.

“Random” doesn’t mean haphazard. Humans think we’re good at constructing random sequences of letters and digits, but actually we’re very bad at it. Try typing 1300 “random” letters on your keyboard. If you do it really randomly, you should get about 1300÷26 = 50 of each letter. (Note: about 50 of each, not exactly 50. To determine whether a particular sample of text is unreasonably far from random letters, see Testing Goodness of Fit to a Model.) But if you’re like most people, the distribution will be very different from that: some letters will occur many more than 50 times, and others many less.

So how do you construct a random sample? You need a frame, plus a random-number generator or random-number table.

If you have a table of random numbers, the table will come with instructions for use. I’ll show you how do it with the TI-83/84, but you could also do it with Excel’s RANDBETWEEN( ) function, or with any other software that generates pseudo-random numbers. (The Web site random.org provides true random numbers based on atmospheric noise.)

Seeding the Random-Number Generator

Random numbers from software or a calculator aren’t really random, but what we call pseudo-random numbers. That means that they are generated by deterministic calculations designed to mimic randomness pretty well but not perfectly. To help them do a better job, you need to “seed” the random number sequence, meaning that you give it a unique starting point so that your sequence of random numbers is different from other people’s.

You seed the random numbers only once. To do this:

1. Turn on the calculator and press [CLEAR].
2. Come up with a number through some means other than choosing it. For instance, select the first number you see in the newspaper or in a book that you let fall open where it will. Type this number into the calculator. (Eyes closed, I tapped the financial page with a pen and used the number that the pen touched.)
3. Press [STO→], which shows on your screen as →.
4. Press [MATH] [◄] [1] to paste rand to the screen. Press [ENTER].

Again, you need to seed random numbers only once on your calculator.

Selecting Members of the Sample

For this you need to know the size of the population, which is the number of individuals in your frame. You will generate a series of random numbers between 1 and the population size, as follows:

1. Press [MATH] [◄] [5] to paste randInt( to your screen.
2. Press [1] [,], enter the population size, and press [)] [ENTER] to generate the first random number. In my case the population size was 20,147 and my first random number was 4413, so the first member of my sample will be the 4413th individual, in order, from the sampling frame.
3. Press [ENTER] to generate the next random number. (The randInt function may or may not be displayed again, depending on your calculator model and settings.) In my case, the next random number is 4949, so the 4949th individual in my frame becomes the second member of my sample.
4. Continue pressing [ENTER] until you have your desired sample size. If you get a duplicate random number, simply ignore it and take the next one. (If your calculator has [8] randIntNoRep, use it instead of plain randInt to prevent duplicates from appearing in the first place.)

1B2.  Almost as Good: Systematic Samples

Definition: A systematic sample with k = some number is one where you take every kth individual from a representative subset of the population to make up your sample.

Example 5: Standing outside the grocery store all day, you survey every 40th person. That is a systematic sample with k=40.

If properly taken, a systematic sample can be treated like a random sample. Then why do I call it almost as good? Because you have to make one big assumption: that the variable you’re surveying is independent of the order in which individuals appear. In the grocery-store example, you have to assume that shoppers in the time period when you take your survey are representative of all shoppers. That may or may not be true. For example, a high proportion of Wegmans shoppers at lunch time are buying prepared foods to eat there or take back to work. At other times, the mix of groceries purchased is likely to be different.

Taking a Systematic Sample

1. Estimate the number of individuals you will be sampling from, and call this N. (Here your sampling frame is smaller than the population.) In the grocery-store example, estimate how many shoppers will pass the point where you will stand during the time you’re standing there. If you estimate 1200 shoppers during the six hours when you’ll take your survey, then N=1200.

   If you’re pretty unsure of N, you may need to observe that spot without taking the survey, just to get a preliminary count.

2. Decide how large a sample you want. Divide N by your desired sample size, rounding down, and call the result k. If you want 95 grocery shoppers in your sample, then k = N/95 = 1200/95 = 12.63 → k=12.

   If your estimate of N is uncertain, you’ll want to reduce k a bit. This will increase your sample size, but a sample that’s too large (within reason) is better than one that’s too small.

3. If you have never seeded the random-number generator, do it now. See Seeding the Random-Number Generator, above.
4. Take a random number from 1 to k to determine which person will be first in your sample. To do this, press [MATH] [◄] [5] to paste randInt(, then [1] [,]. Enter the value of k and press [)] [ENTER].

   Caution: It’s 1 to k, not 1 to N. If you need to survey every 12th person, then you use randInt(1,12). For determining where to start in the first 12 people, randInt(1,95) and randInt(1,1200) are both wrong.

   At right you see an illustration with k=12. The calculator has determined that I will start with the 2nd person and take every 12th person after that: 2, 14, 26, 38, 50, and so on.

5. If you reach your desired sample size sooner than expected, keep going for the originally planned time. Why? Because you don’t know whether the individuals that appear early are different from those that appear late. The good news is that the larger sample will give you more accurate results, always a good thing.

1B3.  Good but Hard: Cluster Samples

Sometimes a true random sample is possible but unreasonably difficult. For example, you could use census records to take a random sample of 1000 adults in the US, but that would mean doing a lot of travel. So instead you take a cluster sample.

Example 6: You want to have 600 representative Americans try your new neck pillow to gauge your potential market. Travel to 600 separate locations across the country would be ridiculously expensive, so you randomly select 30 census tracts and then randomly select 20 individuals within each selected census tract.

A cluster sample makes one big assumption: that the individuals in each cluster are representative of the whole population. You can get away with a slightly weaker assumption, that the individuals in all the selected clusters are representative of the whole population. But it’s still an assumption. For this and other technical reasons, a cluster sample cannot be analyzed in all the same ways as a random sample or systematic sample. Analysis of cluster samples is outside the scope of this course.

1B4.  Stratified Samples

Sometimes you can identify subgroups of your population and you expect individuals within a subgroup to be more alike than individuals of different subgroups. In such a case, you want to take a stratified sample.

Definition: If you can identify subgroups, called strata (singular: stratum), that have something in common relative to the trait that you’re studying, you want to ensure that your sample has the same mix of those groups as the population. Such a sample is called a stratified sample.

Example 7: You’re studying people’s attitudes toward a proposed change in the immigration laws for a Presidential candidate. You believe that some races are more likely to favor loosening the law and others are more likely to oppose it. If the population is 66% non-Hispanic white, 14% Hispanic, 12% black, 4% Asian, and so on, your sample should have that same composition.

A stratified sample is really a set of mini-samples grouped together.

Example 8: You want to survey attitudes towards sports at a college that is 45% male and 55% female, and you want 400 in your sample. You would take a sample of 45%×400 = 180 male students and 55%×400 = 220 female students to make up your sample of 400. Each mini-sample would be taken by a valid technique like a random sample or systematic sample.

1B5.  Census

Definition: A census is a sample that contains every member of the population.

In many situations, it’s impossible or highly inconvenient to take a census. But with the near-universal computerization of records, a census is practical in many situations where it never used to be.

Example 9: At the push of a button, a librarian can get the exact average number of times that all library books have been checked out, with no need for sampling and estimation. An apartment manager can tell the exact average number of complaints per tenant. And so forth.

A census is the only sample that perfectly represents the population, because it is the whole population. If you can take a census, you’ve reduced a problem of inferential statistics to one of descriptive statistics. But even today, only a minority of situations are subject to a census. For instance, there’s no way to test a drug on every person with the condition that the drug is meant to treat. It’s totally impractical to interview every potential voter and determine his or her preferences. And so forth.

1B6.  Bogus Samples

Any sample where people select the individual members is a bogus sample. That means every sample where people select themselves, and every sample where the interviewer decides whether to include or exclude individual members.

Why is that bad? Remember, a proper sample is a smaller group that is representative of the population. No sample will represent the population perfectly, but you do the best you possibly can.

The good samples listed above can go bad if you make various kinds of mistakes, mistakes (“Statistical Errors”, later in this chapter), but a sample that doesn’t depend on the workings of chance is always wrong and cannot be made right. The textbooks will give you names for the types of bad samples — convenience sample, opportunity sample, snowball sample, and so on — but why learn the names when they’re all bogus anyway?

So goodbye to Internet polls and petitions, letter-writing campaigns, “the first 500 volunteers”, and every other form of self-selected sample. If people select themselves for a sample, then by definition they are not representative because they feel more strongly than the people who didn’t select themselves. You can make statements about the people who selected themselves, but that tells you nothing about the much larger number who didn’t select themselves. (More about this in Simon 2001 [see “Sources Used” at end of book], Web Polls.)

Goodbye also to any kind of poll where the pollster selects the individual people. If you set up a rule that depends on the workings of chance and then follow it, that’s okay. But if you decide on the spur of the moment who gets included, that’s bogus.

Why is it bad to just approach people as you see them? Because studies show that you are more likely to approach people that you perceive to be like you, even if you’re not aware of that. Ask yourself if you are truly equally likely to select someone of a different race or sex from yourself, someone who is dressed much richer or poorer than you, someone who seems much more or much less attractive, and so forth. Unless you’re Gandhi, the honest answer is “not equally likely”. It doesn’t make you a bad person, just a bad pollster like everyone else. If you tend to pick people who are more like you, your sample is not representative of the population.

The same principle applies to studies of non-humans. Here the investigator’s intrinsic biases may be less clear, but unless you choose your sample based on chance you can never be sure that those biases didn’t “skew it up”.

1C.  Data and Variables

1C1.  What Are Data? What Are Variables?

Example 10: You record the birth weights of babies born in a certain hospital during the year. The variable is “birth weight”.

Example 11: In April, you ask all the members of your sample whether they had the flu vaccine that year and how many days of work or school they lost because of colds or flu. (Can you see at least two problems with that second question? If not, you will after you read about Nonsampling Errors, later in this chapter.) This is bivariate data. One variable is “flu shot?” and the data points are all yes or no; the other variable is “days lost to colds and flu” and the data points are whole numbers.

1C2.  Quantitative or Qualitative?

Sometimes we talk about data types, and sometimes about variable types. They’re the same thing. For instance, “weight of a machine part” is a continuous variable, and 61.1 g, 61.4 g, 60.4 g, 61.0 g, and 60.7 g are continuous data.

Continuous or discrete data? Sometimes when you have numeric data it’s hard to say whether you have discrete or continuous data. But since you’ll graph them differently, it’s important to be clear on the distinction. Here are two examples of doubtful cases: salary and age.

It’s true that your salary can be only a whole number of pennies. But there are a great many possible values, and the distance between the possible values is quite small, so you call salary a continuous variable. Besides, you don’t ask “how many pennies do you make?” but rather “how much do you make?”

What about age? Well, age at last birthday is clearly discrete since it can be only a whole number: “how many years old were you at your last birthday?” But age now, including years and months and days and fractions of days, would be continuous, again because you can subdivide it as finely as desired.

1C3.  Summary Statements

When you see a summary statement, you have to do a little mental detective work to figure out the data type. Always ask yourself, what was the original measurement taken or question asked?

Example 12: “The average salary at our corporation is $22,471.” The original measurement was the salary of each individual, so this is continuous data.

Example 13: “The average American family has 1.7 children.” Don’t let “1.7” fool you into identifying this as a continuous variable! What was the original question or measurement? “How many children are there in your family?” That’s discrete data.

Example 14: “Four out of five dentists surveyed recommend Trident sugarless gum for their patients who chew gum.” Yes, there are numbers in the summary statement, but the original question asked of each dentist was “Do you recommend Trident?” That is a yes/no question, so the data type is categorical.

1D.  Statistical Errors

Summary: In statistics, an error is not necessarily a mistake. This section explores the types of statistical errors and where they come from.

Definition: An error is a discrepancy between your findings and reality. Some errors arise from mistakes, but some are an inevitable part of the sampling process.

1D1.  Sampling Error

Except for a census, no sample is a perfect representation of the population. So the sample mean (average), for example, will usually be a bit different from the population mean. Sampling errors are unavoidable, even if you do everything right when you take a random sample. They’re not mistakes, they’re just part of the nature of things.

Although sampling error cannot be eliminated, the size of the error can be estimated, and it can be reduced. For a given population, a larger sample size gives a smaller sampling error. You'll learn more about that when you study sampling distributions.

1D2.  Nonsampling Errors

Definition: Nonsampling errors are discrepancies between your sample and reality that are caused by mistakes in planning, in collecting data, or in analyzing data.

Nonsampling errors make your sample unrepresentative of the population and your results questionable if not useless. Unlike sampling errors, nonsampling errors cannot be reduced by taking larger samples, and you can’t even estimate the size of most nonsampling errors. Instead, the mistakes must be corrected, and probably a new sample must be taken.

There are many types of nonsampling errors. Different authors give them different names, but it’s much more important for you to recognize the bad practice than to worry about what to name it. In taking your own samples, and in evaluating what other people are telling you about their samples, always ask yourself: what could go wrong here? has anything been done that can make this sample unrepresentative of the population? Here are some of the more common types of nonsampling errors. After you read through them, see how many others you can think of.

Self-Selected Samples

This is almost always bogus. People who select themselves are by definition different from people who don’t, which means they are not representative. It can be very hard to know whether that difference matters in the context of a particular study. Since you can never be sure, it is safest to avoid the problem and not let people select themselves.

But medical studies all use volunteers. (They have to, ethically.) Why doesn’t that make the sample bogus? They’re volunteers, but usually they’re not self-selected volunteers. For example, researchers may ask doctors and hospitals to suggest patients who meet a particular profile; they use probability techniques to select a sample from that pool.

But things are not always simple. For example, some companies or researchers may advertise and pay volunteers to undergo testing. In this case you have to ask very serious questions about whether the volunteers are representative of the general population. Statistical thinking isn’t a matter of black and white, but some pretty sophisticated judgment can be involved. Your take-away is: don’t accept anything at face value, but always ask: What important facts are being left out? What does that do to the credibility of the results?

Sampling Bias

Definition: Sampling bias results from taking a sample in a way that tends to over- or under-represent some subgroup of the population.

Example 15: If you’re doing a survey on student attitudes toward the cafeteria, and you conduct the survey in the cafeteria, you are systematically under-representing students who don’t use the cafeteria. It seems logical that attitudes are more negative among students who don’t use the cafeteria than among students generally, so by excluding them you will report overall attitude as more favorable than it really is.

“Bias” is a good example of the words in statistics that don’t have their ordinary English meaning. You’re not prejudiced against students who dislike the cafeteria. “Bias” in statistics just means that something tends to distort your results in a particular direction.

Example 16: The classic example of sampling bias is the Literary Digest fiasco in predicting that Landon would beat Roosevelt in the 1936 election. The magazine sent questionnaires to all its subscribers, it phoned randomly selected people in telephone books, and it left stacks of questionnaires at car dealerships with instructions to give one to every person who test drove a car. The sample size was in the millions.

This procedure systematically over-represented people who were well off and systematically under-represented poorer people. In 1936 the Great Depression still held sway, and most people did not have the disposable income to subscribe to a fancy magazine, let alone a home telephone; the very thought of buying a car would have struck them as ridiculous or insulting. In that era, the Republicans appealed more to the rich and the Democrats more to the working class. So the net effect of the Literary Digest’s procedure was that it made the country look a lot more Republican than it actually was. Since Landon was a Republican and FDR a Democrat, FDR’s actual support was much greater than shown by the poll, and Landon’s was much less.

Notice that a sample size of millions did not overcome sampling bias. A larger sample size is not an answer to nonsampling errors.

The Digest’s original article can be found in Landon in a Landslide: The Poll That Changed Polling (American Social History Project) [see “Sources Used” at end of book].

While we’re on the subject of presidential elections, different nonsampling errors also led to wrong predictions of a Dewey victory over Truman in 1948. For analyses of both the 1936 and the 1948 statistical mistakes, see Classic Polling Surprises (2004) [see “Sources Used” at end of book] and Introduction to Polling (n.d.) [see “Sources Used” at end of book].

used by permission; source: https://xkcd.com/2618/

Selection Bias

Beyond sampling bias, there are many other bad practices in selecting your sample can bias the results. Wikipedia’s Selection Bias [see “Sources Used” at end of book] has a good rundown of quite a few.

Non-Response Errors

If you’re taking a mail survey, a significant number of people (probably a majority) won’t respond. Are the responders representative of the non-responders, or has a bias been introduced by the non-response? That’s a tough question, and the answer may not always be clear.

For this reason, mail surveys are often coded so that the investigators can tell who did respond, and follow up with those who didn’t. That follow-up can be more mail, a phone call, or a visit.

Even with in-person polls, non-response is a problem: many people will simply refuse to participate in your survey. Depending on what you’re surveying, that could be unimportant or it could be a fatal flaw.

Response Errors

Definition: Response errors occur when respondents give answers that don’t match objective truth or their own actual opinions.

Poorly worded survey questions are a major source of response errors, and lead to biased results or completely meaningless results. There may not be a perfect survey question, but having several people review the questions against a list of possible problems will greatly reduce the level of response errors.

But response errors can never be completely eliminated. For instance, people tend to shade their answers to make themselves look good in their own eyes or in the interviewer’s eyes. Most people rate themselves as better-than-average drivers, for example, which obviously can’t be true. And self-reporting of objective facts is always suspect because memory is unreliable.

Example 17:

• “How often do you read to your child?” (People will tend to award themselves points for good intentions, and they don’t want to look like bad parents.)
• “Do you think immigrants should be allowed to take jobs from honest Americans?” (That’s a leading question. Compare with “Should immigrants be allowed to apply for jobs and pay taxes in the US?” You can see that a given person might give different answers to those two questions.)
• “How much do you spend on food, including groceries and restaurants, in a typical week?” (Most people don’t carry an accurate accounting system in their heads.)
• “Do you agree with the X Party platform on gun control, education, abortion, and taxes?” (Asking too much in one question. If you agree with some but not all, how do you answer?)
• “Do you favor prison reform?” (Too vague — nobody’s against prison reform in principle, but the specifics of a particular policy would make all the difference.)
• “The President has proposed a Federal 30-day waiting period for the purchase of any automatic or semiautomatic weapon. Do you favor this proposal?” (Superficially this looks good: it’s specific, and it’s asking only one thing. But in fact it’s biased by the use of “favor” alone. You should word questions neutrally: “Do you favor or oppose. …” Better yet, “What is your opinion of this proposal?” with options of strongly agree, agree, neutral, disagree, and strongly disagree.
• “In the race for mayor, do you favor candidate A, B, or C?” (Some people are more likely to choose the first alternative because it’s first, and they don’t like to say they have no strong opinion. It is better to vary the order of candidates randomly to avoid a response error in favor of A. Of course, you should also offer “other” and ’not sure” as alternatives.)

Data Errors

These include mistakes by interviewers in recording respondents’ answers, mistakes by investigators in measuring and recording data, and mistakes in entering the recorded data.

Inappropriate Analysis

In the second half of the course you’ll learn a number of inferential statistics procedures. Each one is appropriate in some circumstances and inappropriate in others. If you use the wrong form of analysis in a given situation, or you apply it wrongly, your results will be about as good as the results from using a hammer to drive a screw.

1E.  Observation and Experiment

Summary: There are two main methods of gathering data, the observational study and the experiment. Learn the differences, and what each one can tell you.

1E1.  Observational Study Versus Designed Experiment

Many, many statistical investigations try to find out whether A causes B. To do this, you have two groups, one with A and one without A, or you have multiple groups with different levels of A. You then ask whether the difference in B among the groups is significant. The two main ways to investigate a possible connection are the observational study and the experiment.

The concepts aren’t hard, but there’s a boatload of vocabulary. Let’s get through the definitions first, and then have some concrete examples to show how the terms are used. Please read the definitions first, then read the first example and refer back to the definitions; repeat for the other examples.

Definition: In an observational study, the investigator simply records what happens (a prospective study) or what has already happened (a retrospective study). In an experiment, the investigator takes an more active role, randomly assigning members of the sample to different groups that get treated differently.

Which is better? Well, in an observational study, you always have the problem of wondering whether the groups are different in some way other than what you are studying. This means that an observational study can never establish cause. The best you can do after an observational study is to say that you found an association between two variables.

How do we establish cause, when for ethical or practical reasons we can’t do an experiment? The nine criteria are listed in Causation (Simon 2000b [see “Sources Used” at end of book]) and were first laid down by Sir Austin Bradford Hill in 1965.

Example 18: Over the course of a year, you have parents record the number of minutes they spend every day reading to their child, and at the end of the year you record each child’s performance on standard tests. The explanatory variable is parental time spent reading to the child, and the response variable(s) are performance on the standardized test(s).

Definitions: In an experiment, the experimenter manipulates the suspected cause(s), called explanatory variable(s) or factor(s). A specific level of each factor is administered to each group. The level(s) of the explanatory variable(s) in a given group are known as its treatment.

Example 19: To test productivity of factory workers, you randomly assign them to three groups. One group gets an extra hour at lunch, one group gets half-hour breaks in morning and afternoon, and one group gets six 10-minute breaks spaced throughout the day. The explanatory variable or factor is structuring of break time, and the three levels or treatments are as described.

Definitions: In an experiment, each member of a sample is called a unit or an experimental unit. However, when the experiment is performed on people they are called subjects or participants.

Confounding and Lurking Variables

In Example 18, about reading to children, you find generally that the more time parents spend reading to first graders, the better the children tend to do on standard tests of reading level.

Is the reading time responsible for the improved test scores? You can easily think of other possible explanations. Parents who spend more time reading to their children probably spend more time with them in general. They tend to be better off financially — if you’re working two jobs to make ends meet, you probably have little time available for reading to your children. Economic status and time spent with children in general are examples of lurking variables in this study.

Definition: A hidden variable that isn’t measured and isn’t part of your design but affects the outcome is called a lurking variable.

Example 20: In a large elementary school, you schedule half the second grade to do art for an hour, two mornings a week, with the district’s art teacher. The other half does art for an hour, two afternoons a week, with the same teacher, but they are told at the beginning that all their projects will be displayed and prizes given for the best ones.

Can you learn anything from this about whether the chance to win prizes prompts children to do a better job on art projects? The problem is that there’s not just one difference in treatment here, the promised prizes. There’s also the fact that everyone’s project will be on display. And maybe mornings are better (or worse) for doing art than afternoons. Maybe the teacher is a morning person and fades in the afternoon, or is not a morning person and really shines in the afternoon. Even if there’s a difference in quality of the projects, you can’t make any kind of simple statement about the cause, because of these confounding variables.

In the art example, you wanted to find out whether promising prizes makes children do better art work. But the promise of prizes wasn’t the only difference between the two groups. Time of day and public display are confounding variables built into the design of this experiment. You know what they are, but you can’t untangle their effect from the effect of what you actually wanted to study.

What’s the difference between lurking variables and confounding variables? Both confuse the issue of whether A causes B.

A lurking variable L is associated with or causes both A and B, so any relationship you see between A and B is just a side effect of the L/A and L/B relationships. For example, counties with more library books tend to have more murders per year. Does reading make people homicidal? Of course not! The lurking variable is population size. High-density urban counties have more books in the library and more murders than low-density rural counties.

A confounding variable C is typically associated with A but doesn’t cause it, so when you look at B you don’t know whether any effect comes from A, from C, or from both. For example, after a year with a lot of motorcycle deaths, a state passes a strict helmet law, and the next year there are significantly fewer deaths. Was the helmet law responsible? Maybe, but time is a confounding variable here. Were motorcyclists shocked at the high death toll, so that they started driving more carefully or switched to other modes of transit?

Don’t obsess over the difference between lurking and confounding variables. Some authors don’t even make a distinction. You should recognize variables that make results questionable; that’s a lot more important than what you call them.

That said, if you want to see two more takes on the difference, have a look at Confounding and Lurking Variables (Virmani 2012 [see “Sources Used” at end of book]) and Confounding Variables (Velleman 2005 [see “Sources Used” at end of book]).

Lurking and confounding variables are the boogeyman of any statistical work. Lurking variables are the reason that an observational study can show only association, not causation. In experiments, you have the potential to exclude lurking variables, or at least to minimize them, but it takes planning and extra work, and you need to be careful not to create a design with built-in confounding..

Whenever any experiment claims that A causes B, ask yourself what lurking variables there might be, and whether the design of the study has ruled them out. You can’t take this for granted, because even professional researchers sometimes cut corners, knowingly or unknowingly.

Extended Examples

Example 21: Does smoking cause lung cancer?
Initial studies in the mid-20th century had three or four groups: non-smokers, light smokers, moderate smokers, and heavy smokers. They looked at the number and severity of lung tumors in the groups to see whether there was a significant difference, and in fact they found one.

This was an observational study. Ethically it had to be: if you suspect smoking is harmful you can’t assign people to smoke.

Explanatory variable: smoking level (none, light, moderate, heavy). Levels or treatments don’t apply to an observational study.

Response variable: tumor production

Because this was an observational study, there was no control for lurking variables, and even with a significant result you can’t say from this study that smoking causes lung cancer. What lurking variables could there be? Well, maybe some genetic factor both makes some people more likely to smoke and makes them more susceptible to lung cancer. This is a problem with every observational study that finds an effect: you can’t rule out lurking variables, and therefore you can’t infer causation, no matter how strong an association you find.

Since you can’t do an experiment on humans that involves possibly harming them, how do you know that smoking causes lung cancer? A good explanation is in Causation (Simon 2000b [see “Sources Used” at end of book]).

Example 22: Does aspirin make heart attacks less likely?
Here you can do an experiment, because aspirin is generally recognized as safe. Investigators randomly assigned people to two groups, gave aspirin to one group but not the other, and then monitored the proportion who had heart attacks. They found a significantly lower risk of heart attack in the aspirin group.

This was a designed experiment.

Explanatory variable: aspirin. There were two levels or treatments: yes and no.

Response variable: heart attack (yes/no)

From this experiment, you can say that aspirin reduces the risk of heart attack. How can you be sure there were no lurking variables? By randomly assigning people to the two groups, investigators made each group representative of the whole population. For example, overweight is a risk factor for heart attacks. The random assignment ensures that overweight people form about the same proportion in each group as in the population. And the same is true for any other potential lurking variable. (It helps to have larger samples, and in this study each sample was about 10,000 people.)

Example 23: Does prayer help surgical patients?
Here again, no one thinks prayer is harmful, so ethically the experimenters were in the clear to assign cardiac-bypass patients randomly to three groups: people who knew they were prayed for, people who were prayed for and didn’t know it, and people who were not prayed for. Investigators found no significant difference in frequency of complications between the patients who were prayed for and those who were not prayed for.

This was a designed experiment.

Explanatory variables: receipt of prayer (two levels, yes and no) and knowledge of being prayed for (also two levels, yes and no). There were three treatments: (a) receipt=yes and knowledge=yes, (b) receipt=yes and knowledge=no, (c) receipt=no.

Response variable: occurrence of post-surgical complications (yes/no).

(You can read an abstract of the experiment and its results in Study of the Therapeutic Effects of Intercessory Prayer [Benson 2006 [see “Sources Used” at end of book]]. The full report of the experiment is in Benson 2005 [see “Sources Used” at end of book].)

Because lurking variables can’t be ruled out in an observational study, investigators always prefer an experiment if possible. If ethical or other considerations prevent doing an experiment, an observational study is the only choice. But then the best you can hope for is to show an association between the two variables. Only with an experiment do you have a hope of showing causation.

1E2.  Experimental Techniques

Okay, so you always have to do an experiment if you want to show that A causes B. Let’s look in more detail at how experiments are conducted, and learn best practices for an experiment.

Caution: Design of Experiments is a specialized field in statistics, and you could take a whole course on just that. This chapter can only give you enough to make you dangerous.☺ While you’re planning your first experiment in real life, it’s a good idea to get help from someone senior or a professional statistician.

R. A. Fisher “virtually invented the subject of experimental design” (Upton and Cook 2008, 144 [see “Sources Used” at end of book]), and pioneered many of the techniques that we use today. He was a great champion of planning: Upton & Cook quote him as saying “To call in the statistician after the experiment is done may be no more than asking him to perform a postmortem examination: he may be able to say what the experiment died of.”

Completely Randomized Design

Definitions: Experimenters randomly assign members to the various treatment groups. We say that they have randomized the groups, and this process is called randomization.

Why randomize? Why not just put the first half of the sample in group A and the second half in group B? Because randomization is how you control for lurking variables.

Think about the study with aspirin and heart attacks. You know that different individuals are more or less susceptible to heart attacks. Risk factors include smoking, obesity, lack of exercise, and family history. You want your aspirin group and your non-aspirin group to have the same mix of smokers and non-smokers as the general population, the same mix of obese and non-obese individuals, and so on. Actually it’s harder than that. There aren’t just “smokers” and “non-smokers”; people smoke various amounts. There aren’t just “obese” and “fit” people, but people have all levels of fitness.

It would be very laborious to do stratified samples and get the right proportions for a lot of variables. You’d have to have a huge number of strata. And even if you did do those matchups, taking enormous trouble and expense, what about the variables you didn’t think of? You can never be sure that the samples have the same composition as the population.

It really must be random assignments — you can’t just assign test subjects to groups alternately. Steve Simon (2000a) explains why, with examples, in Alternating Treatments [see “Sources Used” at end of book].

Randomization is the indispensable way out. Instead of trying to match everything up yourself — and inevitably failing — you let impersonal random chance do your work for you.

Are you guaranteed that the sample will perfectly represent the population? No, you’re not. Remember sampling error, earlier in this chapter. Samples vary from the population; that’s just the nature of things. But when you randomize, in the long run most of your samples will be representative enough, even though they’re not perfectly representative.

Randomized Block Design

Notice I said that randomization works in the long run. But in the short run it may not. Suppose you are testing a weight-loss drug on a group of 100 volunteers, 50 men and 50 women. If you completely randomize them, you might end up with 20 men and 30 women in one group, and 30 men and 20 women in the other. (There’s about a 20% chance of this, or a more extreme split.)

Why is this bad? Because you don’t know whether men and women respond differently to the drug. If you see a difference between your 20/30 placebo group and your 30/20 treatment group, you don’t know how much of that is the drug and how much is the difference between men and women. Gender is a confounding variable.

What to do? Create blocks, in this case a block of the 50 men and a block of the 50 women. Then within each block you randomly assign individuals to receive medication or a placebo. Now you can find how the drug affects women and how it affects men. This is called a randomized block design. When you can identify a potentially confounding variable before you perform your experiment, first divide your subjects into blocks according to that variable, and then randomize within each block. Do this, and you have tamed that confounding variable.

R. A. Fisher coined the term “randomized block” in 1926.

In this example, gender would be called a blocking variable because you divide your subjects into blocks according to gender. Now there’s no problem separating the effects of the drug from the effects of gender in your experimental results.

When I talked about complete randomization, I said it would be laborious to take strata of a lot of variables, and that complete randomization was the answer. But here I’m suggesting exactly that for men and women in the weight-loss study. Right about now, you might be telling me, “Make up your mind!”

This is where some judgment is needed in making tradeoffs. Men and women typically have different percentages of body fat, and they are known to respond differently to some drugs. It makes sense that a weight-loss drug could have different response from men and women, and therefore you block on gender. But no other factor stands out as both important and measurable. If you tried to block on motivation, for instance, how would you measure it?

“Block what you can, randomize what you cannot” is a good rule, sometimes attributed to George Box. A variable is a candidate for blocking when it seems like it could make a difference, and you can identify and measure it. For other variables, we depend on randomization, either complete randomization or randomization within blocks.

Matched Pairs

There’s one circumstance where you can be sure that the subgroups are perfectly matched with respect to lurking variables: when you use a matched-pairs design. This is kind of like a randomized block design where each block contains two identical subjects.

Example 24: You want to know whether one form of foreign-language instruction is more effective than another. So you take fifty pairs of identical twins, and assign one twin from each pair to group A and the other twin to group B. Then you know that genetic factors are perfectly balanced between the two groups. And if you restrict yourself to twins raised together, you’ve also controlled for environmental factors.

A special type of matched-pairs design matches each experimental unit to itself.

Example 25: You want to know the effect of caffeine on heart rate. You don’t assemble a sample, give coffee to half of them, and measure the difference in heart rate between the groups. People’s heart rates vary quite a bit, so you would have large variation within each group, and that might swamp the effect you’re looking for.

Instead, you measure each individual’s resting heart rate, then give him or her a cup of coffee to drink, and after a specified time measure the heart rate again. By comparing each individual to himself or herself, you determine what effect caffeine has on each person’s heart rate, and people’s different resting heart rates aren’t an issue.

See also: Experimental Design in Statistics shows how the same experiment would work out with a completely randomized design, randomized blocks, and matched pairs.

Control Group and Placebo

Definition: In an experiment, usually one of the treatments will be no treatment at all. The group that gets no treatment is called the control group.

But “no treatment at all” doesn’t mean just leaving the control group alone. They should be treated the same way as the other groups, except that they get zero of whatever the other groups are getting. If the treatment groups are getting injections, the control group must get injections too. Otherwise you’ve introduced a lurking variable: effects of just getting a needle stick, and in humans the knowledge that they’re not actually getting medicine.

Definition: A placebo is a substance that has no medical activity but that the subjects of the experiment can’t tell from the real thing.

The placebo effect is well known. Sick people tend to get better if they feel like someone is looking after them. So if you gave your treatment group an injection but your control group no injection, you’d be putting them in a different psychological state. Instead, you inject your control group with salt water.

TheProfessorFunk has a fun three-minute YouTube video in the placebo effect (Keogh 2011 [see “Sources Used” at end of book]). Thanks to Benjamin Kirk for drawing this to my attention.

You might think placebos would be unnecessary when experimenting on animals. But if you’ve ever had a pet, you know that some animals are stressed by getting an injection. If the control group didn’t get an injection, you’d have those differing stress levels as a lurking variable. So you administer a placebo.

Example 26: Sometimes, for practical or ethical reasons, you have to get a little bit creative with a control group. Here’s an excellent example from Wheelan (2013, 238) [see “Sources Used” at end of book]:

Suppose a school district requires summer school for struggling students. The district would like to know whether the summer program has any long-term academic value. As usual, a simple comparison between students who attend summer school and those who do not would be worse than useless. The students who attend summer school are there because they are struggling. Even if the summer school program is highly effective, the participating students will probably still do worse in the long run than the students who were not required to take summer school. What we want to know is how the struggling students perform after taking summer school compared with how they would have done if they had not taken summer school. Yes, we could do some kind of controlled experiment in which struggling students are randomly selected to attend summer school or not, but that would involve denying the control group access to a program that we think would be helpful.

Instead, the treatment and control groups are created by comparing those students who just barely fell below the threshold for summer school with those who just barely escaped it. Think about it: the [group of all] students who fail a midterm are appreciably different from [the group of all] students who do not fail a midterm. But students who get a 59 percent (a failing grade) are not appreciably different from those students who get a 60 percent (a passing grade).

Double Blind

Some people do better if they think they’re getting medicine, even if they’re not. To avoid this placebo effect, the standard technique is the double blind.

Definitions: In a double-blind experiment, neither the test subjects nor those who administer the treatments know what treatment each subject is getting. In a single-blind experiment, the test subjects don’t know which treatment they’re getting, but the personnel who administer the treatments do know.

Okay, given that people’s thoughts influence whether they improve, a single blind makes sense. If you let someone know they’re not getting medicine in a trial, they’re less likely to improve. But why isn’t that enough? Why is a double blind necessary?

For one thing, there’s always the risk that a doctor or nurse might tell the subject, accidentally or on purpose. But beyond that, if you’re treating someone who has a terrible disease, you might treat them differently if they’re getting a placebo that if they’re getting real medicine, even if you don’t realize you’re doing it. Why take the risk of introducing another lurking variable? Better to use a double blind and just rule out the possibility.

You might wonder how it’s done in practice. In a drug trial, for instance, each test subject is assigned a code number, and the drug company then packages pills or vaccines with a subject’s code number on each. The doctors and nurses who administer the treatments just match the code number on the pill or vaccine to each subject’s code number. Of course all the pills or vaccines look alike, so the workers who have contact with the subjects don’t know who’s getting medicine and who’s getting a placebo. And what they don’t know, they can’t reveal.

1F.  Sharp Points

1F1.  Rounding and Significant Digits

You’ll be dealing with numbers through most of this course. Handle them right, and you won’t get burned! There are three issues here: how many digits to round to, how to round to that number of digits, and when to do your rounding.

How Many Digits?

There are a lot of rules for how many digits you should round to, but we’re not going to be that rigorous in this course. Instead, you’ll use common sense supplemented by a few rules of thumb. What’s common sense? Avoid false precision, and avoid overly rough numbers.

The rules are important, but we have only so much time, and you’ve probably learned them in your science courses. If you want to, look up “significant figures” or “significant digits” in the index of pretty much any science textbook, or look at Significant Figures/Digits [see “Sources Used” at end of book].

Example 27: When you fill your car’s gas tank, the pump shows the number of gallons to three decimal places. You can also describe that as the nearest thousandth of a gallon. How much gas is that? Convert it to teaspoons (Brown 2018 [see “Sources Used” at end of book]): (0.001 gal) × (4 qt/gal) × (4 cup/qt) × (16 Tbsp/cup) × (3 tsp/Tbsp) ≈ 0.8 tsp. You can bet there’s several times that much in the hose when the pump shuts off. Three decimal places at the gas pump is false precision a/k/a spurious accuracy. That third decimal place is just noise, statistical fluctuations without real significance.

On the other hand, suppose the pump showed only whole gallons. This is too rough. You can go along pumping gas for no extra charge (bad for the merchant), and then abruptly the cost jumps by several dollars (bad for you).

Here are some rules of thumb to supplement your common sense. These are not matters of right and wrong, but conventions to save thinking time:

• Round averages and other statistics to one more decimal place than the original data. If you’ve surveyed families for the number of children in each household, you have a bunch of whole numbers, which have zero decimal places. Your average should have one decimal place.
• Round probabilities to four decimal places unless you show them as exact fractions.
• Round z-scores (Chapter 3 and later) and other test statistics to two decimal places.

How to Round Numbers

Round in one step. Say you have a number 1.24789, and you want to round it to one decimal place. Draw a line — mentally or with your pencil — at the spot where you want to round: 1.2|4789. If the first digit to the right of that line is a 0, 1, 2, 3, or 4, throw away everything to the right of the line. It doesn’t matter what digits come after that first digit. Here, the first digit to the right of the line is a 4, so you throw away everything to the right of the line: 1.24789 rounded to one decimal place is 1.2.

Rounding in multiple steps, 1.24789 → 1.2479 → 1.248 → 1.25 → 1.3, is wrong. (Why? Because 1.24789 is 0.05211 units away from 1.3, but only 0.04789 units away from 1.2.) You must round in one step only.

As you know, if the first digit to the right of the line is a 5, 6, 7, 8, or 9, you raise the digit to the left of the line by one and throw away everything to the right of the line. To one decimal place, 1.27489 is 1.2|7489 → 1.3.

You may need to “carry one”. What is 1.97842 to one decimal place? 1.9|7842 needs you to increase that 9 by one. That means it becomes a zero and you have to increase the next digit over: 1.9+0.1 = 2.0. Therefore, 1.97842 rounded to one decimal place is 2.0.

When to Round Numbers

Here’s the Big No-No: Never do further calculations with rounded numbers. What’s the right way? Round only after the last step in calculation.

Example 28: True story: In Europe, average body temperature for healthy people was determined to be 36.8°C, as repeated in A Critical Appraisal of 98.6°F (Mackowiak, Wasserman, Levine 1992 [see “Sources Used” at end of book]). Rounding to the nearest degree, the average human body temperature is 37°C. So far so good.

But in the US, thermometers for home use are marked in degrees Fahrenheit. Some nimrod converted 37°C using the good old formula 1.8C + 32 and got 98.6°F, and that’s what’s marked on millions of US thermometers as “normal” temperature. If you’ve got one of those, ask for your money back, because it’s wrong.

Why is it wrong? The person who did the conversion committed the Big No-No and did further calculations with a rounded number. For a correct calculation, use the unrounded number, 36.8. (Okay, 36.8 was probably rounded from 36.77 or 36.82 or something. But the point is that it’s the least rounded number available.) 1.8 × 36.8 + 32 = 98.24 → 98.2, and that is the average body temperature for healthy humans.

Randall Munroe shows what can happen if you carry rounding-in-the-middle-of-a-calculation to extremes:

used by permission; source: https://xkcd.com/2585/

1F2.  Powers of 10 from Your Calculator

When a calculation results in a number lower than about 0.0005, your calculator will usually present it in the dreaded scientific notation, like this example. Be alert for this! Your answer is not 1.99 (or however you want to round it). Your answer is 1.99×10^-4.

How do you convert this to a decimal for reading by ordinary humans? (And yes, you should usually do that — definitely, if your work will be read by non-technical people.)

The exponent (the number after the E minus) tells you how many zeroes the decimal starts with, including the zero before the decimal point. 1.99×10^-4 is 0.000 199 or 0.0002.

When a decimal starts with a bunch of zeroes, especially if the decimal is long, many people use spaces to separate groups of three digits. This makes the decimal easier to read.

1F3.  Show Your Work!

Don’t just write down answers; show your work. This is in your own best interest:

• It helps you organize your thoughts, so that you’re less likely to make a mistake.
• If your work is substantially correct, you may get partial credit even if your final answer was wrong.
• Your instructor probably won’t give full credit for a bare answer with nothing to back it up. (My own practice is to write WTCF for “where’d this come from?”)

  “But,” I hear you object, “in the real world, all that matters is getting the right answer.” True enough, but there’s a difference between being in the real world and preparing for the real world. Part of your study is to develop thought and work habits that ensure you will get the right answer when there’s nobody around to check you. You expose your process now, so that problems can be corrected.

How do you show your work?

The general idea is to show enough that someone familiar with the course content can follow what you did.

When evaluating a formula, write down the formula, then on a line below show it with the numbers replacing the letters. Your calculator can handle very complicated formulas in one step, so your next line will be your last line, containing the final answer and any rounding you do. Example:

SEM = σ/√n

SEM = 160/√37

SEM = 26.30383797 → SEM = 26.3

You’ll be using a lot of the menus and commands on the TI-83 or TI-84. Here are some tips:

• Show all command arguments. If you’re using randInt to get five random integers from 1 to 100, write down randInt(1,100,5). That’s the only way your instructor will know that you know how to use that function. If you think the command is randInt(5,100), now is the time to correct that misunderstanding.
• Abbreviate repetitive information. When you put a column of numbers into list 1, you don’t have to write down all the numbers and say “L1”. Instead, just write “x’s in L1” (use the actual column description if it’s not “x’s”).
• Focus on commands, not keystrokes. If you’re doing 1-VarStats L1,L2, write that. For pity’s sake, don’t write all the keystrokes, [STAT] [►] [1] [2nd] [L1] [,] [2nd] [L2] [ENTER]. I put them in this book because you’re just learning them. But someone familiar with the course knows how to get the command, and I hate to think of all the time and paper you could waste.
• Show inputs first, then outputs. Many students show their answer, then as an afterthought they write down the command. Write down the command before you enter it in the calculator. When writing down the outputs, you can omit any that are the same as the inputs.

1F4.  Optional:  ∑ Means Add ’em Up

You’ll find that your calculator does the complicated stuff for you, but here and there I’ve scattered formulas in BTW paragraphs in case you want to peek behind the curtain.

Stats formulas usually need to do the same thing to every member of a data set and then add up the results. The Greek letter ∑, a capital sigma, indicates this. This summation notation makes formulas easier to write — easier to read, too, once you get used to it.

Some examples:

• ∑x = sum of all data points. (x means a data point. If you had to write this out the long way, it would be x₁ + x₂ + x₃ + … + x_n, where n is the size of your data set.)
• ∑x² = square each data point and add up the squares. (∑ is an addition operator, so powers and multiplication happen before the summation.)
• ∑xf = multiply each unique data point by the number of times it occurs, and add up the results. (f means frequency or repetition count.)
• ∑x²f = square each unique data point and multiply by the number of times it occurs, then add up the results.
• ∑(x − x̅)² = take each data point and subtract the average of the whole sample, square the result, and add up all the squares. (x̅ is the average of a sample. The parentheses tell you that you don’t square the average, you square the differences.)

What Have You Learned?

Chapter 2 WHYL →  

Exercises for Chapter 1

Write out your solutions to these exercises, showing your work for all computations. Then check your solutions against the solutions page and get help with anything you don’t understand.

Caution! If you don’t see how to start a problem, don’t peek at the solution — you won’t learn anything that way. Ask your instructor or a tutor for a hint. Or just leave it and go on to a different problem for now. You may find when you return to that “impossible” problem that you see how to do it after all.

Solutions → 

Summary: To make sense out of a mass of raw data, make a graph. Non-numeric data want a bar graph or pie chart; numeric data want a histogram or stemplot. Histograms and bar graphs can show frequency or relative frequency.

Graph Paper
for Free: Why buy an expensive pad of graph paper, especially if you only need a few sheets? You can print your own for free using Incompetech’s Plain Graph Paper PDF Generator and at Math Worksheets Land’s Graph Paper. Both are sources not just for the ordinary square grid, but for various specialty graph papers.

2A.  Non-Numeric Data

Any graph of non-numeric data needs to show two things: the categories and the size of each. Probably you’re already familiar with the two most common types, which are the bar graph and pie chart.

The sizes of categories can be shown as raw counts, called frequencies, or percentages, called relative frequencies. (Relative frequencies can also be shown as decimals, but I think most people respond better to “20%” than “.20”.)

How do you decide whether to show frequencies or relative frequencies? This is a stylistic choice, not a matter of right and wrong. Your choice depends on what’s important, what point you’re trying to make. If your main concern is just with the individuals in your sample, go with frequencies. But if you want to show the relationship of the parts to the whole, show relative frequencies.

2A1.  Bar Graph

Example 1: In fall 2012, the Pew Research Center (2013a) [see “Sources Used” at end of book] surveyed American adults on their habits of reading to their children. The survey included 434 adults who had at least one child under age 12, and the results are shown in the table.

(Remember, you can’t call the data numeric just because you see numbers in a summary statement. You have to go back to the individual data points, which are categorical: “every day”, “a few times a week”, and so on. If the Pew Center had asked “how many days a week do you read to your child?” and got answers 0, 1, 2, 3, 4, 5, 6, and 7, that would be a set of numeric data.)

Your bar chart or bar graph must follow these rules:

• The bars have equal width and equal spacing; they do not touch. Each bar is labeled with its category below the axis.
• Typically for non-numeric data, there’s no One True Order for the categories. Try to find an order that feels natural. If you prefer, you can order the categories from the tallest to the shortest bar; that is called a Pareto chart.
• The frequency or relative-frequency axis (usually the vertical axis) starts at 0, and you need to show the 0 label. That axis is a number line, so tick marks are equally spaced and represent consistent numbering. (The frequency 0 goes next to the horizontal axis. Don’t offset it downward.)
• The height or length of each bar is proportional to the number or percentage of individuals in that category. You can write frequencies or percentages at the top of every bar, but this is optional because you’re labeling your frequency axis.
• The frequency axis always needs a title. The category axis may or may not need a title, depending on whether the graph title and category names make the chart easy enough to understand.

Usually the category axis is horizontal, so the frequency axis and the bars are vertical. But you can also make a horizontal bar chart, where the category axis is vertical and the frequency axis and bars are horizontal.

You can make a bar graph by hand, or use software such as Microsoft Excel. If you make a bar graph by hand, use graph paper and draw the axes and bars with a straightedge — wobbly bars make you look like you had a liquid lunch.

Here’s my bar graph for parents reading to children:

A couple of comments on best practices:

• Notice that I made one square on the vertical axis equal 10 people, or five squares equal 50. That way when I have numbers like 113 or 39 I know how high to draw my bars. If you pick three or four squares per 50 people, you have a much harder job to draw the bars at the correct heights because you have to figure things like “if 50 is 3 squares, then 113 must be 113/(50/3) = about 6.8 squares.” Always pick “nice” numbers for your numeric scales.
• Notice also that I drew horizontal lines at the major milestones. These “gridlines” help the reader assess the heights of the bars more accurately.

Optional:  Bar Graph in Excel

Getting some kind of bar graph out of Excel is easy. But then there’s a lot of fiddling around to reverse some of Excel’s rather strange format choices. Here are instructions for Excel 2010. If you have Excel 2007, 2013, or 2016, you’ll find that they’re pretty similar.

1. Get your categories into one column and your frequencies into the next column. The first row of each column should be the column headings from the table. Don’t enter a total row.
2. With your mouse, highlight all rows and columns of your chart. (It doesn’t matter whether you include the column heads.) Click the Insert tab and then Charts » Column, and select the first 2-D column chart.
   
3. Right-click the useless legend at the right, “Series1” or “Number of Parents”, and select Delete.
4. When you right-clicked the legend, three Chart Tools tabs appeared. On the Layout tab of the ribbon click Chart Title » Above Chart. Click into the chart title and type a better one. (Maybe Excel already gave your chart a title, but “Number of Parents” is the proper title of the frequency axis, not the whole chart.)
5. Click Axis Titles » Primary Vertical Axis » Rotated Title. Click on the words “Axis Title” that appear in the chart, and type the new title “Number of Parents” for your frequency axis.
6. If your category axis needs a title, click Axis Titles » Primary Horizontal Axis » Title Below Axis and enter the axis title.
7. For some reason, the chart has tick marks between the categories. Right-click one of them, select Format Axis, and change Major tick mark type to None. That gives the chart you see here.
   
8. You may have to tweak the formatting of the graph further; here are some suggestions. (If you try something and don’t like the result, press Ctrl-Z to undo the change.)
   • If the category names are long, try shifting them to vertical alignment: right-click on any of them and select Format Axis » Alignment. In Text direction, select Rotate 90°.
   • You may need to resize the whole graph to improve spacing or to make the bars’ heights show better contrast. Look carefully at the frame and you’ll see handles in each corner and the middle of each side. To resize the graph, drag any of the handles.
   • To change fonts of the axis labels or titles, click the element, then click the Home tab in the ribbon. Change font or font size as desired.

If you prefer a horizontal bar chart, it’s easy to make the change. Click into the chart area, then on the Design tab on the ribbon click Change Chart Type » Bar and select the first one.

Okay, well, nothing is that easy! Excel puts the categories in backwards order, so right-click the category axis and select Format Axis » Axis Options » Categories in reverse order. Still on the Axis Options dialog, click Horizontal axis crosses at maximum category.

Bar Graph with Relative Frequencies

The frequency bar graph tells us about the 434 individuals in the Pew Research Center’s sample. But why collect that sample except for what it can tell us about how often parents in general read to their children?

You know from Sampling Error in Chapter 1 that the proportions in the population are probably not the same as the sample, but probably not very far off either. So you compute those proportions and then redraw your graph to show percentages instead of raw counts.

First, total all the frequencies to get the sample size n = 434. (In this case n is given already, but often it isn’t.) Then convert each frequency into a relative frequency. The formula, if you need one, is f/n. For example, 9 parents never read to their under-12 children. The relative frequency is f/n = 9/434 = 0.021 or 2%: 2% of parents never read to their children. Enter that and the other relative frequencies in the table, as shown at right.

You may see some bar graphs with relative frequencies as decimals. There’s nothing wrong with that for technical audiences, but general audiences usually respond better to percentages.

Your relative frequencies may not add up to exactly 100% (or 1.0000), because of rounding. Don’t change any of the numbers to force a total.

Once you have your relative frequencies, you can make your bar graph. Choose round numbers for the tick marks on your relative frequency axis, for example every 5% or every 10%. I won’t inflict another of my sketches on you, but you can see a finished relative-frequency bar graph below.

Optional:  Relative Frequencies in Excel

To my surprise, I found that Excel doesn’t include relative-frequency bar graphs in its repertoire. You have to enter some formulas to compute the relative frequencies, and then create the graph from them. (Of course you could compute the relative frequencies yourself and enter them in Excel as numbers, but whenever possible I like to be lazy and make the computer do the work.)

1. Enter the categories in a column, leave a blank column, and enter the frequencies. If you already have the categories and frequencies in adjacent columns, right-click on the letter at the top of the frequency column and select Insert.
2. Click into the cell below the last frequency, and type “=sum(” (without the quotes). Then with your mouse select the frequencies. Finally, type a closing parenthesis and hit the Enter key.
3. In the address box just above the first column of the spreadsheet, type a unique name such as TOTPARENTS and press the Enter key. This makes it easier to refer to this total cell.
4. Click into the empty relative-frequency cell for the first category. Type an = sign, then click on the first frequency cell. (In the illustration, the relative-frequency cells are in column B and the frequency cells are in column C.) Type /TOTPARENTS (including / mark for division) and press the Enter key.
5. Grab the “handle” at the lower right of the cell you just typed into, and drag it down to fill the Relative Frequency column.
6. Click the % sign in the ribbon to change the decimals to percentages. (The % sign is near the middle of the ribbon, on the Home tab.)

Now highlight the category and relative-frequency columns, click the Insert tab and the first 2-D column chart, and tweak the graph as you did before. Your result should be something like the one you see here.

On this chart, neither axis really needs a label. The percent signs reinforce the message in the chart title that the bars show relative frequencies. And the category names together with the chart title tell the reader exactly what is being represented.

It’s a judgment call where to place tick marks on the relative-frequency axis, and you really need to look at the data to make a decision. Four categories are under 10%, so it makes sense to show the 5% line and help the reader get a sense of the relative sizes. Of course, if you show 5% then you have to show every 5% increment up to the top of the graph.

Side-by-Side Bar Graph

You may want to compare two populations: men and women, for instance, or one year versus another year. To do this, a side-by-side bar graph is ideal. A side-by-side bar graph has two bars for each category, and a legend shows the meaning of the bars.

The two populations you’re comparing are almost never the same size. Therefore side-by-side graphs almost always show relative frequencies rather than frequencies.

Example 2: In Educational Attainment, the Census Bureau (2014) [see “Sources Used” at end of book] showed the educational attainment of the population in selected years 1940 to 2012. I chose the years 1992 and 2012 and prepared this graph to show the change over that 20-year period.

What do you see? Comparing 2012 to 1992, the proportion of the population with no college (the first four categories) declined, and the proportion with some college or a college degree increased. You should be able to see why this has to be a relative-frequency chart: in a frequency chart, the larger population in 2012 would make all the bars taller than the 1992 bars, and you’d be hard put to see any kind of trend.

Stacked Bar Graph

Example 3: Another way to compare two populations is the stacked bar graph. In the side-by-side bar graph, above, each group of bars was one category, and each bar within a group was a population. With the stacked bar graph, you have one bar for each population, and one piece of that bar for each category. (A stacked bar graph is kind of like an unrolled pie chart.)

Here’s a stacked bar graph for the same data set:

What do you see? Look first at the legend that lists the categories, then at the two bars. The top two segments represent some college. In 1992, about 56% of adults had no education beyond high school. But in 2012, only about 42% had a high-school diploma or less, meaning that 58% had at least some college. The proportions of college and no college were reversed in those 20 years.

You can also see that, though the group with four years of high school shrank, it didn’t shrink as much as the group with college grew. In other words, it’s not just more high-school graduates going on to college, it’s a higher proportion of the population entering high school. All the categories without a high-school diploma shrank. In 1992, 20% of adults had less than a high-school diploma and 80% were high-school graduates; in 2012, only about 12% had less than a high-school diploma and 88% had graduated from high school.

What’s the best way to compare two populations? The answer depends on what you’re trying to show. The side-by-side graph seems to be better at showing how each category changed, and the stacked graph is usually better at showing the mix, especially if you want to group the categories mentally. In the side-by-side graph, you can easily see the decline in adults with a fourth-grade education or less, but the shift to a college-educated population is much harder to see. It’s just the opposite with the stacked graph.

As always, get clear in your own mind what you’re trying to show, and then select the type of graph that shows that most clearly.

Did you notice that this stacked bar graph shows relative frequencies? (Maybe you didn’t notice, because it seems like the natural way to go.) A stacked bar graph could show frequencies instead of relative frequencies, if you want to emphasize the different sizes of the populations, but then it becomes harder to compare the mix in the populations.

When you make a stacked bar graph in Excel, there’s no need to pre-compute the percentages. Just select the third type of 2-D column chart, 100% Stacked Column.

2A2.  Making a Table from Scratch

Example 4: In the first example, you were given a table of categories and counts. But more likely you’ll just have a mass of data points, like this:

Before you can make any kind of graph, you need a table to summarize the data. You’re probably tempted to count the number of shovels, the number of balls, and so on, but it’s way too easy to make mistakes that way. Why? Because you have to go over the data set multiple times, and you may count something twice or miss something.

The better procedure is to tally the categories in a table. It’s a win-win: the procedure is faster, and you’re less likely to make a mistake.

Simply go through the data, one item at a time. If you’re seeing a given category for the first time, add it to your list with a tally mark; if that category is already in your table, just add a tally mark. Here’s my table of tallies after going through the first two columns of data:

Please complete your tallies on your own before you look at mine.

After you’ve tallied all the data, count the tallies in each category and total the counts. Of course the total should equal your sample size n. Here’s my complete table:

Always check the total of your frequencies. If it matches the sample size, that’s no guarantee everything is correct; but if it doesn’t match, you know something is wrong.

Once you’ve got your table, you can make a graph by following the procedures above. If you’re publishing the table itself, give just the category names and sizes and the total, but leave out the tallies.

2A3.  Pie Chart

Where a bar graph tends to emphasize the sizes of categories in relation to each other, a pie chart tends to emphasize the categories as divisions of the whole. This distinction is not hard and fast; it’s just a matter of emphasis.

To make a pie chart, you need a compass, or something else that can draw a circle, and you need a protractor. The angle of each segment of the pie will be 360°×f/n, where f is the frequency of the category and n is the sample size — in other words, it’s 360° times the relative frequency, whether you’re showing frequencies or relative frequencies on the pie chart. But in practice, if you’re going to make a pie chart you’ll use Excel or some other software.

Optional:  Pie Chart in Excel

Excel can draw a pie chart for you, but you have to make a bunch of tweaks before it’s usable. There’s one bit of good news: with a pie chart, unlike a bar graph, Excel can compute relative frequencies automatically. I’ll show you how to do that for the data about parents reading to children, for which we made a bar graph earlier.

1. Highlight the categories and frequencies, but not the total. Click the Insert tab and then Pie, and choose the first 2-D pie. You see the result at right.

   Many people stop there, but this is an absolutely horrible design. Readers have to keep looking back and forth to match up the colors, and often there are similar colors. Color-blind people are really screwed, and if you print the chart on a black-and-white printer it’s hopeless. Fortunately you can fix this!

2. You’re going to put the category names with the pie segments, so right-click the legend (the list of categories at the right) and select Delete.
3. Click on the “Number of Parents” title and type in a better one, such as “How Often Parents Read to Children”. (Don’t type the quotes in the title, of course.)
4. In the ribbon, on the Layout tab, click Data Labels » More Data Label Options. Under Label Contains, select Category Name, select either Value or Percentage, and select Show Leader Lines. Under Label Position, select Best Fit. Click Close.
5. You may want to resize the graph to make the labels less crowded, depending on the sizes of the segments. Drag a handle with your mouse, as you did before.

2B.  Numeric Data

2B1.  Histogram for Numeric Data

How can you draw a picture of numeric data? The answer is a histogram.

The term “histogram” was coined by Karl Pearson in lectures some time before 1895.

Example 5: Let’s use the lengths of some randomly selected iTunes songs:

How do you make sense of this? As you might expect, the first step is to make a table. But you don’t want to treat each number as its own category, because that would produce a really uninteresting graph. Instead you create categories, except for numeric data you call them classes. The rules for classes are very simple:

• The classes must cover all the data points.
• They must all be the same width.
• There must be no gaps between classes.

Notice that the rules don’t tell you how many classes there must be, or what width a class must have. That’s where your discretion comes in. You want to pick class boundaries that are “nice” numbers, and you don’t want too many classes or too few. In practice, five to nine classes is usually about the right number.

How does that apply to the iTunes songs? Take a look at the data. The lowest number seems to be 113, and the highest is 837. That gives a range in “nice” numbers of about 100–850. If you set class width to 100 you have eight classes, so that seems about right.

Now go ahead and make your tally marks to create the table. Instead of category names, you use class boundaries. You already know how to make tally marks, so I’ll just give you the results:

Even though the 700–799 class has no data points, it’s still a class and it will occupy the same width in the histogram as any other class. A bar with zero height shows in the histogram as a gap, and that’s good because it emphasizes that there’s something unusual about the point in the 800–899 class (which was 837 seconds).

If the class width is 100, how come the class bounds are 100–199 and not 100–200? In fact, some authors do write these class bounds as 100–200, 200–300, and so on, with the understanding that if a number is right on the boundary it goes in the upper class. All authors agree that the class width is the difference between the lower bounds of two consecutive classes, not the difference between lower and upper bounds of one class. So whether you write 100–199 for the first class or 100–200, the class width is 200 minus 100, which is 100.

Once you have the table, the histogram is straightforward. You can draw the histogram by hand or use Excel. I’ll show Excel later, but here’s my hand-made histogram for the iTunes data.

Notice that you label the data bars on their edges: 100, 200, …, 900, not 100–199, 200–299, …. Label the left edge of each bar, and also the right edge of the last bar. The right edge of the last bar is always one class width more than the left edge, so even if you’ve got 800–899 in your table the last bar’s edges are 800 and 900.

Like all histograms, this one is good at showing the shape of the data (skewed right; see below), the center (somewhere in the upper 200s to 300s), and the spread (from 100ish to 800ish seconds, or about two minutes to 13 minutes). In Chapter 3 you’ll learn how to measure center and spread numerically, but there’s always a place for a picture to help people grasp a data set as a whole.

This data set also shows an outlier, located somewhere in the 800–899 class. Not every data set will have an outlier, of course, and a rare sample might have more than one. When an outlier occurs, your first move is to go back to your original data sheets and make sure that it’s not simply a mistake in entering your data. If it’s a real data point, then you can ask what it means. In this case, the message is pretty simple: tunes generally run up to about 11 or 12 minutes (700 seconds), but the occasional one can be several minutes longer.

Histogram Versus Bar Graph

A histogram is similar to a bar graph, but with the following differences:

For both histogram and bar graph, the frequencies must start at 0. However, in a histogram the data axis typically doesn’t start at zero. You just leave some space between the frequency axis and the first bar, and the scale of the data axis is considered to start at the first bar.

Relative-Frequency Histogram

Though I don’t show it here, you could make a relative-frequency histogram, the same way you made a relative-frequency bar chart. The relative frequencies range from 0 for the 700–799 class to 20/50 = 40% for the 200–299 class.

Optional:  Histogram in Excel

Believe it or not, out of all the chart types in Excel, the standard histogram was not included until Excel 2016. If you’ve got Excel 2016, click Insert » Recommended Charts and select the histogram.

In Excel 2013 and earlier, to make a histogram you have to combine a column chart and a scatterplot (Middleton) [see “Sources Used” at end of book], or download additional software. You can follow the detailed instructions in that document, or you can download the free Better Histogram add-in from TreePlan Software to do the job. (It works in Excel 2010 and later.) through 2016.) If you’re using Better Histogram:

1. Enter all the original numbers in a column in Excel.
2. Double-click the downloaded ZIP file, and within it double-click Better-Histogram-2007. You will have to enable macros.
3. Click the Add-Ins tab in the ribbon, and then Better Histogram.
   • Data Range: Click the “_” button at right and highlight your numbers.
   • Start Value: The lower bound of your first class, not the lowest number in the data.
   • Step Value: Your class width.
   • Stop Value: The right-hand edge of the last class, which in this example is 900 (not 899).
4. Better Histogram will create a new sheet in your workbook with a frequency table and histogram. Click on the chart title and enter a new title. Click on the horizontal axis title and either delete it or change it to more appropriate text. The result is shown at right.
5. Optional: You might wish to jazz up the chart visually. If so, click on the Design tab of Excel’s ribbon and choose a design. Color is fine, but don’t choose different colors for different bars because that can make bars look larger or smaller than they actually are. Here’s what I got from clicking the blue theme.

2B2.  Ungrouped Discrete Data

To make sense of most data sets, you need to group the data into classes. But sometimes your data have only a few different values. In such cases, you probably want to skip the grouping and just have one histogram bar for each different response. The height of the bar tells you how often that response occurred, as usual.

Example 6: A state park collected data on the number of adults in each vehicle that entered the park in a given time interval:

3 1 1 3 3 3 0 7 3 1    3 6 4 5 3 2 3 4 2 3
0 2 2 4 8 3 3 1 3 3    3 4 1 5 2 2 6 3 4 2

There are only nine different values, so it seems a little silly to group them. Instead, just tally the occurrences, as shown at right.

Label ungrouped data under the centers of the bars, just like categorical data, not under the edges. Some authors still make the bars touch because the data are numeric, and others keep the bars separated because the data are ungrouped. I prefer the second approach, but I’ll accept the other. Here’s my histogram:

Caution: This particular data set has at least one occurrence of every value between min and max. But suppose it didn’t; suppose there were no vehicles with 7 adults? In that case, you would draw the histogram exactly the same, except that the bar above “7” would have zero height. The horizontal axis for numeric data must always have a consistent scale for its whole length, so you never close up any gaps.

Optional:  Ungrouped Discrete Histogram in Excel

You can graph ungrouped discrete data in Excel, if you wish. The key is to fool Excel into treating the data like categorical data:

1. Type the unique values in one column. But as you type each number, type an apostrophe (') first. Don’t put 0, 1, 2 and so on in the cells, but '0, '1, '2. The apostrophe won’t appear, but it tells Excel to treat the numbers like text. (You may notice that Excel left justifies those numbers.)
2. Type the frequencies in a second column.
3. Highlight the numbers in both columns, and on the Insert tab click Column. Select the first 2-D column.
   
4. Make all the same adjustments you made for the bar graph, above.

   By the way, you might notice that the tick marks on the vertical axis are every two cars on this graph, but they were every five cars on my hand-drawn histogram. One is not better than the other; it’s a stylistic choice.

5. Optional: If you want to make the bars touch, right-click on the graph, select Format Data Series, and under Series Options change Gap Width to 0%. Then click Border Color and select Solid Line with a color of white.

2B3.  Shapes of Data Sets

You should know the names of the most common shapes of numeric data. Why? It’s easier to talk about data that way, and — as you’ll see in the next chapter — you treat different-shaped distributions a little differently.

The first question is whether the data set is symmetric or skewed. The histogram of a symmetric data set would look pretty much the same in a mirror; a skewed data set’s histogram would look quite different in a mirror.

If a distribution is skewed, you say whether it’s skewed left or skewed right. A distribution that is skewed left, like the first one below, has mostly high scores, and a distribution that is skewed right, like the second one below, has mostly low scores. The direction of skew is away from the bulk of the data, toward the long skinny tail, where there are few data points.

Skewed left or negatively skewed	Skewed right or positively skewed

Example 7: Scores on a really easy test would be skewed left: most people get high scores, but a few get low or very low scores.

Lifespan in developed countries is skewed left: there are relatively few infant and child deaths, and most people live into their 60s, 70s, or 80s. (The first graph in Calculus Applied to Probability and Statistics [Waner and Costenoble 1996] [see “Sources Used” at end of book] illustrates this.)

People’s own evaluation of their driving skills and safety are left skewed: few people rate themselves below average and most rate themselves above average. Illusory Superiority [see “Sources Used” at end of book] cites a study by Svenson showing this “Lake Wobegon effect”.

Example 8: People’s departure times after a concert would be skewed right: most people leave shortly before or after the performers finish, but a few straggle out for some time afterward. Skewed-right distributions are more common than skewed-left distributions.

Salaries at almost any corporation are another good example of a distribution that is skewed right: most people make a modest wage, but a few top people make much more.

There are several types of symmetric distributions, but here are the two you’ll meet most often. A uniform distribution is one where all possible values are equally likely to occur. The normal distribution has a precise definition, which you’ll meet in Chapter 7, but for now it’s enough to say that it’s the famous bell curve, with the middle values occurring the most often and the extreme values occurring much less often.

You’ll notice that both of the examples below are “bumpy”. That’s usual. In real life you pretty much never meet an exact match for any distribution, because there are always lurking variables, measurement errors, and so on. And even if a population does perfectly follow a given distribution, like the probability distributions you’ll meet in Chapter 6, still a sample doesn’t perfectly reflect the population it came from: sampling error is always with us. When we say that a data set follows such-and-such a distribution, we mean it’s a close match, not a perfect match.

Uniform	Normal (“bell curve”)

Example 9: Winning lottery numbers are uniformly distributed. (In the short term some numbers occur more often than others, but over the long run they tend to even out.)

The results of rolling one die many times are uniformly distributed. (But the results of rolling two dice are not uniformly distributed: 7 is the most likely, 2 and 12 are tied for least likely, and the other numbers are intermediate.)

The normal distribution or bell curve occurs very often, and in fact many natural and industrial processes produce normal distributions. This happens so often that we often just say or write ND for “normal distribution” or “normally distributed”.

Example 10: Men’s and women’s heights follow separate normal distributions. People’s arrival times at an event are ND. IQ scores, and scores on most tests, are ND. The amount of soda in two-liter bottles is ND. Your commute times on a given route are ND.

2B4.  Stem Plot

Suppose you have a discrete data set with few repetitions. An ungrouped histogram would have most bars at the same low height; a grouped histogram might show a pattern but you’d lose the individual data points.

If your discrete data set isn’t too large (n < 100, give or take), and the range isn’t too great, you can eat your cake and have it too. The stemplot, also known as a stem-and-leaf diagram, is a mutant hybrid between a histogram and a simple list of data.

The idea is that you take all the digits of each data point except the last digit and call that the stem; the last digit is the leaf. For example, consider scores of 113 and 117. They are two leaves 3 and 7 on a common stem 11 (meaning 110).

To construct a stemplot, you look over your data set for the minimum and maximum, then write the stems in a column, from lowest to highest. Just like with a histogram, there are no gaps, so if you have data in the 50s and the 70s but not in the 60s you still need a stem of 6.

However, your stems probably won’t start at 0. Start them with the lowest data point that actually occurred, and end them with the highest data point that actually occurred.

The stemplot was invented by John Tukey in 1970.

Example 11: Here is a set of IQ scores from 50 randomly-selected tenth graders:

 99  77  83 111 141      89  98  84  93 124
110  73  96  60 102      87 123 120 100  95
100  90 104  85 129      81 119 112 103  76
108  91  94 114 108      92  96  94  88 101
117 106 103 105 113      97 106 109  80 116

To make your stemplot, eyeball the data for the minimum and maximum, which are 60 and 141. Write the stems, 6 to 14, in a column at the left of your paper, starting several lines below the top. Then draw a vertical line just to the right of them.

Now go through the data points, one by one, and add each leaf to the proper stem. During this process, you might find a value outside what you thought were the min and max. That’s no problem. Just add the stem and then the leaf. (Again, the stems can’t have gaps, so if your first stem is 6 and you come across a data point 47, you have to add stems 4 and 5, not just 4.)

Finally, add a title and a legend or key to your stemplot. Here is the result:

            IQ Scores
 6 | 0
 7 | 7 3 6
 8 | 3 9 4 7 5 1 8 0
 9 | 9 8 3 6 5 0 1 4 2 6 4 7
10 | 2 0 0 4 3 8 8 1 6 3 5 6 9
11 | 1 0 9 2 4 7 3 6
12 | 4 3 0 9
13 |
14 | 1
                    key: 11 | 7 = 117

If you lie down and look at this sideways, it looks like a histogram. But the bonus is that you can still see all the actual data points within the groupings of 60–69, 70–79, etc.

A stemplot is great at showing shape, center and spread of distributions plus outliers, but most data sets don’t lend themselves to a stemplot. If your data set is too large, your leaves will run off the edge of the page. If your data set is too sparse — if the range is large for the number of data points — most of your stems won’t have leaves and the plot won’t really show any patterns in the data. But when you have a moderate-sized data set and the data range is moderate, a stemplot is probably better than a histogram because the stemplot gives more information.

One last touch is sorting the leaves. I don’t think that’s important enough to take the extra effort in a homework problem or on a quiz, but if you’re going to be presenting your stemplot to other people then you probably want to sort the leaves. Here’s the same stemplot with sorted leaves:

            IQ Scores
 6 | 0
 7 | 3 6 7
 8 | 0 1 3 4 5 7 8 9
 9 | 0 1 2 3 4 4 5 6 6 7 8 9
10 | 0 0 1 2 3 3 4 5 6 6 8 8 9
11 | 0 1 2 3 4 6 7 9
12 | 0 3 4 9
13 |
14 | 1
                    key: 11 | 7 = 117

A glance at this stemplot shows you quite a lot. The data set is normally distributed, the center is around 100 points, the spread is 60–141, and there’s an outlier at 141.

2C.  Bad Graphs

You now know how to make good graphs, so be on the lookout for bad graphs. Sometimes they’re bad just because whoever drew them didn’t know any better, or didn’t think. But some people may deliberately try to deceive you with a graph.

Example 12: File this one under “what were they thinking?” The left-hand graph doesn’t have a title, so you don’t know what “Yes” and “No” mean. You have to look back and forth between the graph and the legend, and anyone with red-green color blindness probably won’t be able to see which segment is which. Oh yes — what percentages of the sample answered “Yes” and “No”? You can guess that it’s around a third versus two thirds, but that’s not very precise.

The right-hand graph cures those problems. It’s now crystal clear which segment is Yes and which is No, and what proportion of the sample gave each answer. This actually lets you show more information in less space, a win-win. (Of course you wouldn’t use a vague term like “Opinions” — that’s just there to remind you to give your graph a title.)

Example 13: There’s no telling whether this one is deliberate deception or just incompetent graphing. An oatmeal company, which shall remain nameless, wanted to show that eating oatmeal for four weeks reduces cholesterol. The first graph makes a strong case — until you look at the scale on the vertical axis. (Don’t even think about wasting your time on a graph with no vertical scale.)

The scale doesn’t start at zero, so it makes differences look much bigger than they are. Your frequency or relative frequency scale must always start at zero (and you must show the zero). The second graph is properly drawn, and now you can see that the drop in cholesterol is only a slight one.

Example 14: It’s all very well to create visual interest, but not if it makes the reader misinterpret the graph.

In the left-hand graph, you can tell from the scale that B is supposed to be three times as large as A, but since it’s three times as high and three times as wide it’s actually nine times as large, giving the reader a distorted impression of the amount of difference. Even if your “bars” are pictures, they still have to be the same width. The corrected version is shown at right. (It’s still not quite correct, though, because 0 is not shown on the vertical axis.)

2D.  Really Good Graphs

If you follow the rules in this chapter, you’ll make good, professional graphs. But there are plenty of other ways to make good graphs, depending on the data you’re trying to show.

There’s a classic picture book that can give you lots of good ideas. Edward Tufte’s The Visual Display of Quantitative Information has been around since 1983, and no one has yet done it any better. (Tufte has produced newer editions.)

Example 15: One famous graph in Tufte’s book is particularly stunning. Charles Minard wanted to present a lot of time-series data about Napoleon’s disastrous campaign in Russia in the winter of 1812–1813: where battles took place, numbers of casualties, temperature, and so forth. He elected to make a kind of stylized map showing just the rivers and the cities where events happened. (Niemen at the left is the Niemen River, Russia’s western border at the time. Moscow, “Moscou” in French, is as far east as Napoleon got.) Across that, Minard showed the army strength as a broad swath at the start that shrank to almost nothing by the end of the retreat westward. Below are dates of events, temperatures, and precipitation. It’s a huge amount of information on one piece of paper.

This tiny rendition doesn’t do it justice, but if you click on it visit http://upload.wikimedia.org/wikipedia/commons/2/29/Minard.png you’ll see it at a better size. (Your browser may still reduce it to fit on your screen. Try clicking into the picture and you should see it at original size, though you’ll have to scroll around to see the details. It sounds like a lot of effort, but I promise you it’s worth it. Or just get the book, because it has plenty more!)

Example 16: Here’s one I ran across in my reading. It’s not the graph of the century like Minard’s, but it’s a cut above the usual. In Bear Attacks: Their Causes and Avoidance (2002), Stephen Herrero had the problem of contrasting bears’ diet in spring, summer, and fall. (Of course in winter they’re not eating.)

He could have drawn three pie charts, or a stacked bar graph, but instead he came up with a great alternative. (You can click on the picture to enlarge it.) (A larger version is at https://BrownMath.com/swtpic/chap02_beardiet.jpg.) Each component of diet is clearly labeled right in the graph, not in some legend off to the side, and the contrasting backgrounds make it a little more interesting visually. A stacked bar graph would convey the same information, but I like this presentation because it suggests that “spring”, “summer”, and “fall” are not completely separate but rather transition one into the next.

The vertical axis is clearly labeled, too. There’s no doubt what the numbers are (as opposed to some units of weight, for instance, or something more esoteric like pounds of feed per hundreds of pounds of bear).

He probably could have left off the title off the category axis — after all, we know that the seasons are seasons, and the graph title also conveys that information. But that’s a minor point. My only real quibble with this graph is that the overall graph title at the bottom is too small.

What Have You Learned?

Overview: With numeric data, the goal of descriptive stats is to show shape, center, spread, and outliers.

Chapter 3 WHYL → ← Chapter 1 WHYL

Exercises for Chapter 2

Write out your solutions to these exercises, showing your work for all computations. Then check your solutions against the solutions page and get help with anything you don’t understand.

Caution! If you don’t see how to start a problem, don’t peek at the solution — you won’t learn anything that way. Ask your instructor or a tutor for a hint. Or just leave it and go on to a different problem for now. You may find when you return to that “impossible” problem that you see how to do it after all.

Solutions → 

3A.  Measures of Center

3A1.  The Three M’s: Mean, Median, Mode

There are three common measures of the center of a data set: mean, median, and mode.

The formula for the mean is x̅ = ∑x/n or μ = ∑x/N, meaning that you add up all the numbers in the data set and then divide by sample size or population size.

Example 1: You’re interviewing at a company. You ask about the average salary, and the interviewer tells you that it’s $100,000. That sounds pretty good to you. But when you start work, you find that everybody you work with is making $10,000. What went wrong here?

The interviewer told the truth, but left out a key fact: Everybody but the president makes below the average. Eight employees make $10,000 each, the vice president makes $50,000, and the president makes $870,000. Yes, the mean is (8×10,000 + 50,000 + 870,000)/10 = $100,000, but that’s not representative because the president’s salary is an outlier. It pulls the mean away from the rest of the data, and skews the salary distribution toward the right. This graph tells the sad tale:

There was your mistake. Salaries at most companies are strongly skewed right, so most employees make less than the average. When a data set is skewed, the mean is pulled toward the extreme values. (A data set can be skewed without outliers, but when there are outliers the data set is almost certain to be skewed.)

You should have asked for the median salary, not the average (mean) salary. There are 10 employees, and 50% of 10 is 5, so the median is less than or equal to five data points and greater than or equal to five data points. The fifth-highest and sixth-highest salaries are both $10,000, so the median is $10,000.

The median is more representative than the mean when a data set is skewed. The mean is pulled toward an extreme value, but the median is unaffected by extreme values in the data set. We say that the median is resistant.

Example 2: What is the median of the data set 8, 15, 4, 1, 2? Put the numbers in order: 1, 2, 4, 8, 15. There are five numbers, and 50% of 5 is 2.5. You need the number that is above 2 data points and below 2 data points; the median is 4.

Example 3: What is the median of the data set 7, 24, 15, 1, 7, 45? There are six data points, and in order they are 1, 7, 7, 15, 24, 45. 50% of 6 is 3; you need the number that is above 3 data points and below 3 data points. It’s clear that the median is between 7 and 15, but where exactly? When the sample size is an even number, the median is the average of the two middle numbers. Therefore the median for this data set is the average of 7 and 15, (7+15)/2 = 11.

3A2.  Mean, Median, Mode, and the Shape of a Data Set

When a distribution is symmetric, the mean and median are close together. If it’s unimodal, the mode is close to the mean and median as well.

But have you ever taken a course that was graded on a curve, and one or two “curve wreckers” ruined things for everyone else? What happened? Their high scores raised the class average (mean), so everybody else’s scores looked worse. The class scores were skewed right: low scores occurred frequently, and high scores were rare. (You can see shapes of skewed distributions in Chapter 2.)

When a distribution is skewed, the mean is pulled toward the extreme values. The median is resistant, unaffected by extreme values. And you can reverse that logic too: if the mean is greater than the median, it must be because the distribution is skewed right. From the median to the mean is the direction of skew.

Skewed left, mean < median (usually)		Skewed right, mean > median (usually)

For heaven’s sake, don’t memorize that! Instead, just draw a skewed distribution and ask yourself approximately where the mean and median fall on it.

Karl Pearson gives the rule median = (2×mean + mode)/3 for moderately skewed distributions. For more about this, see Empirical Relation between Mean, Median and Mode [see “Sources Used” at end of book].

Caution! All the statements in this section are a rule of thumb, true for most distributions. The logic holds for almost every unimodal continuous distribution, and for discrete distributions with a lot of different values. But it tends to break down on discrete distributions that have only a few different values. For more about this, see von Hippel 2005 [see “Sources Used” at end of book].

3B.  Summary Numbers on the TI-83 …

Summary: The 1-VarStats command gives you mean, median, and much more for any data set. If you have just a plain list of numbers, enter the name of that list on the command line. If you have a frequency distribution, enter the name of the data list and the name of the frequency list on the command line.

Excel: Excel can do these computations. This isn’t an Excel course, but if you’re an Excel head you can figure out how to get this information. One way is with the Data Analysis tool, part of the Analysis Toolpak add-in that comes with Excel (though you may have to enable it). Another way is to click in a blank cell, click Formulas » More Functions » Statistical and select the appropriate worksheet function.

3B1.  … from a List of Numbers

Example 4: Professor Marvel had a statistics class of fifteen students, and on one quiz their scores were

10.5   13.5   8   12   11.3   9   9.5   5   15   2.5   10.5   7   11.5   10   10.5

Your TI-83 or TI-84 can give you the mean, median, and other numbers that summarize this data set.

1. If you have any partial commands visible, press [CLEAR].
2. Press [STAT] [ENTER] to get into the edit screen for statistics lists. You can use any list, but let’s use L1 this time. (If you don’t see L1, and pressing the left arrow doesn’t bring it into view, press [STAT] [5] [ENTER] [STAT] [ENTER].)
3. Cursor to the L1 label at the top — not the top number, the column heading — and press [CLEAR] [ENTER] to clear the list.
4. Enter your numbers, pressing [ENTER] after each one.
5. After entering the last number, check all the numbers carefully and make any needed corrections. If you duplicated a number, press [DEL] to remove it; if you left out a number, press [2nd DEL makes INS] to open a space for it.
6. Press [STAT] [►] [1] to select 1-VarStats.
   • If you have a newer TI-84 with the “wizard” interface selected, a little menu will appear. Identify the list that contains your data: [2nd 1 makes L1]. For a simple list of numbers like this one, there is no frequency list, so press [DEL].
   • If you have an older calculator or you’ve turned off the “wizard’ interface, the calculator will paste 1-VarStats to the home screen. On the same line, identify the list that contains your data: [2nd 1 makes L1].
7. After writing down the complete command on your paper — 1-VarStats L1 — press [ENTER] to execute it.

The results screen is shown below. A down arrow on the screen says that there is more information if you press [▼], and an up arrow says that there is more information if you press [▲].

Look first at the bottom of the screen. Always check n first — if it doesn’t match your sample or population size, the other numbers are big sacks of bogosity. In this case a quick count of the original data set shows 15 numbers, which is the right quantity. (Of course, this check can’t determine if you miskeyed any numbers. Only double and triple checking can protect you from that kind of mistake.)

What are you seeing on this screen?

• x̅ is the mean. The calculator doesn’t know whether a given data set is a sample or a population, so it can’t guess whether to display x̅ or μ. This data set is quiz scores from the complete class, so it’s a population and you write down μ = 9.7. Always use the right symbols, even if the calculator doesn’t.

  A word about rounding: The rules for significant digits and rounding are beyond the scope of this course, but beware of being ridiculously precise. (For example, most gasoline pumps are calibrated in 0.001 gallon units. But 0.001 gallon is two tablespoons, and there’s considerably more gas than that in the hose, so that precision is just silly.)

  A good rule of thumb is to report sample statistics and population parameters to one more decimal place than the original data. Then why did I say μ = 9.7 instead of 9.72, since the original data have one decimal place? That’s a valid question, and my answer is that 9.72 would not be wrong but it feels overly precise when there are only fifteen data points, most are whole numbers, and the rest are a whole number plus ½.

• ∑x and ∑x² are the sum of the original data and the sum of the squares of the original data. You won’t use them in this course, and there’s no reason to write them down.
• Sx and σx are the standard deviation, computed by two methods. Choose Sx with a sample and σx with a population. More when we get to Measures of Spread!

  Since this data set is a population, select σx = 3.057929583 and write down σ = 3.1.

  The name standard deviation was created in 1893 by Karl Pearson. (We might wish that he had chosen something with fewer than six syllables.) He assigned the symbol σ to the standard deviation of a population in 1894.

• n is the sample size or population size. Since you have a population, you write down N = 15, not n=15.
• minX and maxX are the smallest and largest data points.
• Q1 and Q3 are the first and third quartiles, which we’ll get to under Measures of Position.
• Med is the median, which you met earlier. Med = 10.5.
• The minimum, Q1, median, Q3, and maximum are together called the five-number summary. I’ll have more to say about the five-number summary later in this chapter.

Showing your work and your results, you write down:

1-VarStats L1

μ = 9.7

σ = 3.1

N = 15

min = 2.5

Q1 = 8

Med = 10.5

Q3 = 11.5

max = 15

3B2.  … from an Ungrouped Distribution

Example 5: Your TI-83 or TI-84 can also compute statistics of a frequency distribution. Let’s try it with the data from Chapter 2 for number of adults in vehicles entering the park.

Enter the data values in one statistics list, such as L1. Enter the frequencies in a second list, such as L2. Press [STAT] [►] [1] to select 1-VarStats.

• In the “wizard” interface, enter [2nd 1 makes L1] for List and [2nd 2 makes L2] for FreqList.
• In the non-“wizard” interface, press [2nd 1 makes L1], then [,] (comma), then [2nd 2 makes L2] and [ENTER]. The data list must come first and the frequency list second.

  Caution — rookie mistake: Students often leave off the frequency list. Your calculator is pretty good, but it can’t read your mind. The only way it knows that you have a frequency distribution is if you give it both the frequency list and the data list.

Either way, write down the complete command on your paper: 1-VarStats L1,L2.

Here are the results:

Again, look at n first. That protects you from the rookie mistake of leaving off the frequency list. If n is wrong, redo your 1-VarStats command and this time do it right.

These forty vehicles are obviously not all the vehicles that enter the park, so they are a sample, not a population. You therefore write down the statistics as follows:

1-VarStats L1,L2

x̅ = 3.0

s = 1.7 (from Sx = 1.73186575)

n = 40

min = 0

Q1 = 2

Med = 3

Q3 = 4

max = 8

Weighted Average

Sometimes you take an average where some data points are more important than others. We say that they are weighted more heavily, and the mean that you compute in this way is called a weighted average or weighted mean.

You’re intimately familiar with one example of a weighted average: your GPA or grade point average.

Example 6: The NHTSA’s Corporate Average Fuel Economy or CAFE Rule (NHTSA 2008) [see “Sources Used” at end of book] specifies a corporate average of 34.8 mpg (miles per gallon) for passenger cars. Let’s keep things simple and suppose that ZaZa Motors makes three models of passenger car: the Behemoth gets 22 mpg, the Ferret gets 35 mpg, and the Mosquito gets 50 mpg. Does ZaZa meet the standard?

To answer that, you can’t just average the three models: (22+36+50)/3 = 36 mpg. Suppose the company sells one Mosquito and the rest are Behemoths and a sprinkling of Ferrets? You have to take into account the number of cars of each model sold. In effect, you have a frequency distribution with mpg figures and repetition counts. Let’s suppose these are the sales figures:

Put the miles per gallon in L1 and the frequencies in L2. (How do you know it’s not the other way around? You’re trying to find an average mpg, so the mpg numbers are your data.) You should find:

1-VarStats L1,L2

μ = 32.3 mpg

N = 370,000 passenger cars

Even though two of the three models meet the standard, the mix of sales is such that ZaZa Motors’ CAFE is 32.3 mpg, and it’s not in compliance.

The formula for the mean of a grouped distribution and the formula for a weighted average are the same formula: μ = ∑xf/N for a population or x̅ = ∑xf/n for a sample. Either way, take each data value times its frequency. Add up all those products, and divide by the population size or sample size. For the notation, see ∑ Means Add ’em Up in Chapter 1.

3B3.  … from a Grouped Distribution

In a grouped frequency distribution, one number called the class midpoint stands for all the numbers in the class.

Definition: The class midpoint for a given class equals the lower boundary plus half the class width. This is half way between the lower class boundary of this class and the lower class boundary of the next class.

Example 7: Let’s revisit the lengths of iTunes songs from the ungrouped histogram in Chapter 2. What is the midpoint of the 300 to 399 class?

The class width equals the difference between lower boundaries: 400−300 = 100. Half the class width is 50, so the midpoint is 300+50 = 350. You could also compute the class midpoint as (300+400)/2 = 350.

However, it is wrong to take (300+399)/2 = 349.5 as class midpoint or 399−300 = 99 as class width. Don’t use the upper boundary in finding the class midpoint.

Of course you don’t have to compute every class midpoint the long way. Once you have the midpoint of the first class, (100+200)/2 = 150, just add the class width repeatedly to get the rest: 250, 350, … 850. The grouped frequency distribution, with the class midpoints, is shown at right.

What good is the class midpoint? It’s a stand-in for all the numbers in its class. Instead of being concerned with the nine different numbers in the 100 to 199 class, twenty different numbers in the 200 to 299 class, and so on, we pretend that the entire data set is nine 150s, twenty 250s, and so on. This means you get approximate statistics, but you get them with a lot less work.

Is this legitimate? How good is the approximation? Usually, quite good. In most data sets, a given class holds about equally many data points below the class midpoint and above the class midpoint, so the errors from making the approximation tend to balance each other out. And the bigger the data set, the more points you have in each class, so the approximation is usually better for a larger data set.

Procedure: Enter the class midpoints in one statistics list, such as L1. Enter the frequencies in another list, such as L2. Enter the command 1-VarStats L1,L2 and write down the complete command on your paper.

Again, avoid the rookie mistake: include the class-midpoint list and the frequency list in your command.

The results screens are below. As usual, before you look at anything else, check that n matches the size of the data set. 50 is correct, so that’s one less worry.

There’s a problem with the second screen, though. Your calculator knows you have a frequency distribution, because you gave two lists to the 1-VarStats command. But it doesn’t have the original data, so it doesn’t know the true minimum (lowest data point). When you read minX=150, you interpret that to mean that the lowest data point occurs in the class whose midpoint is 150; in other words, the minimum is somewhere between 100 and 199. Your knowledge of the rest of the five-number summary has the same limitation. For instance, the median isn’t 250; all you know is that it occurs somewhere between 200 and 299.

Because of these limitations, you don’t do anything with the second results screen from a grouped distribution. The mean and standard deviation don’t have this problem: they’re approximate, but the approximation is good enough. (n is exact, not an approximation.)

These 50 iTunes songs are obviously not all the songs there are, not even all the songs in any particular person’s iTunes library. They are a sample, not a population. Therefore you write down your work and results like this:

1-VarStats L1,L2

x̅ = 316 (or you could write 316.0)

s = 145.1

n = 50

3C.  Measures of Spread

There are four common measures of the spread of a data set: range, interquartile range or IQR, variance, and standard deviation. (You may also see spread referred to as dispersion, scatter, variation, and similar words.)

3C1.  Range and IQR (Interquartile Range)

Definition: The range of a data set is the distance between the largest and smallest members.

Example 8: If the largest number in a data set is 100 and the smallest is 20, the range is 100−20 = 80, regardless of what numbers lie between them and what shape the distribution might have.

Caution: The range is one number: 80, not “20 to 100”.

Obviously the range has a problem as a measure of spread: It uses only two of the numbers. Since only the two most extreme numbers in the data set get used to compute the range, the range is about as far from resistant as anything can be.

In favor of the range is that it’s easy to compute, and it can be a good rough descriptor for data sets that aren’t too weird. The interquartile range has something of the same idea, but it is resistant.

You’ll learn about percentiles and quartiles in the next section, Measures of Position, but for now let’s just take a quick non-technical example.

Example 9: Consider the data set 1, 2, 3, 3, 3, 4, 5, 8, 11, 11, 15, 23. There are twelve numbers, and the middle 50% (six numbers) are 3, 3, 4, 5, 8, 11. The interquartile range is 11−3 = 8.

3C2.  Standard Deviation

The IQR is a better measure of spread than the range, because it’s resistant to the extreme values. but it still has the problem that it uses only two numbers in the data set. Isn’t there some measure of spread that uses all the numbers in the data set, as the mean does? The answer is yes: the variance and the standard deviation use all the numbers.

Your calculator gives you the standard deviation, as you saw above. The variance is important in a theoretical stats course, but not so much in this practical course. We’ll measure spread with the standard deviation almost exclusively. (To save wear and tear on my keyboard and your printer, I’ll often use the abbreviation SD.)

If you’d like to know how the variance and SD are computed, read the “BTW” section that follows. Otherwise, skip down to “What Good Is the Standard Deviation, Anyway?”

To see how the variance is computed, let’s go back to Professor Marvel’s quiz scores. We computed the mean as 9.7, or to use the unrounded value, μ = 9.72. (Never round numbers if you’re going to use them in further calculation; that’s the Big No-no.)

If you want to devise a measure of spread, it seems reasonable to consider spread from the mean, so try subtracting the mean from each quiz score and then adding up all those deviations. You get zero, so obviously “sum of deviations” isn’t a useful measure of spread.

But with the next column you strike gold. Squaring all the deviations changes the negatives to positives, and also weights the larger deviations more heavily. This is progress! Now divide the total of squared deviations by the population size and you have the variance: σ² = 140.2640/15 = 9.3509. (σ is the Greek letter sigma.)

(When computing the variance of a sample, you divide by n−1 rather than n. The reasons are technical and are explained in Steve Simon’s articles Degrees of Freedom (1999a) [see “Sources Used” at end of book] and Degrees of Freedom, Part 2 (2004) [see “Sources Used” at end of book].

The variance is quite a good measure of spread because it uses all the numbers and combines their differences from the mean in one overall measure. But it’s got one problem. If the data are dollars, the squared deviations will be in square dollars, and therefore the variance will be in square dollars. What’s a square dollar? (No, I don’t know either.) You want a measure of spread that is in the same units as the original data, just like the mean and median are. The simplest solution is to take the square root of the variance, and when you do that you have the standard deviation (SD), σ = √(140.2640/15) = 3.05793, which rounds to 3.1. And because the standard deviation is in the same units as the original data, it can be used as a yardstick, as you’ll see below.

For lovers of formulas, here they are. The standard deviation of a population, σ, has population size N on the bottom of the fraction; the standard deviation of a sample, s, has sample size n minus 1 on the bottom of the fraction. If you’re not familiar with the ∑ notation (sigma or summation), ∑x² means square every data value and add the squares; ∑x²f means square every data value, multiply by the frequency, and add those products. For the notation, see ∑ Means Add ’em Up in Chapter 1.

Formulas for Standard Deviation
	of a List of Numbers	of a Frequency Distribution
When the data set is the whole population
When the data set is just a sample

Why are there two formulas on each row under “list of numbers”? The first formula is the definition, and the second is a shortcut for faster computations. Of course they’re mathematically equivalent; you could prove that if you wanted to.

Sir Ronald Fisher coined the term variance in 1918. He used the symbol σ² for the variance of a population, since Pearson had already assigned σ to the standard deviation, and the variance is the square of the SD.

What Good Is the Standard Deviation, Anyway?

The standard deviation will be the key to inferential statistics, starting in Chapter 8, but even within the realm of descriptive statistics there are some applications. In addition to this section, you’ll see an application in z-Scores, below.

Working with the quiz scores on your TI-83 or TI-84, you found that the population mean was μ = 9.7 and the population SD was σ = 3.1. What does this mean?

Just as a concept, the standard deviation gives you an idea of the expected variation from one member of the sample or population to the next. The SD in this example is about a third of the mean, so you expect some variation but not a lot. But can you do better than this? Yes, you can!

The Empirical Rule for Normal Distributions

You can predict what percentage of the data will be within a certain number of standard deviations above or below the mean. In a normal distribution, 68% of the data are between one SD below the mean and one SD above the mean (μ±σ), 95% are within two SD of the mean (μ±2σ), and 99.7% are within three SD of the mean (μ±3σ).

This is the Empirical Rule or 68–95–99.7 Rule. Caution! It’s good for normal distributions only.

You’ll notice that the 68%, 95%, and 99.7% of data occur within approximately one, two, and three SD of the mean. More accurate figures are shown in the pictures, but for now we’ll just use the simple rule of thumb. You’ll learn how to make precise computations in Chapter 7.

It’s not a traditional part of the Empirical Rule, but another useful rule of thumb is that, in a normal distribution, about 50% of the data are within 2/3 of a SD above and below the mean.

Example 10: Adult women’s heights are normally distributed with μ = 65.5″ and σ = 2.5″. (By the way, different sources give different values for human heights, so don’t be surprised to see different figures elsewhere in this book.) How tall are the middle 95% of women?

Solution: The middle 95% of the normal distribution lies between two SD below and two SD above the mean. 2σ = 2×2.5 = 5″, and 65.5±5 = 60.5″ to 70.5″, so 95% of women are 60.5″ to 70.5″ tall.

Actually there are two interpretations. You can say that 95% of women are 60.5″ to 70.5″ tall, or you can say that if you randomly select one woman the probability that she’s 60.5–70.5″ tall is 95%. Any probability statement can be turned into a proportion statement, and vice versa. You’ll learn about this in Interpreting Probability Statements in Chapter 5.

Example 11: What fraction of women are 65.5″ to 68″ tall?

Solution: 68−65.5 = 2.5, so 68″ is one standard deviation above the mean. You know that 68% of a normal distribution is within μ±σ. You also know that the normal distribution is symmetric, so 68%/2 = 34% of women are within one SD below the mean, and 34% are within one SD above the mean. Therefore 34% of women are 65.5″ to 68″ tall.

You can combine the three diagrams above and show data in regions bounded by each whole number of standard deviations, like this:

Where do these figures come from? For example, how do we know that about 13.5% of the population is between one and two standard deviations below the mean in a normal distribution? Well, 95% is between two SD below and two SD above the mean. Half of 95% is 47.5%, so 47.5% of the population is between the mean and two SD below the mean. Similarly, about 68% is between one SD below and one SD below, so 68/2 = 34% is between the mean and one SD below. But if 47.5% is between μ−2σ and μ — call it Region A — and 34% is between μ−σ and μ, then the part of Region A that is not in the 34% is the part between μ−2σ and μ−σ, and that must be 47.5−34 = 13.5%. If you had an afternoon to kill, you could work out the other seven percentages.

With this diagram, you can work Example 11 more easily, directly reading off the 34% figure for women between mean height and one SD above the mean. You can also work more complicated examples, like this one.

Example 12: If you randomly select a woman, how likely is it that she’s taller than 70.5″?

Solution: 70.5−65.5 = 5.0, so 70.5″ is two SD above the mean. From the diagram, you see that 2.35+0.15 = 2.5% of the population is more than two SD above the mean. Answer: a randomly selected woman has a 2.5% of being more than 70.5″ tall.

Optional:  Chebyshev’s Inequality

If you have a normal distribution, the Empirical Rule tells you how much of the population is in each region. What if you don’t have a normal distribution?

As you might expect, the portions of the population in the various regions depends on the shape of the distribution, but Chebyshev’s Inequality (or Chebyshev’s Rule) gives you a “worst case scenario” — no matter how skewed the distribution, at least 75% of the data are within 2 SD of the mean, and at least 89% are within 3 SD of the mean.

More generally, within k SD above and below the mean, you will find at least (1−1/k²)·100% of the data. (If you plug in k = 1, you’ll find that at least 0% of the data lie within one SD of the mean. Distributions where all the data are more than one SD away from the mean are unusual, but they do exist.)

Example 13: For the quiz scores, two standard deviations is 2×3.0579 = 6.1, so you expect at least 1−1/2² = 1−¼ = 75% of the quiz scores to be within the range 9.7±6.1 = 3.6 to 15.8. Remember that this is a worst case. In fact, 14 of the 15 numbers (93%) are within those limits.

3D.  Measures of Position

3D1.  Percentiles

Percentiles are most often used in measures of human development, like your child’s performance on standardized tests, or an infant’s length or weight.

Example 14: Your daughter takes a standardized reading test, and the report says that she is in the 85th percentile for her grade. Does this make you happy or sad? Solution: 85% of her grade read as well as she does, or less well; only 15% read better than she does. Presumably this makes you happy.

Example 15: Consider the data set 1, 4, 7, 8, 10, 13, 13, 22, 25, 28. (To find percentiles, you have to put the data set in order.)

(a) What is the percentile rank of the number 13? Solution: There are ten numbers in the data set, and seven of those are ≤13. Seven out of ten is 70%, so the percentile rank of 13 is 70, or “13 is at the 70th percentile”, or P70 = 13.

(b) Find P60 for this data set. Solution: What number is greater than or equal to 60% of the numbers in the data set? Counting up six numbers from the beginning, you find … 13 again. So 13 is both P60 and P70.

(Anomalies like this are usual when you have small data sets. It really doesn’t make sense to talk about percentiles unless you have a fairly large data set, typically a population like all third graders or all six-week-old infants.)

3D2.  Quartiles

One fourth is 25% and three fourths is 75%, so Q1 = P25 and Q3 is P75. (I chose a definition of percentiles that makes this happen. Some authors use different definitions, which may give slightly different results.)

What, no Q2? There is a Q2, but two quarters is one half, or 50%, so the second quartile is better known as the median: 50% of the data are less than or equal to the 50th %ile, alias Q2, alias the median.

The quartiles and the median divide the data set into four equal parts. We sometimes use the word quartile in a way that reflects this: the “bottom quartile” means the part of the data set that is below Q1, and the “`upper quartile” or “top quartile” means the part of the data set that is above Q3.

Q1 and Q3 are part of the five-number summary (later in this chapter). From Measures of Spread, you already know that they’re used to find the interquartile range, and later in this chapter you’ll use the IQR to make a box-whisker plot.

Just like percentiles, quartiles are defined slightly differently by different authors. Dr. Math gives a nice, clear rundown of different ways of computing quartiles in Defining Quartiles in The Math Forum (2002) [see “Sources Used” at end of book]. I follow Moore and McCabe’s method, which is also used by your TI-83 or TI-84.

3D3.  z-Scores

You’ll use z-scores more than any other measure of position. (Remember that every measure of position measures the position of one data point within the sample or population that it is part of.)

How do you find out how many SD a number is above or below the mean of its data set? You subtract the mean, and then divide the result by the SD.
z-score within a sample:      z-score within a population:
Either way, it’s

When you compute a z-score, the top and bottom of the fraction are both in the same units as the original data, and therefore the z-score itself has no units. z-scores are pure numbers.

What good are z-scores? You’ll use them in inferential statistics, starting in Chapter 9, but you can also use them in descriptive statistics.

For one thing, a z-score gives you economy in language. Instead of saying “at least 75% of the data in any distribution must lie between two standard deviations below the mean and two standard deviations above the mean”, you can say “at least 75% of the data lie between z = ±2.”

A z-score helps you determine whether a measurement is unusual. For instance, how good is an SAT verbal score of 300? Scores on the SAT verbal are ND with mean of 500 and SD of 100, so z = −2. The Empirical Rule tells you only 2½% of students score that low or lower.

And z-scores are also good for comparing apples and oranges, as the next example shows.

Example 16: You have two candidates for an entry-level position in your restaurant kitchen. Both have been to chef school, but different schools, and neither one has any experience. Chris presents you with a final exam score of 86, and Fran with a final exam score of 67. Which one do you hire?

At first glance, you’d go with the one who has the higher score. But wait! Maybe Fran with the 67 is actually better, and just went to a tougher school. So you ask about the average scores at the two schools. Chris’s school had a mean score of 76, and Fran’s school had a mean score of 59. Assuming that the students at the two schools had equal innate ability, Fran went to a tougher school than Chris.

Chris scored 10 points above the school average, while Fran scored only 8 points above the school average. Now do you hire Chris? Not yet! Maybe there was more variability in Chris’s class, so 10 points above the average is no big deal, but there was less variability in Fran’s, so 8 points above the mean is a big deal. So you dig further and find that the standard deviations of the two classes were 8 and 4. At this point, you make a table:

The z-scores tell you that Fran stands higher in Fran’s class than Chris stands in Chris’s class. Assuming that the two classes as a whole were of equal ability, Fran is the stronger candidate.

3E.  Five-Number Summary

Definition: The five-number summary of a data set is the minimum value, Q1, median, Q3, and maximum value (in order).

The five-number summary combines measures of center (the median) and spread (the interquartile range and the range). A plot of the five-number summary, called a box-whisker diagram (below), shows you shape of the data set.

On the TI-83 or TI-84, the five-number summary is the second output screen from 1-VarStats. Caution! Remember that the second screen is meaningful only for a simple list of numbers or an ungrouped distribution, not for a grouped distribution. To produce a five-number summary, you need all the original data points.

Example 17: Here is the second output screen from the quiz scores earlier in this chapter. The five-number summary is 2.5, 8, 10.5, 11.5, 15.

The median is 10.5, meaning that half the students scored 10.5 or below and half scored 10.5 or above.

The interquartile range is Q3−Q1 = 11.5−8 = 3.5. Half of the students scored between 8 and 11.5.

3E1.  Outliers

Example 18: Here again are the quiz scores from earlier in this chapter:

10.5   13.5   8   12   11.3   9   9.5   5   15   2.5   10.5   7   11.5   10   10.5

Find the outliers, if any.

The five-number summary, above, gave you the quartiles: Q1 = 8 and Q3 = 11.5. The interquartile range is 11.5−8 = 3.5, and 1.5 times that is 5.25. The fences are 8−5.25 = 2.75 and 11.5+5.25 = 16.75. All the data points but one lie within the fences; only 2.5 is outside. Therefore 2.5 is the only outlier in this data set.

You can find outliers more easily by using your TI-83 or TI-84; see below.

Why do you care about outliers? First off, an outlier might be a mistake. You should always check all your data carefully, but check your outliers extra carefully.

But if it’s not a mistake, an outlier may be the most interesting part of your data set. Always ask yourself what an outlier may be trying to tell you. For example, does this quiz score represent a student who is trying but needs some extra help, or one who simply didn’t prepare for the quiz?

What do you do with outliers? One thing you definitely don’t do: Don’t just throw outliers away. That can really give a false picture of the situation.

But suppose you have to make some policy decision based on your analysis, or run a hypothesis test (Chapters 10 and 11) and announce whether some claim is true or false?

One way is to do your analysis twice, once with the outliers and once without, and present your results in a two-column table. Anyone who looks at it can judge how much difference the outliers make. If you’re lucky, the two columns are not very different, and whatever decision must be made can be made with confidence.

But maybe the two columns are so different that including or excluding the outliers leads to different decisions or actions. In that case, you may need to start over with a larger sample, change your data collection protocol, or call in a professional statistician.

For more on handling outliers, see Outliers (Simon 2000d) [see “Sources Used” at end of book].

3E2.  Box-Whisker Diagrams

The five-number summary packs a lot of information, but it’s usually easier to grasp a summary through a picture if possible. A graph of the five-number summary is called a boxplot or box-whisker diagram.

The box-whisker diagram was invented by John Tukey in 1970.

A box-whisker diagram has a horizontal axis, which is the number line of the data, and the number line need not start at zero. Either the axis or the chart as a whole needs a title, but there’s usually no need for a title on both. There is no vertical axis.

For the graph itself, first identify any outliers and mark them as squares or crosses. Then draw a box with vertical lines at Q1, the median, and Q3. Lastly, draw whiskers from Q3 to the greatest value in the data set that isn’t an outlier, and from Q1 to the smallest value in the data set that isn’t an outlier.

Example 19: Let’s look at a box-whisker plot of those same quiz scores, which were

10.5   13.5   8   12   11.3   9   9.5   5   15   2.5   10.5   7   11.5   10   10.5

The five-number summary is reproduced at right. You recall from the previous section that there is one outlier, 2.5, so the smallest number in the data set that isn’t an outlier is 5.

Here’s a plot that I made with StatTools from Palisade Corporation:

Box-Whisker Plot, and Shape of a Data Set

The box-whisker plot is almost as good as a histogram for showing you the shape of a distribution. If one whisker is longer than the other, and especially if there are outliers on the same side as the long whisker, the distribution is skewed in that direction. If the whiskers are about the same length and there are no outliers, but one side of the box is longer than the other, that usually indicates skew in that direction as well.

Example 20: In the boxplot of quiz scores, just above, you see an outlier on the left side, and the left side of the box is longer than the right. That indicates that the distribution is left skewed.

Box-Whisker Plot on TI-83/84/89

You can use your TI-83 or TI-84 to make a box-whisker plot. The calculator comes with that ability — see Box-Whisker Plots on TI-83/84 — but it’s easier to use MATH200A Program part 2. See Getting the Program for instructions on getting the program into your calculator.

(If you have a TI-89, see Box-Whisker Plots on TI-89.)

To make a box-whisker plot with the program, begin by entering the numbers into a statistics list, such as L1. (If you have an ungrouped frequency distribution, put the numbers in one list and the frequencies in a second list. You need the original data for a boxplot, so you can’t make a boxplot of a grouped frequency distribution.)

Now press [PRGM]. If you can see MATH200A in the list, press its menu number; otherwise, use the [▼] or [▲] key to get to MATH200A, and press [ENTER].

With the program name on your home screen, press [ENTER] (again) to run the program, and yet again to dismiss the title screen. You’ll then see a menu. Press [2] for box-whisker plot.

The program asks whether you have one, two, or three samples. Select 1, since that’s what you have.

The program wants to know whether you have a plain list of numbers or a grouped frequency distribution. Since you have a plain list, choose 1.

The program needs to know which list holds the numbers to be plotted.

Finally, the program presents the box-whisker plot.

Finding Outliers with the TI-83/84/89 or Excel

When you have a box-whisker plot on your screen, whether you used MATH200A part 2 or the calculator’s native commands, if you see any outliers press [TRACE] and then [◄] or [►] to find which data points are outliers.

(For the TI-89, see Box-Whisker Plots on TI-89. If you prefer to use Excel to find outliers, see Normality Check and Finding Outliers in Excel.)

Five-Number Summary from TI-83/84/89 Boxplot

After pressing the [TRACE] key, you can get the five-number summary by pressing [◄] or [►] repeatedly. If there are outliers at the left, use the lowest one for the minimum (first number in the five-number summary); if there are outliers at the right, use the highest one for the maximum (last number in the five-number summary).

What Have You Learned?

Overview: With numeric data, the goal of descriptive stats is to show shape, center, spread, and outliers.

Chapter 4 WHYL → ← Chapter 2 WHYL

Exercises for Chapter 3

Write out your solutions to these exercises, showing your work for all computations. Then check your solutions against the solutions page and get help with anything you don’t understand.

Caution! If you don’t see how to start a problem, don’t peek at the solution — you won’t learn anything that way. Ask your instructor or a tutor for a hint. Or just leave it and go on to a different problem for now. You may find when you return to that “impossible” problem that you see how to do it after all.

Solutions → 

Intro: When you get two numbers from each member of the sample (bivariate numeric data), you make a plot to look for a relationship between them. If a straight line seems like a good fit for the plotted points, we say that they follow a linear model. In this chapter, you’ll learn when to use a linear model, and how to find the best one.

4A.  Mathematical Models

The chapter intro talks about points following a “linear model”. But what is a linear model, and what does it mean to follow one? Well, since a linear model is one kind of mathematical model, let’s talk a little bit about mathematical models.

You know what a model is in general, right? A copy of the original, usually smaller and with unimportant details left out. Think of model airplanes, or architect’s models of buildings.

A mathematical model is like that. Real Life is Complicated,™ and mathematical models help us manage those complications.

Definition: A mathematical model is a mathematical description of something in the real world. An object or process or data set follows a model if the calculations you do with the model match reality closely enough to be useful.

You’ve already met one model in Chapter 3: the grouped frequency distribution. Instead of dealing with all the data points, you do calculations using the midpoint of each class. That gives you approximate mean and SD, but the approximation is close enough to be useful.

The MathIsFun site has a nice example of modeling the space inside a cardboard box, going beyond the h×w×l formula; see Mathematical Models.

You’ll meet plenty more models in this book: probability models in Chapter 5, several discrete models in Chapter 6, and the normal model in Chapter 7.

But in this chapter we’re concerned with the linear model.

The linear model is a good one if it describes the data well enough to let you make useful calculations.

4B.  Scatterplot, Correlation, and Regression on TI-83/84

4B1.  Step 0. Setup

The calculator will remember these settings when you turn it off: next time you can start with Step 1.

4B2.  Step 1. Make the Scatterplot

Before you even run a regression, you should first plot the points and see whether they seem to lie along a straight line. If the distribution is obviously not a straight line, don’t do a linear regression. (Some other form of regression might still be appropriate, but that is outside the scope of this course.)

Let’s use this example from Sullivan (2011, 179) [see “Sources Used” at end of book]: the distance a golf ball travels versus the speed with which the club head hit it.

Turn off other plots.	[Y=] Cursor to each highlighted = sign or Plot number and press [ENTER] to deactivate.
Set the format screen.	Press [2nd ZOOM makes FORMAT]. Just select everything in the left column.
Enter the numbers in two statistics lists.	[STAT] [1] selects the list-edit screen.   Cursor onto the label L1 at top of first column, then [CLEAR] [ENTER] erases the list. Enter the x values.   Cursor onto the label L2 at top of second column, then [CLEAR] [ENTER] erases the list. Enter the y values.
Set up the scatterplot.	[2nd Y= makes STAT PLOT] [1] [ENTER] turns Plot 1 on.   [▼] [ENTER] selects scatterplot.   [▼] [2nd 1 makes L1] ties list 1 to the x axis.   [▼] [2nd 2 makes L2] ties list 2 to the y axis.   (Leave the square as the selected mark for plotting.)
Plot the points. I have the grid turned on in some of these pictures, but earlier I told you to turn it off. That’s simplest. If you want the grid, you can turn it on, but then you’ll have to adjust the grid spacing for almost every plot. To adjust grid spacing, press [WINDOW], set Xscl and Yscl to appropriate values for your data, and press [GRAPH] to see the result.	[ZOOM] [9] automatically adjusts the window frame to fit the data.
Check your data entry by tracing the points.	[TRACE] shows you the first (x,y) pair, and then [►] shows you the others. They’re shown in the order you entered them, not necessarily from left to right.

A scatterplot on paper needs labels (numbers) and titles on both axes; the x and y axes typically won’t start at 0. Here’s the plot for this data set. (The horizontal lines aren’t needed when you plot on graph paper.)

When the same (x,y) pair occurs multiple times, plot the extra ones slightly offset. This is called jitter. In the example at the right, the point (6,6) occurs twice.

If the data points don’t seem to follow a straight line reasonably well, STOP! Your calculator will obey you if you tell it to perform a linear regression, but if the points don’t actually fit a straight line then it’s a case of “garbage in, garbage out.”

For instance, consider this example from DeVeaux, Velleman, Bock (2009, 179) [see “Sources Used” at end of book]. This is a table of recommended f/stops for various shutter speeds for a digital camera:

If you try plotting these numbers yourself, enter the shutter speeds as fractions for accuracy: don’t convert them to decimals yourself. The calculator will show you only a few decimal places, but it maintains much greater precision internally.

You can see from the plot at right that these data don’t fit a straight line. There is a distinct bend near the left. When you have anything with a curve or bend, linear regression is wrong. You can try other forms of regression in your calculator’s menu, or you can transform the data as described in DeVeaux, Velleman, Bock (2009, ch 10) [see “Sources Used” at end of book] and other textbooks.

4B3.  Step 2. Perform the Regression

Your input screen should look like this, for the “wizard” and non-wizard interfaces:

Write down the slope a, the y intercept b, the coefficient of determination R², and the correlation coefficient r. (A decent rule of thumb is four decimal places for slope and intercept, and two for r and R².)

a = 3.1661, b = −55.7966

R² = 0.88, r = 0.94

Now let’s take a look in depth at each of those.

Correlation Coefficient, r

“Several sets of (x,y) [pairs], with the correlation coefficient for each set. Note that correlation reflects the noisiness and direction of a linear relationship (top row), but not the slope of that relationship (middle), nor many aspects of nonlinear relationships (bottom).”
source: Correlation and Dependence [see “Sources Used” at end of book]

Look first at r, the coefficient of linear correlation. r can range from −1 to +1 and measures the strength of the association between x and y. A positive correlation or positive association means that y tends to increase as x increases, and a negative correlation or negative association means that y tends to decrease as x increases. The closer r is to 1 or −1, the stronger the association. We usually round r to two decimal places.

Karl Pearson developed the formula for the linear correlation coefficient in 1896. The symbol r is due to Sir Francis Galton in 1888.

For real-world data, 0.94 is a pretty strong correlation. But you might wonder whether there’s actually a general association between club-head speed and distance traveled, as opposed to just the correlation that you see in this sample. Decision Points for Correlation Coefficient, later in this chapter, shows you how to answer that question.

Be careful in your interpretation! No matter how strong your r might be, say that changes in the y variable are associated with changes in the x variable, not “caused by” it. Correlation is not causation is your mantra.

It’s easy to think of associations where there is no cause. For example, if you make a scatterplot of US cities with x as number of books in the public library and y as number of murders, you’ll see a positive association: number of murders tends to be higher in cities with more library books. Does that mean that reading causes people to commit murder, or that murderers read more than other people? Of course not! There is a lurking variable here: population of the city.

When you have a positive or negative association, there are four possibilities: x might cause changes in y, y might cause changes in x, lurking variables might cause changes in both, or it could just be coincidence, a random sample that happens to show a strong association even though the population does not.

used by permission; source: https://xkcd.com/552/ (accessed 2021-11-15)

If correlation is not causation, then how can we establish causation? For example, how do we know that smoking causes lung cancer in humans? Obviously we can’t perform an experiment, for ethical reasons. Sir Austin Bradford Hill laid down nine criteria for establishing causation in a 1965 paper, The Environment and Disease: Association or Causation? [see “Sources Used” at end of book] Short summaries of the “Bradford Hill criteria” are many places on the Web, including Steve Simon’s (2000b) Causation [see “Sources Used” at end of book].

Regression Line, ŷ = ax+b

Write the equation of the line using ŷ (“y-hat”), not y, to indicate that this is a prediction. b is the y intercept, and a is the slope. Round both of them to four decimal places, and write the equation of the line as

ŷ = 3.1661x − 55.7966

(Don’t write 3.1661x + −55.7966.)

These numbers can be interpreted pretty easily. Business majors will recognize them as intercept = fixed cost and slope = variable cost, but you can interpret them in non-business contexts just as well.

The slope, a or b₁ or m, tells how much ŷ increases or decreases for a one-unit increase in x. In this case, your interpretation is “the ball travels about an extra 3.17 yards when the club speed is 1 mph greater.” The slope and the correlation coefficient always have the same sign. (A negative slope would mean that y decreases that many units for every one unit increase in x.)

The intercept, b or b₀, says where the regression line crosses the y axis: it’s the value of ŷ when x is 0. Be careful! The y intercept may or may not be meaningful. In this case, a club-head speed of zero is not meaningful. In general, when the measured x values don’t include 0 or don’t at least come pretty close to it, you can’t assign a real-world interpretation to the intercept. In this case you’d say something like “the intercept of −55.7966 has no physical interpretation because you can’t hit a golf ball at 0 mph.

Here’s an example where the y intercept does have a physical meaning. Suppose you measure the gross weight of a UPS truck (y) with various numbers of packages (x) in it, and you get the regression equation ŷ = 2.17x+2463. The slope, 2.17, is the average weight per package, and the y intercept, 2463, is the weight of the empty truck.

Coefficient of Determination, R²

The last number to look at (third on the screen) is R², the coefficient of determination. (The calculator displays r², but the capital letter is standard notation.) R² measures the quality of the regression line as a means of predicting ŷ from x: the closer R² is to 1, the better the line. Another way to look at it is that R² measures how much of the total variation in y is predicted by the line.

In this case R² is about 0.88, so your interpretation is “about 88% of the variation in distance traveled is associated with variation in club-head speed.” Statisticians say that R² tells you how much of the variation in y is “explained” by variation in x, but if you use that word remember that it means a numerical association, not necessarily a cause-and-effect explanation. It’s best to stick with “associated” unless you have done an experiment to show that there is cause and effect.

There’s a subtle difference between r and R², so keep your interpretations straight. r talks about the strength of the association between the variables; R² talks about what part of the variation in the y variable is associated with variation in the x variable, and how well the line predicts y from x. Don’t use any form of the word “correlated” when interpreting R².

Only linear regression will have a correlation coefficient r, but any type of regression — fitting any line or curve to a set of data points — will have a coefficient of determination R² that tells you how well the regression equation predicts y from the independent variable(s). Steve Simon (1999b) gives an example for non-linear regression in R-squared [see “Sources Used” at end of book].

In straight-line regression, R² is the square of r, so if you want a formula just compute r and square the result.

4B4.  Step 3. Display the Regression Line

Show line with original data points.

[GRAPH]

What is this line, exactly? It’s the one unique line that fits the plotted points best. But what does “best” mean?

The same four points on left and right. The vertical distance from each measured data point to the line, y−ŷ, is called the residual for that x value. The line on the right is better because the residuals are smaller.
source: Dabes and Janik (1999, 179) [see “Sources Used” at end of book]

For each plotted point, there is a residual equal to y−ŷ, the difference between the actual measured y for that x and the value predicted by the line. Residuals are positive if the data point is above the line, or negative if the data point is below the line.

You can think of the residuals as measures of how bad the line is at prediction, so you want them small. For any possible line, there’s a “total badness” equal to taking all the residuals, squaring them, and adding them up. The least squares regression line means the line that is best because it has less of this “total badness” than any other possible line. Obviously you’re not going to try different lines and make those calculations, because the formulas built into your calculator guarantee that there’s one best line and this is it.

Carl Friedrich Gauss developed the method of least squares in a paper published in 1809.

4B5.  Optional:  Display the Residuals

I would like you to know the material in this section, but it's not part of the MATH200 syllabus so I don’t require it. No homework or quiz problems will draw from this section. You will, however, need to calculate individual residuals; see Finding Residuals, below.

“No regression analysis is complete without a display of the residuals to check that the linear model is reasonable.”

DeVeaux, Velleman Bock (1999, 227) [see “Sources Used” at end of book]

The residuals are automatically calculated during the regression. All you have to do is plot them on the y axis against your existing x data. This is an important final check on your model of the straight-line relationship.

Turn off other plots.	Press [Y=]. Cursor to the highlighted = sign next to Y1 and press [ENTER]. Cursor to PLOT1 and press [ENTER].
Set up the plot of residuals against the x data.	Set up Plot 2 for the residuals. Press [2nd Y= makes STAT PLOT] [▼] [ENTER] [ENTER] to turn on Plot 2. Press [▼] [ENTER] to select a scatterplot.   The x’s are still in L1, so press [2nd 1 makes L1] [ENTER]. In this plot, the y’s will be the residuals: press [2nd STAT makes LIST], cursor up to RESID, and press [ENTER] [ENTER].
Display the plot.	[ZOOM] [9] displays the plot.

You want the plot of residuals versus x to be “the most boring scatterplot you’ve ever seen.” (DeVeaux, Velleman, Bock 2009, 203) [see “Sources Used” at end of book] “It shouldn’t have any interesting features, like a direction or shape. It should stretch horizontally, with about the same amount of scatter throughout. It should show no bends, and it should have no outliers. If you see any of these features, find out what the regression model missed.”

Don’t worry about the size of the residuals, because [ZOOM] [9] adjusts the vertical scale so that they take up the full screen.

If the residuals are more or less evenly distributed above and below the axis and show no particular trend, you were probably right to choose linear regression. But if there is a trend, you have probably forced a linear regression on non-linear data. If your data points looked like they fit a straight line but the residuals show a trend, it probably means that you took data along a small part of a curve.

Here there is no bend and there are no outliers. The scatter is pretty consistent from left to right, so you conclude that distance traveled versus club-head speed really does fit the straight-line model.

Residual Plot Showing Problems

Refer back to the scatterplot of f/stop against shutter speed. I said then that it was not a straight line, so you could not do a linear regression. If you missed the bend in the scatterplot and did a regression anyway, you’d get a correlation coefficient of r = 0.98, which would encourage you to rely on the bad regression. But plotting the residuals (at right) makes it crystal clear that linear regression is the wrong type for this data set.

This is a textbook case (which is why it was in a textbook): there’s a clear curve with a bend, variation on both sides of the x axis is not consistent, and there’s even a likely outlier.

Optional Advanced:  Residuals and R²

I said in Step 2 that the coefficient of determination measures the variation in measured y that’s associated with the variation in measured x. Now that you understand the residuals, I can make that statement more precise and perhaps a little easier to understand.

The set of measured y values has a spread, which can be measured by the standard deviation or the variance. It turns out to be useful to consider the variation in y’s as their variance. (You remember that the variance is the square of the standard deviation.)

The total variance of the measured y’s has two components: the so-called “explained” variation, which is the variation along the regression line, and the “unexplained” variation, which is the variation away from the regression line. The “explained” variation is simply the variance of the ŷ’s, computing ŷ for every x, and the “unexplained” variation is the variance of the residuals. Those two must add up to the total variance of the measured y’s, which means that as percentages of the variation in y they must add to 100%. So R² is the percent of “explained” variation in the regression, and 100%−R² is the percent of “unexplained” variation.

and

Now I can restate what you learned in Step 2. R² is 88% because 88% of the variance in y is associated with the regression line, and the other 12% must therefore be the variance in the residuals. This isn’t hard to verify: do a 1-VarStats on the list of measured y’s and square the standard deviation to get the total variance in y, s²_y = 59.93. Then do 1-VarStats on the residuals list and square the standard deviation to get the “unexplained” variance, s²_e = 7.12. The ratio of those is 7.12/59.93 = 0.12, which is 1−R². Expressing it as a percentage gives 100%−R² = 12%, so 12% of the variation in measured y’s is “unexplained” (due to lurking variables, measurement error, etc.).

4C.  How to Find ŷ from a Regression on TI-83/84

Summary: The regression line represents the model that best fits the data. One important reason for doing the regression in the first place is to answer the question, what average y value does the model predict for a given x? This page shows you two methods of answering that question.

4C1.  Method 1: Trace on the Regression Line Graph (preferred)

You can make predictions while examining the graph of the regression line on the TI-83/84 or TI-89.

Advantages to this method: aside from being pretty cool, it avoids rounding errors, and it’s very fast for multiple predictions.

Enter the x value.

Press the black-on-white numeric keys including [(−)] and decimal point if needed.   As soon as you press the first number, you’ll see a large X= appear at the bottom left of the screen. Enter any additional digits and press [ENTER].   The TI-83/84 displays the predicted average y value (ŷ) at the bottom right and puts a blinking cursor at that point on the regression line.

Caution: ŷ = 267.1 yards is the predicted or expected average distance for a club-head speed of 102 mph. But that does not mean any particular golf ball hit at that speed will travel that exact distance. You can think of ŷ as the average travel distance that you’d would expect for a whole lot of golf balls hit at that speed.

Extrapolation: Just Say No (Usually)

Caution: A regression equation is valid only within the range of actual measured x values, and a little way left and right of that range. If you try to go too far outside the valid range, the calculator will display ERR:INVALID.

It’s not just being cranky. The line describes the points you measured, so it’s usable between your minimum and maximum x values and maybe a little way outside those limits. But unless you have very solid reasons why the same straight-line model is good beyond that range, you can’t extrapolate.

Take a look at this graph of men’s and women’s winning times in the Olympic 100-meter dash from 1928 to 2012, which I made from data compiled by Mike Rosenbaum [see “Sources Used” at end of book]. (The women’s 100 m dash became an Olympic event in 1928.)

From this you can reasonably guess that if women had run in the 1924 Olympics, the winner would have finished in around 12.2 or 12.3 seconds. And the 2016 winner will probably finish in around 11.5 seconds. But the further you go outside your measured data, the more riskier your predictions.

Will men’s and women’s times generally continue to decrease? Probably: training will get better, nutrition will improve, global communications will make it less likely that a stellar runner goes undiscovered. But will the decrease follow a straight line? Certainly not! Think about it for a minute. If times keep decreasing on a straight line, eventually they’ll cross the x axis and go negative. Runners will finish the race before they start it! So obviously the straight-line model breaks down — the only question is where. You don’t know, and you can’t know. All you know is that it’s not safe to extrapolate.

Bogus extrapolations give statistics a bad name and make people say “you can prove anything with statistics.” Here’s an example. I’ve just extended the two trend lines to “prove” that after the 2160 Olympics women will run the 100 meters faster than men. Pretty clearly, the linear model breaks down before then.

It’s not safe to extrapolate to earlier times, either. The intercepts tell you that in the year zero, the fastest man in the world took 31.6 seconds to run 100 m, and the fastest woman took 44.7 seconds. Does that seem believable?

4C2.  Method 2: Use Calculated Regression Equation (if necessary)

But what if you don’t still have the regression line on your calculator, for instance if you’ve done a different regression? In that case, you can go back to your written-down regression equation and plug in the desired x value.

Advantage of this method: You already know how to substitute into equations.  Disadvantages: depending on the specific numbers involved, you may introduce rounding errors. Also, since you’re entering more numbers there’s an increased chance of entering a number wrong.

Example: To find the predicted average y value for x = 102, go back to the regression equation that you wrote down, and substitute 102 for x:

ŷ = 3.1661x − 55.7966

ŷ = 3.1661*102 − 55.7966

ŷ = 267.1456 → 267.1

In this example, the rounding error was very small, and it disappeared when you rounded ŷ to one decimal place. But there will be problems where the rounding error is large enough to affect the final answer, so always use the trace method if you can.

Again, please observe the Cautions above. With this method, the calculator won’t tell you when your x value is outside a reasonable range, so you need to be aware of that issue yourself.

4C3.  Finding Residuals

Each measured data point has an associated residual, defined as y−ŷ, the distance of the point above or below the line. To find a residual, the actual y comes from the original data, and the predicted average ŷ comes from one of the methods above.

Example: Find the residual for x = 102.
Solution: From the original data, y = 264. From either of the methods above, ŷ = 267.1. Therefore the residual is y−ŷ = 264−267.1 = −3.1 yards.

If a given x value occurs in more than one data point, you have multiple residuals for that x value.

4D.  Decision Points for Correlation Coefficient

Summary: After you compute the linear correlation coefficient r of your sample, you may wonder whether this reflects any linear correlation in the population. By comparing r to a critical number or decision point, you either conclude that there is linear correlation in the population, or reach no conclusion. You can never conclude that there’s no correlation in the population.

This page gives a simple mechanical test, but a proper statistical test exists. The optional advanced handout Inferences about Linear Correlation explains how decision points are computed and the theory behind the test. You need to learn about t tests before you can understand all of it, but right now you can use the Excel spreadsheet that you’ll find there. Or you can use MATH200B Program part 6 to do the computations.

4D1.  Procedure

The decision points are used to answer the question “From the linear correlation r of my sample, can I rule out chance as an explanation for the correlation I see? Can I infer that there is some correlation in the population?”

To answer that question, temporarily disregard the sign of r. This is the absolute value of r, written | r |. Then compare | r | to the decision point, and obtain one of the only three possible results:

Here’s a table of decision points (also known as critical values of r) for various sample sizes.

(If your sample size is not shown, either refer to the Excel workbook or use the next lower number that is shown in the table. Example: n = 35 is not shown, and therefore you will use the decision point for n = 30.)

4D2.  Examples

You survey 50 randomly selected college students about the number of hours they spend playing video games each week and their GPA, and you find r = −0.35. You look up n = 50 in the table and find 0.279 as the decision point. |r|>d.p. (0.35 > 0.279). You conclude that for college students in general, video game play time is negatively associated with GPA, or that GPA tends to decrease as video-game playing increases.

You randomly select 21 college students. For the amount they spend on textbooks and their GPA, you find r = +0.20. n=21 isn’t in the table of decision points, so you select 0.444, the decision point for n=20. |r|≤d.p. (0.20 ≤ 0.444). Therefore, you are unable to make any statement about an association between textbook spending and GPA for college students in general.

4D3.  Interpretation

Be very careful with your interpretation, and don’t say more than the statistics will allow.

The question was simply whether there is some correlation in the population, not how much. The population might have stronger or weaker correlation than your sample; all you know is that it has some. (Though you won’t learn how to do it in this course, it is possible to estimate the correlation coefficient of the population.)

If you conclude there is some correlation in the population, it’s probable, not certain. From a completely uncorrelated population, there’s still one chance in 20 of drawing a sample with | r | greater than the decision point. Because 1/20 is .05, we say that .05 is the significance level.

Even if you conclude that there is some correlation in the population, that’s the start of your investigation, not the end. If there’s a correlation in the population, you can’t just assume that one variable drives the other: correlation is not causation. Steve Simon’s (2000b) Causation [see “Sources Used” at end of book] gives some hints for investigating causation, using smoking and lung cancer as an example.)

Finally, note that there’s no way to reach the conclusion “there’s no correlation in the population." Either there (probably) is, or you can’t reach any conclusion. This will be a general pattern in inferential statistics: either you reach a conclusion of significance, or you don’t reach any conclusion at all. (As you’ll see in Chapter 10, you can conclude “something is going on”, you can fail to reach a conclusion, but you can never conclude “nothing is going on”. Lack of evidence for is not evidence against.)

4E.  Optional:  Scatterplot, Correlation, and Regression in Excel

4E1.  Plot the Points

Here again are the data:

1. Enter the x-y pairs in rows or columns; row or column heads are optional.
2. With your mouse, highlight the data but not the headers. Click Insert. In the Charts section, click Scatter and choose the first scatterplot type.
3. Right-click the useless “Series1” legend and click Delete.
4. This time I got lucky, but sometimes Excel puts too much white space at the left or bottom of the chart. If this happens to you, right-click the axis numbers and select Format Axis. Change Minimum to Fixed and type in a sensible value.
5. In the Excel ribbon, click Layout » Axis Titles » Primary Horizontal Axis Title » Title Below Axis and type the axis title. Include units if any. In this case, you have club-head speed in miles per hour.
6. Click Axis Title » Primary Vertical Axis Title » Rotated Title and type the axis title, including units if any. In this case, you have distance traveled in yards.
7. Click Chart Title » Above Chart and type your chart title.
8. For a neater appearance, you can right-click the horizontal axis, select Format Axis, and change Major tick mark type to None. Repeat for the vertical axis. Your chart should look like this:

   

4E2.  Show the Regression Line

1. In the Excel ribbon, click Layout. In the Analysis group, click Trendline » More Trendline Options.
2. In the dialog box that appears, click Trendline Options at the left. At the top right, select Linear. At the bottom right, select Display Equation on chart and Display R-squared value on chart.
3. Click and drag the regression equation and R² value so that they’re not covering any data points or any part of the line. Then right-click them and select Format Trendline Label. Click Fill at the left, then at the right click Solid Fill and change the color to white. (This keeps the gridlines from running through the text.) If you wish, click Border Color » Solid Line. Here’s the result:

   

4E3.  Show the Correlation Coefficient

Excel won’t put r on the chart, but you can compute it in a worksheet cell:

1. Click into an empty worksheet cell. Type =CORREL( including the = sign and opening parenthesis.
2. Highlight your y list with your mouse — numbers only, not the header — and type a comma.
3. Highlight your x list with your mouse — again, just the numbers. Type a closing parenthesis and press [ENTER].

(You can get the slope, y intercept, or R² into the worksheet by following the above procedure but substituting SLOPE, INTERCEPT, or RSQ for CORREL.)

4E4.  Predict the Average y

Like your calculator, Excel can find the ŷ value (predicted average y) for any x.

Caution: A regression equation is valid only within the range of actual measured x values, and a little way left and right of that range. If you go outside that range, Excel will happily serve up garbage numbers to you.

On average, how far do you expect a golf ball to travel when hit at 102 mph?

1. Type your x value, 102, in an empty cell.
2. Click into an empty worksheet cell. Type =FORECAST( including the = sign and opening parenthesis.
3. Click into the cell that holds your x value, and type a comma.
4. Highlight your y list with your mouse — numbers only, not the header — and type a comma.
5. Highlight your x list with your mouse — again, just the numbers. Type a closing parenthesis and hit [ENTER]. You’ll see the predicted average distance, 267.1 yards.

The prediction formula, like all Excel formulas, is “live”: if you type in a new x Excel will display the corresponding ŷ. If this doesn’t happen, in the Excel ribbon click Formulas » Calculation Options » Automatic.

What Have You Learned?

Chapter 5 WHYL → ← Chapter 3 WHYL

Exercises for Chapter 4

Write out your solutions to these exercises, showing your work for all computations. Then check your solutions against the solutions page and get help with anything you don’t understand.

Caution! If you don’t see how to start a problem, don’t peek at the solution — you won’t learn anything that way. Ask your instructor or a tutor for a hint. Or just leave it and go on to a different problem for now. You may find when you return to that “impossible” problem that you see how to do it after all.

7 A scatterplot is shown at right. Would the value of r be strongly positive, near zero, or strongly negative? Briefly explain your answer.

Solutions → 

Intro: By now you know: There’s no certainty in statistics. When you draw a sample from a population, what you get is a matter of probability. When you use a sample to draw some conclusion about a population, you’re only probably right. It’s time to learn just how probability works.

If you’re learning independently, you can skip the sections marked “Optional” and still understand the chapters that follow. If you’re taking this course with an instructor, s/he may require some or all of those sections. Ask if you’re not sure!

For easy reference, tables used in more than one problem are duplicated at the end of this document.

5A.  Probability Basics

5A1.  What Is Probability?

Example 1: Ten thousand doctors took aspirin every night for six years, and 104 of them had heart attacks. The relative frequency is 104/10000 = 1.04%, so the probability of heart attack is 1.04% for doctors taking aspirin nightly.

Each doctor represents a trial, and the outcome of each trial is either “heart attack” or “no heart attack”. The group of 10,000 trials is a probability experiment.

Definition: An event is a group of one or more possible outcomes of a trial. Usually those outcomes are related in some way, and the event is named to reflect that.

Example 2: If you draw a card from a deck without looking, there are 52 possible outcomes (assuming the jokers have been removed). “Ace” is an event, representing a group of four outcomes, and the probability of that event is 4/52 or 1/13. “Spade” is an event, representing a group of 13 outcomes, so its probability is 13/52 or 1/4. “Ace of spades” is both an outcome and an event, with a probability of 1/52.

Write probabilities as fractions, decimals, or percentages, like this:

P(event) = number

Example 3: On a coin flip, P(heads) = 0.5, read as the probability of heads is 0.5. “P(0.5)” is wrong. Don’t write P(number); always write P(event) = number.

All probabilities are between 0 and 1 inclusive. A probability of 0 means the event is impossible or cannot happen; a probability of 1 means the event is a certainty or will definitely happen. Probabilities between 0 and 1 are assigned to events that may or may not happen; the more likely the event, the higher its probability.

Definition: When an event is unlikely — when it has a low probability of occurring — you call it an unusual event. Unless otherwise stated, “unlikely” means that the probability is below 0.05.

This will be an important idea in inferential statistics.

5A2.  Where Do You Get Probabilities?

Pure thought is enough to give many probabilities: the probability of drawing a spade from a deck of cards, the probability of rolling doubles three times in a row at Monopoly, the probability of getting an all-white jury pool in a county with 26% black population. Any such probability is called a theoretical probability or classical probability.

Theoretical probabilities come ultimately from a sample space, usually with help from some of the laws for combining events. (I’ll tell you about both of these later in this chapter.)

Example 4: A standard die (used in Monopoly or Yahtzee) has six faces, all equally likely to come up. Therefore you know that the probability of rolling a two is 1/6.

On the other hand, some probabilities are impossible to compute that way, because there are too many variables or because you don’t know enough: the probability that weather conditions today will give rise to rain tomorrow, the probability that a given radium nucleus will decay within the next second, the probability that a given candidate will win the next election, the probability that a driver will have a car crash in the next year. To find the probability of an event like that, you do an experiment or rely on past experience, and so it is called an experimental probability or empirical probability.

Example 5: The CDC says that the incidence of TB in the US is 5 cases per 100,000 population. 5/100,000 = 0.005%. Therefore you can say that the probability a randomly selected person has TB is 0.005%.

These two terms describe where a probability came from, but there’s no other difference between experimental and theoretical probabilities. They both obey the same laws and have the same interpretations.

You probably don’t need formulas, but if you want them here they are:

5A3.  Interpreting Probability Statements

Every probability statement has two interpretations, probability of one and proportion of all. You use the interpretation that seems most useful in a given situation.

Example 6: For doctors taking aspirin nightly, P(heart attack in six years) = 1.04%. The “probability of one” interpretation is that there’s a 1.04% chance any given doctor taking aspirin will have a heart attack. The “proportion of all” interpretation is that 1.04% of all doctors taking aspirin can be expected to have heart attacks.

Which interpretation is right? They’re both right, but in a given situation you should use the one that feels more natural.

5A4.  Law of Large Numbers

You know that P(boy) is about 50% for live births, but you’re not surprised to see families with two or three girls in a row. Probability is long-term relative frequency; it can’t predict what will happen in any particular case.

This is expressed in the law of large numbers: as you take more and more trials, the relative frequency tends toward the true probability.

The law of large numbers was stated in 1689 by Jacob Bernoulli.

Example 7: For just a few babies, say the four children in one family, it’s quite common to find a proportion of boys very different from 50%, say one in four (25%) or even zero in four. But consider a class of thirty statistics students. The proportion may still be different from 50%, but a very different proportion (more than 70%, say, or less than 30%) would be unusual. And when you look at all babies born in a large hospital in a year, experience tells you that the proportion will be very close to 50%. The more trials you take, the closer the relative frequency is to the true probability — usually.

But the Law of Large Numbers says that the relative frequency tends to the true probability. Probability can’t predict what will happen in any given case. The idea that a particular outcome is “due” is just wrong, and it’s such a classic mistake that it has a name. The Gambler’s Fallacy is the idea that somehow events try to match probabilities.

Example 8: I’ve just flipped a coin a few times, and the results are shown at the right. The first flip was a tail, and after that flip the relative frequency (rf) of heads is 0. The next flip is a head, and after two flips I’ve had one head out of two trials, so the rf is 0.5. The third flip is also a head, so now the rf is 2/3 or about 0.6667. At this point someone might say, “you’re due for a tail, to move the rf back toward the true probability of 0.5.” That’s the Gambler’s Fallacy.

The coin doesn’t know what it did before, and it doesn’t try to make things “right”. In my trials, the fourth flip moves the rf of heads further from 0.5, and the fifth flip moves it further still. True, the sixth flip moves the rf of heads closer to 0.5, but it could just as well have moved it further away, even if the coin is perfectly fair.

I stopped after six trials. I know that if I went on to do ten trials, or a hundred, or a thousand, over time the proportion of heads would almost always move closer to 0.5 — not necessarily on any particular flip, but in the long run.

Subconsciously you expect random events not to show a pattern, but you may see patterns along the way. For example, if you flip a fair coin repeatedly, inevitably you will see a run of ten heads or ten tails — about twice in every thousand sequences of ten. If you flip the coin once every two seconds, you can expect to see a run of ten flips the same about once every 17 minutes, on average.

Here are two more examples of patterns cropping up in processes that are actually random:

• Clustering Illusion [see “Sources Used” at end of book] at Wikipedia.
• The “hot hand” illusion in basketball: see Gilovich, Vallone, Tversky (1985) [see “Sources Used” at end of book].

Example 9: You have flipped a coin 999 times, and there were 499 heads and 500 tails. What’s the probability of a head on the next flip?

Solution: It is 50%, the same as on any other flip. The Law of Large Numbers tells you that over time you tend to get closer and closer to 50% heads, but it doesn’t tell you anything at all about any particular flip. If you think that the coin is somehow “due for a head”, you’ve fallen into the Gambler’s Fallacy.

5A5.  Sample Space

At bottom, probability is about counting. Empirical probability is the number of times something did happen, divided by the number of trials. Classical probability is similar, but it makes use of a list or table of all possible outcomes, called a sample space. Technically a sample space is just a list of all possible outcomes, but it’s only useful if you make it a list of all possible equally likely outcomes.

For repeated independent trials — flipping multiple coins, rolling multiple dice, making successive bets at roulette, and so on — the size of the sample space will be the number of outcomes in each trial, raised to the power of the number of trials. For example, if you want to compute probabilities for the number of girls in a family of four children, your sample space will have 2⁴ = 16 entries.

Example 10: If you roll two dice, what’s the probability you’ll roll a seven? You could list the sample space as

S = { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }

but the outcomes are not equally likely. There’s only one way to get a twelve, for instance (double sixes), but there are several ways to get a seven (1–6, 2–5, and so on). So it’s much more useful to list your sample space with equally likely outcomes.

When constructing a sample space, be systematic so that you don’t leave any out or list any twice. Here, you’re rolling two dice, and each die has six equally likely results, so you have 6×6 = 36 equally likely outcomes in your sample space. How can you be systematic? List the outcomes in some regular order, like the picture below. Each row lists all the possibilities with the same outcome for the first die; each column lists all the possibilities with the same outcome for the second die.

image courtesy of Bob Yavits, Tompkins Cortland Community College

Once you have a sample space of equally likely outcomes, finding the probability is simple. There are six ways to roll a seven: 6-1, 5-2, 4-3, 3-4, 2-5, 1-6. There are 36 possible outcomes, all equally likely. Therefore the probability of rolling a seven is 6/36 or 1/6 or about 0.1667. In symbols, P(7) = 6/36 or P(7) = 1/6.

Example 11: Find the probability of rolling craps (two, three, or twelve).

Solution: There’s one way to roll a two, two ways to roll a three, and one way to roll a twelve. P(craps) = (1+2+1)/36 = 4/36 or 1/9.

5A6.  Probability Models

Often, it’s not practical to construct a sample space and compute probabilities from it. Instead, you construct a probability model. Probability models are yet another kind of mathematical model as introduced in Chapter 4.

Definition: A probability model is a table showing all possible outcomes and their probabilities. Every probability must be 0 to 1 inclusive, and the total of the probabilities must be 1 or 100%.

A probability model can be theoretical or empirical.

Example 12: Construct a probability model for the number of heads that appear when you flip two coins.

Solution: Start by constructing the sample space. Remember that you need equally likely events if you are going to find probabilities from the sample space. The first coin can be heads or tails, and whatever the first coin is, the second coin can also be heads or tails. So the sample space has 2×2 = 4 outcomes:

S = { HH, HT, TH, TT }

There are four equally likely outcomes, so the denominator (bottom number) on all the probabilities will be 4. The possible outcomes are no heads (one way), one head (two ways), and two heads (one way). The probability model is shown at right. Often a total row is included, as I did, to show that the probabilities add up to 1.

That was an easy example, so easy that you could just as well work from the sample space. But think about more complex situations, especially with empirical (experimental) probabilities. Constructing a sample space may be impractical, but a probability model is relatively easy to create.

Example 13: (adapted from Sullivan 2011, page 235 problem 40): The CDC asked college students how often they wore a seat belt when driving. 118 answered never, 249 rarely, 345 sometimes, 716 most of the time, 3093 always. Construct a probability model for seat-belt use by college students when driving.

Solution: Probability of one is proportion of all, so to get the probabilities you simply calculate the proportions. Sample size was (118+249+345+716+3093) = 4521. The proportions or probabilities are then simply 118/4521, 249/4521, and so on. The probability model is shown at the right.

Comments: Don’t push this model too far. In this sample, 68.4% of college students reported that they always use a seat belt when driving. There’s no uncertainty about that statement; it’s a completely accurate statistic (summary number for a sample). But can you go further and say that 68.4% of college students always wear a seat belt when driving? No, for two reasons.

First, this is a sample. Even if it’s a perfect random sample, it’s still not the population. There’s always sample variability. A different sample of college students would most likely give different answers — probably not very different, since this was a large sample, but almost certainly not identical. Second, and more serious, this survey depended on self reporting: students weren’t observed, they were just asked. When people report their behavior they tend to shade their responses in the direction of what’s socially approved or what they would like to think about themselves (response bias). How many of those “always” responses should have been “most of the time” or “sometimes”? You have no way to know.

5B.  Combining Probabilities

You can find probabilities of simple events by making sample spaces and counting. But life isn’t usually that simple. To find probabilities of more interesting (and complex) events, you need to use rules for combining probabilities.

The rules are the same whether your original probabilities are theoretical or experimental.

5B1.  Probability “or” for Disjoint Events

Definition: When two events can’t both happen on the same trial, they are called mutually exclusive events or disjoint events.

Example 14: You select a student and ask where she was born. “Born in Cortland” and “born in Ithaca” are mutually exclusive events because they can’t both be true for the same person.

Comment: Obviously it’s possible that neither is true. Disjoint events could both be false, or one might be true, but they can’t both be true in the same trial.

Example 15: You select a student and ask his major. “Major in physics” and “major in music” are non-disjoint events because they could be true of the same student. (It doesn’t matter whether they are both true of the student you asked. They are non-disjoint because they could both be true of the same student — think about double majors.)

Rule: For disjoint events, P(A or B) = P(A)+P(B)

Example 16: You draw a card from a standard 52-card deck. What’s P(ace or face card)? (A face card is a king, queen, or jack.)

Solution: Are the events “ace” and “face card” disjoint? Yes, because a given card can’t be both an ace and a face card. Therefore you can use the rule:

P(ace or face card) = P(ace) + P(face card)

But what are P(ace) and P(face card)? A picture may help.

used by permission; source: http://www.jfitz.com/cards/ accessed 2012-09-26

Now you can see that the deck of 52 cards has four aces and twelve face cards. Therefore

P(ace) = 4/52 and P(face card) = 12/52

Since the events are disjoint,

P(ace or face card) = P(ace) + P(face card)

P(ace or face card) = 4/52 + 12/52 = 16/52

Reminder: When you need to compute probability of A or B, always ask yourself first, are the events disjoint? Use the simple addition rule only if the events are disjoint. If events are non-disjoint — if it’s possible for both to happen on the same trial — you have to use the general rule, below.

Take a look at this table of marital status in 2006, from the US Census Bureau. It’s known as a contingency table or two-way table, because it classifies each member of the sample or population by two variables — in this case, sex and marital status.

Example 17: What’s the probability that a randomly selected person is widowed or divorced?

Solution: Are those events disjoint? Yes, because a given person can’t be listed in both rows of the table. (You might argue that a given person can be both widowed and divorced in his or her lifetime, and that’s true. But the table shows marital status at the time the survey was made, not over each person’s lifetime. The “Widowed” row counts those whose most recent marriage ended with the death of their spouse.) Therefore

P(widowed or divorced) = P(widowed) + P(divorced)

How do you find those probabilities? Remember that probability of one = proportion of all. Find the proportions, and you have the probabilities.

P(widowed or divorced) = 13.9/219.7 + 22.8/219.7

P(widowed or divorced) = 36.7/219.7 ≈ 0.1670

Example 18: Find the probability that a randomly selected man is widowed or divorced.

Solution: Disjoint events? Yes, a given man can’t be in both rows of the table. Again, the probabilities are the proportions, but now you’re looking only at the men:

P(widowed or divorced) = P(widowed) + P(divorced)

P(widowed or divorced) = 2.6/106.2 + 9.7/106.2

P(widowed or divorced) = 12.3/106.2 ≈ 0.1158

Now let’s look at a couple of examples of probability “or” for non-disjoint events.

Example 19: Find P(seven or club).

Solution: Are the events “seven” and “club” disjoint? No, because a given card can be both a seven and a club. You can’t use the simple addition rule.

The next section shows you a formula, but in math there’s usually more than one way to approach a problem. Here you can look back at the picture look at the picture (reprinted on the last page) and count from the sample space. There are thirteen clubs, plus the sevens of spades, hearts, and diamonds, for a total of 16. (You don’t count the seven of clubs when counting sevens, because you already counted it when counting clubs.) And therefore P(seven or club) = 16/52.

Example 20: Find P(woman or divorced).

Solution: Disjoint events? No, a given person can be both. So what do you do? The same thing as in the preceding example: you count up all the women, and all the divorced people who aren’t women, and divide by the number of people:

P(woman or divorced) = 113.5/219.7 + 9.7/219.7 = 123.2/219.7 ≈ 0.5608

5B2.  Probability “or” for All Events

Look back at P(seven or club). Those are not disjoint events, so you can’t just add P(seven) and P(club). But what did you do, when counting? You counted the clubs, then you counted the sevens that aren’t clubs. In other words, just adding P(seven) and P(club) would be wrong because that would double count the overlap.

With 52 cards, it’s easy enough just to count. But that’s not practical in every problem, so there’s a rule: go ahead and double count by adding the probabilities, then fix it by subtracting the part you double counted.

Rule: P(A or B) = P(A) + P(B) − P(A and B)

This general addition rule works for all events, disjoint or non-disjoint. (If two events are disjoint, they can’t happen at the same time, P(A and B) is 0, and the general rule becomes the same as the simple rule.)

Let’s redo the last two examples with this new general rule, to see that it gives the same answers.

Example 19 again: Find P(seven or club).

P(seven or club) = P(seven) + P(club) − P(seven and club)

Caution: P(seven and club) doesn’t mean “all the sevens and all the clubs”. It means the probability that one card will be both a seven and a club — in other words, it means the seven of clubs.

P(seven or club) = 4/52 + 13/52 − 1/52

P(seven or club) = 16/52

Example 20 again: Using the table, table of marital status (reprinted on the last page), find P(woman or divorced).

Solution:

P(woman or divorced) = P(woman) + P(divorced) − P(woman and divorced)

P(woman or divorced) = 113.5/219.7 + 22.8/219.7 − 13.1/219.7

P(woman or divorced) = 123.2/219.7 ≈ 0.5608

5B3.  Probability “not” — Complements

About two thirds of students who register for a math class complete it successfully. What’s the probability that a randomly selected student who registers for a math class will not complete it successfully? Of course you already know it’s 1−(2/3) = 1/3. Let’s formalize this.

Definitions: Two events are complementary if they can’t both occur but one of them must occur. If A is an event from a given sample space, then the complement of A, written A^C or not A, is the rest of that sample space.

Describing a complement usually involves using the word “not”. Complementary events (can’t both happen, but one must happen) are a subcategory of disjoint events (can’t both happen).

Example 21: The complement of the event “the student completes the course successfully” is the event “the student does not complete the course successfully.” Obviously the complement need not be a simple event. The complement of “the student completes the course successfully” is “the student never shows up, or attends initially but stops attending, or withdraws, or earns an F, or takes an incomplete but never finishes”, or probably other outcomes I haven’t thought of.

Rule: P(A^C) = 1 − P(A)

This comes directly from the definition, and the rule for “or”. A and A^C can’t both happen, so they’re disjoint and P(A or A^C) = P(A)+P(A^C). But one or the other must happen, so P(A or A^C) = 1. Therefore P(A)+P(A^C) = 1, and P(A^C) = 1−P(A).

Example 22: In rolling two dice, “doubles” and “not doubles” are complementary events because they can’t both happen on the same roll, but one of them must happen. “Boxcars” (double sixes) and “snake eyes” (double ones) can’t both happen, so they’re disjoint; but they are not complementary because other outcomes are possible.

The complement rule is useful on its own, but it really shines as a labor-saving device. Very often when a probability problem looks like a lot of tedious computation, the complement is your friend. This really sticks out with “at least” problems (later), but here are a few simpler examples.

Example 23: The color distribution for plain M&Ms is shown at right. What’s the probability that a randomly selected plain M&M is any color but yellow?

Solution: You could add the probabilities of the five other colors, but of course it’s easier to say

P(Yellow^C) = 1 − P(Yellow)

P(Yellow^C) = 100% − 14% = 86%

Example 24: Referring again to the table of marital status, status (reprinted on the last page), what’s the probability that a randomly selected person is not currently married?

Solution: Since the four marital statuses are disjoint, you could add the probabilities for widowed, divorced, and never married. But it’s easier to take the complement of “married”:

P(not currently married) = P(married^C)

P(not currently married) = 1 − P(married)

P(not currently married) = 1 − 127.7/219.7

P(not currently married) = 0.4188

5B4.  Probability “and” for Independent Events

Definition: Two events are called independent events if the occurrence of one doesn’t change the probability of the other.

Example 25: When you play poker, being dealt a pair in this hand and a pair in the next are independent events because the deck is shuffled between hands. But in casino blackjack, according to Scarne on Cards (Scarne 1965, 144 [see “Sources Used” at end of book]), four decks are used and they aren’t necessarily shuffled between hands. Therefore, getting a natural (ace plus a ten or face card) in this hand and a natural in the next are not independent events, because the cards already dealt change the mix of remaining cards and therefore change the probabilities.

That’s also an example of sampling with replacement (poker) and sampling without replacement (casino blackjack).

Samples drawn with replacement are independent because the sample space is reset to its initial condition between draws. Samples drawn without replacement are usually dependent because what you draw out changes the mix of what is left. However, if you’re drawing from a very large group, the change to the proportions in the mix is very small, so you can treat small samples from a very large group as independent.

Independent events are not disjoint, and disjoint events are not independent. If two events A and B are disjoint, then if A happens B can’t happen, so its probability is zero. One of two disjoint events happening changes the probability of the other, so they can’t be independent.

Rule: For independent events, P(A and B) = P(A) × P(B)

Example 26: In Monopoly, you get an extra roll if you roll doubles, but if you roll doubles three times in a row you have to go to jail. What’s the probability you’ll have to go to jail on any given turn?

Solution: Refer to the picture of the dice. picture of the dice (reprinted on the last page). There are six ways out of 36 to get doubles, so P(doubles) = 6/36 or 1/6. Each roll is independent, so the probability of doubles three times in a row is (1/6)×(1/6)×(1/6) or (1/6)^3 = 1/216, about 0.0046. If you play a lot of Monopoly, you’ll go to jail, because of doubles, between four and five times per thousand turns.

Example 27: The first traffic light on your morning commute is red 40% of the time, yellow 5%, and green 55%. What’s the probability you’ll hit a green all five mornings in any given week?

Solution: Are the five days independent? Yes, because where you hit that light in its cycle on one morning doesn’t influence where you hit it on the next day. The probability of green is 55% each day regardless of what happens on any other day. Therefore, the probability of five greens on five successive mornings is 55%×55%×55%×55%×55% or (0.55)⁵ ≈ 0.0503. About one week in twenty, that light should be green for you all five mornings.

Example 28: Refer again to the table of marital status. status (reprinted on the last page). What’s the probability that a randomly selected person is female and widowed?

Solution: In a two-way table, for probability “and”, you don’t worry about formulas or independence because everything is already laid out for you. 11.3 million persons are female and widowed, out of 219.7 million. Therefore:

P(female and widowed) = 11.3/219.7 ≈ 0.0514.

Example 29: Earlier in this section, I said that samples drawn without replacement are usually dependent, but you can treat them as independent when drawing a small sample from a very large group. Here’s an example. If you randomly select three women, what’s the probability that all three are widowed?

Solution: From the preceding example, the probability that any one woman is widowed was 11.3/219.7. Because three women is a small sample against the millions of women in the census, and the sample is random, you can treat them as independent. If you randomly select one woman out of millions, the mix of marital status in the remaining women is so nearly unchanged that you can ignore the difference. Therefore, the probability that all three women are widowed is

(11.3/219.7) × (11.3/219.7) × (11.3/219.7) = (11.3/219.7)³ ≈ 0.0001.

5B5.  Probability “at least” for Independent Events

There’s no special rule for “at least”, but textbook writers (and quiz writers) love this type of problem, so it’s worth looking at. “At least” problems usually want you to combine several of the probability rules.

Example 30: Think back to that traffic light that’s green 55% of the time, yellow 5%, red 40%. What’s the probability that you’ll catch it red at least one morning in a five-day week?

Solution: You could find the probability of catching it red one morning (five separate probabilities for five separate mornings), or two mornings (ten different ways to hit two mornings out of five), or three, four, or five mornings. This would be incredibly laborious. Remember that the complement is your friend. What’s the complement of “at least one morning”? It’s “no mornings”. So you can find the probability of getting a red on no mornings, subtract that from 1, and have the desired probability of hitting red on at least one morning.

P(at least one red in five) = 1 − P(no red in five)

But the status of the light on each morning is independent of all the others, so

P(no red in five) = P(no red on one)⁵

What’s the probability of no red on any one morning? It’s 1 minus the probability of red on any one morning:

P(no red on one) = 1 − P(red on one) = 1−0.4

Now put all the pieces together:

P(no red on one) = 1 − P(red on one) = 1−0.4

P(no red in five) = [ P(no red on one) ]⁵ = (1−0.4)⁵

P(at least one red in five) = 1 − P(no red in five) = 1 − (1−0.4)⁵ ≈ 0.9222

About 92% of weeks, you hit red at least one morning in the week.

Be careful with your logic! You really do need to work things through step by step, and write down your steps. Some students just seem to subtract things from 1, and multiply other things, and hope for the best. That’s not a very productive approach.

One thing that can help you with these “at least’ and “at most” problems is to write down all the possibilities and then cross out the ones that don’t apply, or underline the ones that do apply. For “at least one red in five”, you have 0 1 2 3 4 5 or 0 | 1 2 3 4 5. Either way, with this enumeration technique, taught to me by Benjamin Kirk, you can see that the complement of “at least one” is “none”.

A common mistake is computing 1−0.4⁵ for P(none), instead of the correct (1−0.4)⁵. “None are red” means “all are not-red”, every one of the five is something other than red. Remember that all are not is different from not all are. In ordinary English, people often say “All my friends can’t go to the concert” when they really mean “Some of my friends can go, but not all of them can go.” In math you have to be careful about the distinction. Here’s an example.

Example 31: For the same situation, what’s the probability that you’ll hit a red light no more than four mornings in a five-day week? (This could also be asked as “at most four mornings” or “four mornings at most”.)

Solution: Try enumerating. “At most four out of five” looks like this: 0 1 2 3 4 5 or 0 1 2 3 4 | 5. The previous example was a “none are” or “all are not”, but this one is a “not all are”.

P(≤ 4 out of 5) = 1 − P(5 out of 5)

P(5 out of 5) = 0.4⁵

P(≤ 4 out of 5) = 1 − 0.4⁵ ≈ 0.9898

About 99% of weeks, you hit the red light no more than four mornings of the week.

Example 32: You’re throwing a barbecue, and you want to start the grill at 2 PM. Fred and Joe live on opposite sides of town, and they’ve both agreed to bring the charcoal. The problem is that they’re both slackers. Fred is late 40% of the time, and Joe is late 30% of the time. What’s the probability you’ll start the grill by 2 PM?

Solution: This is another “at least” problem for independent events, though this time the independent events don’t have the same probability. To have charcoal by 2 PM, at least one of them has to show up by then. What’s the probability that at least one will be on time? Again, you could compute the probability that they’re both on time, that Fred’s on time but Joe’s late, and that Fred’s late and Joe’s on time — all of those together will be the probability of charcoal on time. But again, the complement is your friend. The complement of “charcoal on time” is “charcoal late”, which happens only if they’re both late.

P(charcoal on time) = 1 − P(charcoal late)

P(charcoal on time) = 1 − P(Fred late and Joe late)

(Fred and Joe live on opposite sides of town, so whether one is late has no connection with whether the other one is late. The events are independent.)

P(charcoal on time) = 1 − P(Fred late) × P(Joe late)

P(charcoal on time) = 1 − 0.4×0.3 = 0.88

You’ve got an 88% chance of starting the grill on schedule.

Example 33: The space shuttle Challenger exploded shortly after launch in the 1980s, when one of six gaskets failed. After the fact, engineers realized that they should have known the design was too risky, but they didn’t think past “each gasket is 97% reliable.” The trouble was that if any gasket failed, the shuttle would explode. If you were asked to evaluate the design while the plans were still on the drawing board, what would you conclude? (Note: The design makes the six gaskets independent.)

Solution: The shuttle will explode if one or more gaskets fail. Here’s another “at least” problem, so enumerate the case you’re interested in: 0 | 1 2 3 4 5 6.

P(explosion) = P(at least one gasket fails)

The complement of “at least one gasket fails” (hard to compute) is “no gaskets fail” (much easier). What does it mean for no gaskets to fail? All gaskets must hold. Since the gaskets are independent, that’s easy to compute:

P(all six gaskets hold) = 0.97⁶

The answer you want is the complement of the all-hold or zero-fail case:

P(at least one gasket fails) = 1 − P(all six hold) = 1 − 0.97⁶

P(explosion) = P(at least one gasket fails) = 1 −0.97⁶ ≈ 0.1670

Conclusion: There’s about a 17% chance that the shuttle will explode, just considering the gaskets and ignoring all other possible causes of trouble. This is about the same as the odds of shooting yourself in Russian roulette.

5B6.  Conditional Probability

In 2012, the Honda Accord was the most frequently stolen vehicle in the US (Siu 2013 [see “Sources Used” at end of book]). Does that mean that your Honda Accord is more likely to be stolen than another model?

You’re tested for a rare strain of flu, and the result is positive. Your doctor tells you the test is 99% accurate. Does that mean that there’s a 99% chance you have that strain of flu?

In New York City, a rape victim identifies physical characteristics that match only 0.0001% of people. Police find someone with those characteristics and arrest him. Is there only a 0.0001% chance that he’s innocent?

These are examples of conditional probability — the probability of one event under the condition that another event happened. It’s probably the most misunderstood probability topic, but I’m going to demystify it for you.

The definition may seem hard at first. But after you work through the examples you’ll find it makes sense.

Example 34: P(truck | Ford) is the probability that a vehicle is a truck if it’s a Ford, or the probability that a Ford is a truck, or the proportion of trucks among Fords. P(Ford | truck) is the probability that a vehicle is a Ford if it’s a truck, or the probability that a truck is a Ford, or the proportion of Fords among trucks.

Example 35: Let’s look first at the suspected rapist. The prosecutor presents evidence that these physical characteristics are found in only 0.0001% of people. The prosecutor therefore claims that there’s only a 0.0001% chance the suspect is innocent.

But the defense points out that there are over 8 million people in New York City. 0.0001% × 8,000,000 = 8, so the suspect is not a unique individual at all, but one of about eight people who match the eyewitness accounts. Seven of them are innocent. If there’s no evidence beyond the physical match to tie him to the crime, the probability that this defendant is innocent isn’t 0.0001%, it’s 7/8 or 87.5%. (And that’s just in the city. If you consider the metro area, or the US, or the world, there are even more people who match, so any one of them is even more likely to be innocent.)

The prosecutor’s fallacy is the false idea that the probability of a random match equals the probability of innocence. You can also describe this fallacy as “consider[ing] the unlikelihood of an event, while neglecting to consider the number of opportunities for that event to occur”, in the words of “The Prosecutor’s Fallacy” on the Poker Sleuth site (Stutzbach 2011 [see “Sources Used” at end of book]).

It’s an easy mistake to make if you just think about low probabilities. To not make this error, think in whole numbers, as the defense did. 0.0001% is hard to think about; 8 is much easier.

The key to solving conditional-probability problems is your old friend, probability of one equals proportion of all. The probability that this particular matching person is innocent is the same as the proportion of all matching people that are innocent, or the proportion of innocent people among those who match. Probability problems usually get easier when you turn them into problems about numbers of people or numbers of things.

What does this look like in symbols? (Don’t be afraid of symbols! They are your friend, I promise. Words are slippery and confusing, but when you reduce a problem to symbols you make the situation clear and you are half way to solving it.)

In this example, there’s a 0.0001% chance that a random person would match the physical type of the criminal:

P(matching) = 0.0001%

The prosecution wants you to believe that the probability of a matching individual being innocent is the same:

P(innocent | matching) = 0.0001%    (WRONG)

This is a conditional probability, the probability that one thing is true if another thing is true. Formally, the whole expression is “the probability of innocent given matching”. But it’s easier to think of as “the probability that a person who matches is innocent” or “the proportion of matching people who are innocent”.

The symbols help you clarify your thinking. “The probability of a match” and “the probability of innocence among those who match” are different symbols, and they’re different concepts. You’d expect them to be different probabilities.

The defense showed the right way to figure the probability of innocence given a match. 0.0001%×8,000,000 = 8 people match, and 7 of them are innocent. The probability that a matching person is innocent — the probability that a person is innocent given that he matches — is 87.5%.

P(innocent | matching) = 87.5%    (CORRECT)

Notice what happens with if-then probabilities. You’re considering one group within a subgroup of the population, not one group within the whole population. You’ve reduced your sample space — not all people, but all matching people. The bottom number of your fraction comes from the “given that” part of the conditional probability, because P(innocent | matching) is the proportion of matching people that are also innocent.

To explode the prosecutor’s fallacy, you distinguish between a probability in the whole population and a probability in a subgroup. You also have to ask yourself, “which group?” The issue of medical test results is a good example.

Example 36: There’s a rare skin disease, Texter’s Peril (TP), where you become hypersensitive to the buttons on your phone. (Yes, I am making this up.) It affects 0.03% of adults aged 18–30, three in ten thousand. The only cure is to lay off texting for 30 days, no exceptions. Naturally this is about the worst thing that can happen to anyone.

Your doctor has tested you and the test comes up positive. She tells you that the test is 99% accurate. Does that mean you are 99% likely to have TP? You might think so, and sadly many doctors make the same mistake.

You have a positive test result, and you want to know how likely it is that you have Texter’s Peril. In symbols,

P(disease | positive) = ?

Your doctor told you that the test is 99% accurate, meaning that 99% of people who actually have TP get a positive result:

P(positive | disease) = 99%

These are obviously not the same symbol, so the probability you care about, the probability you have the disease, may well be different from 99%. How can you compute it?

Change those probabilities to whole numbers, and make a table. (I got this technique from the book Calculated Risks [Gigerenzer 2002 [see “Sources Used” at end of book]]. The book cites a study showing that doctors routinely confused probabilities when counseling patients about test results.) You’ve already played with a two-way table; now you’re going to make one. It’s a little bit like filling in a puzzle. I hope you like puzzles. ☺

You don’t know the population size, but that’s okay. Just use a large round number, like a million. Start with what you know.

P(disease) = 0.03%

Out of 1,000,000 people, 0.03% = 300 will have TP, and the other 999,700 won’t. That’s the bottom row of the table, the totals row.

P(positive | disease) = 99%

Of the 300 who have actually have TP, 99% = 297 will get a correct positive result, and 3 will get a false negative. That’s the first column of the table.

P(negative | disease^C) = 99%

(In the real world, a given test may not be equally accurate for positives and negatives, but we’ll overlook that to keep things simple.) Out of 999,700 who don’t have TP, 99% = 989,703 will get a correct negative result, and 9,997 will get a false positive. This is the second column of the table, and now you can fill in the column of totals.

Take a look at that table, specifically the “Positive Test” row. Do you see the problem? Most of the people with positive test results actually don’t have Texter’s Peril, even though the test is 99% accurate!

It took a while to get here, but it’s better to be correct slowly than to be wrong quickly. You can now compute the probability of having TP given that you have a positive test result. Once again, probability of one equals proportion of all, so this is really the same as the proportion of people with positive test results who actually have TP:

P(disease | positive) = 297 / 10,294 = 2.89%

The test is 99% accurate, but because TP is rare, most of the positive results are false positives, and there’s under a 3% chance that a positive result means you actually have Texter’s Peril. There’s a 1 − 297/10,294 = 97.11% chance that a positive result is a false positive.

Notice again: With conditional probability, you’re not concerned with the whole population. Rather, you focus on a subgroup within a subgroup. P(disease | positive) is the proportion of people who actually have the disease, within the subgroup that received a positive test result.

Example 37: What’s the chance that a negative is a false negative, that given a negative test result you actually have TP? In symbols,

P(disease | negative) = ?

You’ve already got the table, so this is a piece of cake. Out of a million people, 989,706 test negative and 3 of them have the disease. The probability that a negative is a false negative is

P(disease | negative) = 3/989,706 ≈ 0.000 003

which is essentially nil.

Example 38: A lot of Web sites in 2013 trumpeted the news that the Honda Accord was the most frequently stolen model in the US the year before. And that’s true. Out of 721,053 stolen cars and light trucks in 2012, Hot Wheels 2012 tells us that 58,596 were Honda Accords (NICB 2013 [see “Sources Used” at end of book]).

But many Web sites warned Honda owners that they were most at risk. For instance, Honda Accord, Civic Remain Top Targets for Thieves at cars.com (Schmitz 2013 [see “Sources Used” at end of book]) leads with “If you own a Honda Accord or Civic, or a full-size Ford pickup truck, you might want to take a moment to make sure your auto-insurance payments are up to date. You drive one of the top three most-stolen vehicles in the US.”

Do you see what’s wrong here? Think about it for a minute before reading on.

Yes, a lot of Honda Accords were stolen, because there are a lot of them on the road. Too many news organizations are sloppy and think that the likelihood a stolen car is an Accord is the same as the likelihood that an Accord will be stolen. This is the doctor’s mistake from the previous example, all over again.

Let’s clarify. You have 58,596 Accords out of 721,053 thefts, so the probability that a stolen car was an Accord — the probability that a car was an Accord given that it was stolen — the probability of “if stolen then Accord” — is

P(Accord | stolen) = 58,596/721,053 = 8.13%

But that doesn’t tell you doodley-squat about your chance of having your Accord stolen. That would be the probability of a car being stolen given that it is an Accord, “if Accord then stolen”. The top number of that fraction is still 58,596, but the bottom number is the total number of Accords on the road:

P(stolen | Accord) = 58,596/(total Accords on the road in 2012)

Do you see the difference? They’re both conditional probabilities, but they’re different conditions. “If stolen then Accord” is different from “if Accord then stolen”. The first one is about Accord thefts as a proportion of all thefts, and the second one is about Accord thefts as a proportion of all Accords. Those are different numbers.

To find the chance that an Accord will be stolen, you need the number of Accords on the road in 2012. A press release from Experian (2012) says there were “more than 245 million vehicles on US roads” in 2012, and 2.6% of them were Accords.

P(stolen | Accord) = (stolen Accords)/(total Accords on the road in 2012)

P(stolen | Accord) = 58,596/(2.6% of 245 million)

P(stolen | Accord) = 58,596/6,370,000

P(stolen | Accord) = 0.92%

Yes, over 8% of cars stolen in 2012 were Accords, but the chance of a given Accord being stolen was under 1%. P(Accord | stolen) = 8.13%, but P(stolen | Accord) = 0.92%.

Optional:  Conditional Probability Formula

Rule: P(B | A) = P(A and B) / P(A)  —or—  N(A and B) / N(A)

The “N” alternatives remind you that often it’s easier just to count than to find probabilities and then divide. Either way, when you consider P(B | A), remember that you’re interested in the likelihood of B given that A occurs. It’s the B cases within the A group, not all the B cases.

P(A | B) is not the same as P(B | A). You’ll get the probability right if you remember that the second event, the “given that” event, supplies the bottom number of the fraction.

Example 39: Find P(stolen | Accord), the chance that any one Accord will be stolen. Using the numbers from Example 38,

P(stolen | Accord) = N(Accord and stolen) / N(Accord)

P(stolen | Accord) = 58,596/6,370,000 = 0.92%

Example 40: I draw a card from the deck, and I tell you it’s red. What’s the probability that it’s a heart? If you didn’t know anything about the card, you’d write P(heart) = ¼ because a quarter of the cards in the deck are hearts. But what is the probability given that it’s red?

P(heart | red) = P(heart and red) / P(red)

P(heart and red) is the probability of a red heart. A quarter of the cards in the deck are red hearts, so this is just ¼. P(red) is of course ½ because half the cards in the deck are red.

P(heart | red) = (¼) / (½) = (¼) × 2 = ½

This one is probably easier to do by just counting:

P(heart | red) = N(heart and red) / N(red)

P(heart | red) = 26/52 = ½

Either way, you’re concerned with the sub-subgroup of hearts within the subgroup of red cards. P(heart | red) = ½ — half of the red cards are hearts.

Example 41: You know P(heart | red) = ½: given that a card is red, there’s a ½ probability that it’s a heart. But what is P(red | heart), the probability that a card is red given that it’s a heart? You probably already know the answer, but let’s run the formula:

P(red | heart) = N(red and heart) / N(heart)

P(red | heart) = 13/13 = 1 (or 100%)

Conditional probabilities often come up in two-way tables.

Example 42: Again using the table of marital status (reprinted on the last page), status, what’s the probability that a randomly selected woman is divorced? In other words, given that the person is a woman, what’s the probability that she’s divorced?

Solution: The problem wants P(divorced | woman), the probability that the person is divorced given that she’s a woman.

P(divorced | woman) = N(divorced and woman) / N(woman)

P(divorced | woman) = 13.1/113.5 ≈ 0.1154

Because we have “given woman” or “if woman”, the bottom number is the number of women, 113.5 million.

5B7.  Optional:  Checking Independence

Remember the definition of independent events? A and B are independent if the occurrence of one doesn’t change the probability of the other. Now that you know about conditional probability, you can define independent events in terms of conditional probability:

Definition: Two events A and B are independent if and only if P(A|B) = P(A).

This makes sense. P(A) is the probability of A without considering whether B happened or not, and P(A|B) is the probability of A given that B happened. If B’s occurrence doesn’t change the probability of A, then those two numbers will be equal.

Example 43: Referring again to the table of marital status, status (reprinted on the last page), show that “woman” and “widowed” are dependent (not independent).

Solution:

P(widowed) = 13.9 / 219.7 ≈ 0.0633

P(widowed | woman) = 11.3 / 113.5 ≈ 0.0996

These numbers are different — the probability of “widowed” changes when “woman” is given, or in English the proportion of widowed women is different from the proportion of widowed people. Therefore the events “woman” and “widowed” are not independent.

By the way, if A and B are independent then B and A are independent. So you could just as well compare P(woman) = 113.5/219.7 ≈ 0.5166 to P(woman|widowed) = 11.3/13.9 ≈ 0.8129. Since those are different, you conclude that “woman” and “widowed” are dependent.

5B8.  Optional:  Probability “and” for All Events

When events are not independent, to find probability “and” you need to use a conditional probability. Remember the formula for conditional probability: P(B | A) = P(A and B) / P(A). Multiply both sides by P(A) and you have P(A) × P(B | A) = P(A and B), or:

Rule: For all events, P(A and B) = P(A) × P(B | A)

Example 44: You draw two cards from the deck without looking. What’s the probability that they’re both diamonds?

Solution: Are these independent events? No! P(diamond₁), the probability that the first card is a diamond, is 13/52 because there are 13 diamonds out of 52. But if the first card is a diamond, the probability that the second card is a diamond is different. Now there are only 12 diamonds left in the deck, out of a total of 51 cards. So P(diamond₂ | diamond₁) = 12/51, which is a bit less than 13/52.

P(diamond₁ and diamond₂) = P(diamond₁) × P(diamond₂ | diamond₁)

P(diamond₁ and diamond₂) = (13/52) × (12/51) ≈ 0.0588

5C.  Sequences instead of Formulas

A lot of probability problems can be solved without using formulas, through the technique of sequences. Here’s the procedure:

1. Write down the “winning sequences”, the sequences that lead to the desired outcome.
2. Assign probabilities to each event in each sequence, from start to end.
3. Multiply the probabilities within each sequence, and then add up the probabilities of all the sequences.

Example 45: Suppose a bag contains 6 oatmeal cookies, 4 raisin cookies, and 5 chocolate chip. You are to draw two cookies from the bag without looking (and without replacement, which would be yucky). What is the probability that you will get two chocolate chip cookies?

Solution: To start with, notice that there are 6+4+5 = 15 cookies. There’s only one winning sequence, but this one illustrates an important point: you have to assign each probability in its situation at that point in its sequence.

1. Sequence: CC₁ and CC₂
2. Probabilities: 5/15 and 4/14.
   You compute the probability CC₂ at this point in the sequence: it’s the probability of a second CC if the first cookie was CC. You don’t care about the probabilities if the first cookie was anything else, because the sequence starts with a CC cookie. That means that, when you are looking for the probability of a second CC cookie, the bag now contains only 14 cookies, and only 4 of them are CC.
3. Arithmetic: (5/15)×(4/14) ≈ 0.0952

Example 46: In the same situation, what’s the probability you’ll get one oatmeal and one raisin?

Solution: Even though you don’t care which order they come in, you have to list both orders among your willing sequences. Remember the example of flipping two coins, or the examples with dice: to make probabilities come out right, consider possible orderings.

1. Sequences: (A) O₁ and R₂; (B) R₁ and O₂
2. Probabilities: (A) 5/15 and 4/14; (B) 4/15 and 5/14
3. Arithmetic: (5/15)×(4/14) + (4/15)×(5/14) ≈ 0.1905

Example 47: Consider the same bag of 15 cookies, but now what’s the probability you get two cookies the same?

Solution:

1. Sequences: (A) O₁ and O₂; (B) R₁ and R₂; (C) CC₁ and CC₂
2. Probabilities — again, the probability for the second cookie takes into account the first cookie that was drawn.
   (A) 6/15 and 5/14; (B) 4/15 and 3/14; (C) 5/15 and 4/14
3. Arithmetic: (6/15)×(5/14) + (4/14)×(3/14) + (5/15)×(4/14) ≈ 0.2952

Example 48: Your teacher’s policy is to roll a six-sided die and give a quiz if a 2 or less turns up. Otherwise, she rolls again and collects homework if a 3 or less turns up. You haven’t done the homework for today and you’re not ready for a quiz. What is the probability you’ll get caught?

Solution: Though you could do this with formulas, you’ll get the same answer with less pain by following the method of sequences. The “winning sequences” in this case are the sequences that lead to either a quiz or homework.

1. There are two sequences: (A) quiz (and stop, without deciding about homework); (B) no quiz, but homework
   Notice that you start each sequence from the same starting point. Notice also that you don’t consider the possible sequence “no quiz and no homework” because in that sequence you don’t get caught.
2. P(quiz) = 2/6 = 1/3. P(no quiz) = 1−1/3 = 2/3. P(homework if die roll) = 3/6 = 1/2.
   (A) 1/3 (B) 2/3 and 1/2
3. (1/3) + (2/3)×(1/2) = (1/3)+(1/3)= 2/3
   There’s a 2/3 probability of a quiz or homework.

Sequences let you think through a situation without getting confused about which formula may apply. Sometimes no formula applies. Here’s a famous example.

Example 49: You’re a contestant on Let’s Make a Deal. You have to pick one of three doors, knowing that there’s a new car behind one of them and a “zonk” (something funny but worthless) behind the other two. Let’s say you pick Door #1.

The host, who of course knows where the car is, opens Door #2 and shows you a zonk. He then asks whether you want to stick with your choice of Door #1, or instead take what’s behind Door #3. What should you do, and why?

(I gave specific door numbers to help make this problem less abstract, but the specifics don’t matter. What does matter is that you pick a door at random, and the host reveals that a door you didn’t pick is the wrong one.)

Solution: There’s really no formula for this one, because the host’s actions aren’t governed by probability. Once you realize that, it’s easy.

1. In the long run, 1/3 of contestants will choose the correct door, whichever one it is, and 2/3 will choose one of the two wrong doors. Why? The show’s producers have to make sure that prizes are equally distributed among the three doors over the long haul. If they favored one door over the others, people would notice and would start picking that door.

   Therefore, P(right door) = 1/3 and P(wrong door) = 2/3.

2. If you chose the right door, the host opens one of the two wrong doors, but obviously you would not benefit by switching.
3. If you chose the wrong door, the host opens the other wrong door and offers you the chance to switch doors. The host has eliminated the other wrong door, and the third door must be the winning door. You should switch.

   If you chose the wrong door and switch doors, you will always win because the host has eliminated the other wrong door.

4. The probability that you chose the right door initially, and will lose if you switch, is 1/3. The probability that you chose the wrong door initially, and will win if you switch, is 2/3.

   In the long run, keeping your original choice is the winning strategy 1/3 of the time, and switching is the winning strategy 2/3 of the time.

5. Switching doors doubles your chance of winning.

This is the famous Monty Hall Problem. Monty Hall developed Let’s Make a Deal and hosted the show for many years. There was a lot of controversy (Tierney 1991 [see “Sources Used” at end of book]) about the answer. Many people who should have known better thought that Door #1 and Door #3 were equally likely after Door #2 was opened. But they forgot that this is not a pure probability problem. The host knows where the car is and picks a door to open based on that knowledge, and that makes all the difference.

What Have You Learned?

Chapter 6 WHYL → ← Chapter 4 WHYL

Exercises for Chapter 5

Write out your solutions to these exercises, showing your work for all computations. Then check your solutions against the solutions page and get help with anything you don’t understand.

Caution! If you don’t see how to start a problem, don’t peek at the solution — you won’t learn anything that way. Ask your instructor or a tutor for a hint. Or just leave it and go on to a different problem for now. You may find when you return to that “impossible” problem that you see how to do it after all.

Problem Set 1

Problem Set 2

Solutions → 

Intro: In Chapter 5, you looked at the probabilities of specific events. In this chapter, you’ll take a more global view and look at the probabilities of all possible outcomes of a given trial.

6A.  Random Variables

The random variable is one of the main concepts of statistics, and we’ll be dealing with random variables from now till the end of the course.

In this chapter, you’ll be concerned with discrete random variables. In the next chapter, you’ll look at one particular type of continuous random variable, the normal distribution.

Example 1: You roll three dice. The number of sixes that appear is a random variable, and the total number of spots on the upper faces is another random variable. These are both discrete.

Example 2: You randomly select a household and ask the family income for last year. This is a continuous random variable.

Example 3: You randomly select twelve TC3 students, measure their heights, and take the average. “Height of a student” is a continuous random variable, and “average height in a 12-student sample” is another continuous random variable.

Example 4: You randomly select 40 families and ask the number of children in each. “Number of children in family” is a discrete random variable, and “average number of children in a sample of 40 families” is a continuous random variable.

6B.  Discrete Probability Distributions

Definition: A discrete probability distribution or DPD (also known as a discrete probability model) lists all possible values of a discrete random variable and gives their probabilities. The distribution can be shown in a table, a histogram, or a formula. Like any probabilities, the probabilities in a DPD can be determined theoretically or experimentally.

Example 5: In March 2013, Royal Auto sent me one of those “Win big!” flyers with a fake car key taped to it. The various prizes, and chances of winning, are shown at right.

This is a discrete probability distribution. The discrete variable X is “prize value”, and the five possible values of X are $100,000 down to $5.

Remember the two interpretations of probability: probability of one = proportion of all. From the table, you can equally well say that any person’s chance of winning a $500 prize is 1/250,000 = 0.000 004 = 0.0004%, or that in the long run 0.0004% of all the people who participate in the promotion will win a $500 prize.

A discrete probability distribution must list all possible outcomes. The total probability for all possible outcomes in any situation is 1. Therefore, for any discrete probability distribution, the probabilities must add up to 1 or 100%.

6B1.  Mean and Standard Deviation of a DPD

Definitions: Suppose you do a probability experiment a lot of times. (For the Royal Auto example, suppose bazillions of people show up to claim prizes.) Each outcome will be a discrete value. The mean of the discrete probability distribution, μ, is the mean of the outcomes from an indefinitely large number of trials, and the standard deviation of the discrete probability distribution, σ, is the standard deviation of the outcomes from an indefinitely large number of trials. The mean of any probability distribution is also called the expected value, because it’s the expected average outcome in the long run.

How do you find the mean and SD of a discrete probability distribution? Well, one interpretation of probability is long-term relative frequency, so you can treat a discrete probability distribution as a relative frequency distribution. (You can also think of the probabilities as weights, with the mean as the weighted average.) On the TI-83/84, that means good old 1-Var Stats, just like in Chapter 3.

Example 6: To find the mean and SD of the distribution of winnings in the Royal Auto sweepstakes, put the x’s in one list and the P(x)’s in another list. Caution: When the probability is a fraction, enter the fraction, not an approximate decimal. The calculator will display an approximate decimal, but it will do its calculations on a much more precise value.

After entering the x’s and p’s, press [STAT] [►] [1] and specify your two lists, such as 1-Var Stats L1,L2. (Yes, the order matters: the x list must be first and the P(x) list second.) When you get your results, check n first. In a discrete probability distribution, n represents the total of the probabilities, so it must be exactly 1. If it’s just approximately 1, you made a mistake in entering your probabilities.

The mean of the distribution is μ = $5.03, and the standard deviation is σ = $45.85.

Interpretation: In the long run, the dealership will have to pay out $5.03 per person in prizes. The SD is a little harder to get a grasp on, but notice that it’s more than nine times the mean. This tells you that there is a lot of variability in outcome from one person to the next. In general, the mean tells you the long-term average outcome, and the SD tells you the unpredictability of any particular trial. You can look at the SD as a measure of risk.

A couple of notes about the calculator output: The calculator knows that a DPD is a population, so it gives you σ and not s for the SD. It should give you μ for the mean, but instead it displays x̅, so you need to make the change. I’ve already mentioned that the sum of the probabilities (n) must be exactly 1, not just approximately 1.

6B2.  Comparing DPDs: Parking Choices

Example 7: When visiting the city, should you park in a lot or on the street? On a quarter of your visits (25%), you park for an hour or less, which costs $10 in a lot; for parking more than an hour they charge a flat $14. If you park on the street, you might receive a simple $30 parking ticket (p = 20%), or a $100 citation for obstruction of traffic (p = 5%), but of course you might get neither. Which should you do?

(Adapted from Paulos 2004 [see “Sources Used” at end of book].)

You have two probability models here, one for the outcomes of parking in a lot, and one for street parking. Begin by putting the two models into tables:

The problem leaves out some things that you can figure for yourself. Remember that every probability model includes all outcomes, and the probabilities add up to 1. If there’s a 25% chance of parking up to an hour, there must be a 100−25 = 75% chance of parking more than an hour. And on the street, if you have a 20+5 = 25% chance of getting some kind of ticket, you have a 100−25 = 75% chance of getting neither. The cost of getting neither ticket is zero.

Now you can fill in the empty cells in the tables.

I showed the total probability to emphasize that it’s 1. Never compute the total of the outcomes (x’s), because that wouldn’t mean anything.

How do these tables help you make up your mind where to park? By themselves, they don’t. But they let you compute μ and σ, and that will help you decide.

I placed the x’s and P(x)’s for the parking lot in L1 and L2, and did 1-Var Stats L1,L2. I placed the x’s and P(x)’s for street parking in L3 and L4 and did 1-Var Stats L3,L4. Here are the results:

Lot:      Street:

As always, look first at n. If it’s not exactly 1, find your mistake in entering the probabilities.

Now you can interpret these results. Parking in the lot is a bit more expensive in the long run (μ = $13.00 per day versus μ = $11.00 per day). But there are no nasty surprises (σ = $1.73, little variation from day to day). Parking on the street is much riskier (σ = $23.64), meaning that what happens today can be wildly different from what happened yesterday.

So what should you do? Statistics can give you information, but part of your decision is always your own temperament. If you like stability and predictability — if you are risk averse — you’ll opt for the parking lot. If it’s more important to you to save $2 a day on average, and you can accept occasionally getting hit with a nasty fine, you’ll choose to park on the street.

6B3.  Fair Price of a Game

(“Fair price” is one of those math words that look like English but mean something different. You should expect to pay more than the fair price because the operator of the game — the insurance company or casino or stockbroker — also has to cover selling and administrative expenses.)

There are two ways to compute the fair price:

• Method 1: Ignore the actual price of the game, multiply each prize by its probability, and add up the products.
• Method 2: If you already know the mean of the probability distribution from the player’s point of view, then fair price = actual price + μ.

Example 8: Take a really simple bar game: a stranger offers to pay you $60 if you roll a 6 with a standard six-sided die, but you have to pay him $12 per roll. Find the fair price of this game.

Method 1: The only prize is $60, and you have a 1/6 chance of winning it. $60×(1/6) = $10.

Method 2: Amounts in L1, probabilities in L2; 1-VarStats L1,L2. Verify that n=1, and read off the mean of −$2. The actual price is $12, so the fair price is $12 + (−2) = $10.

Naturally, the two methods always give the same answer. Method 2 is easier if you already know the mean of the probability distribution; otherwise Method 1 is easier.

Example 9: A lottery has a $6,000,000 grand prize with probability of winning 1 in 3,000,000. It also has a $10 consolation prize with probability of winning 1 in 1000. What is the fair price of your $5 lottery ticket?

Solution: You don’t need μ, so Method 1 is easier: multiply each prize by its probability and add up the products. $6,000,000×(1/3,000,000) + $10×(1/1000) → fair price is $2.01.

Why does a lottery ticket that is worth $2.01 actually cost $5.00? In effect, the lottery is paying out about 2.01/5.00 ≈ 40% of ticket sales in prizes. Some of the 60% that the lottery commission keeps will cover the lottery’s own expenses, and the rest is paid to the state treasury. This is actually fairly typical: most lotteries pay out in prizes less than half of what they take in. By contrast, the illegal “numbers game” pays out about 70%, or at least it did in the 1980s in Cleveland. (Don’t ask me how I know that!)

6C.  Bernoulli Trials

In the examples so far of probability models, I’ve had to give you a table of probabilities. But there are many subtypes of discrete probability distribution where the probabilities can be calculated by a formula. The rest of this chapter will look at part of one family, discrete probability distributions that come from Bernoulli trials.

Bernoulli trials are named after Jacob Bernoulli, a Swiss mathematician. He developed the binomial distribution, which you’ll meet later in this chapter.

Example 10: You randomly interview 30 people to find out which party they will vote for in the next election. These are not Bernoulli trials, because there are more than two possible outcomes. (New York State ballots often have six or more parties listed, though some parties just endorse the Republican or Democratic candidate.)

6D.  The Geometric Model

Example 13: According to the AVMA (2014) [see “Sources Used” at end of book] 30.4% of US households own one or more cats. Suppose you randomly select some households.
(a) How likely is it that the first time you find cat owners is in the fifth household?
(b) How likely is it that your first cat-owning household will be somewhere in the first five you survey?

Although you could compute these individual probabilities using techniques from Chapter 5, there’s a specific model called the geometric model that makes it a lot easier to compute. Also, using the geometric model you can get an overview of the probabilities for various outcomes, which you’d miss by computing probabilities of specific events using the previous chapter’s techniques. If trials are independent, and you want the probability of a string of failures before your first success, you’re using a geometric model.

The probability of success on any given trial, p, completely describes a geometric model.

Here’s a picture of part of the geometric model for cat-owning households, with p = 0.304.

How do you read this? The horizontal axis is x, the number of the trial that gives your first success, and the vertical axis is P(x), the probability of that outcome.

For example, there’s a hair over a 30% chance that you’ll find cat owners in your first household, P(1) = 30.4%. There’s about a 21% chance that the first household won’t own cats but the second household will, P(2) ≈ 21%. Skipping a bit to x = 6, there’s just about a 5% chance that the first five households won’t have cats but the sixth will, P(6) ≈ 5%. And so forth.

x = 1 is always the most likely outcome, and larger x values are successively less and less likely. This is true for every geometric distribution, not just this particular one with p = 0.304.

The geometric model never actually ends. The probabilities eventually get too small to show in the picture, but no matter what x you pick, the probability is still greater than 0.

6D1.  Computing Probabilities

Your TI-83/84 calculator has two menu selections for the geometric model:

• geometpdf(p,x) answers the question “what’s the probability that my first success will come at trial number x?”
• geometcdf(p,x) answers the question “what’s the probability that my first success will come at or before trial number x?” (The “c” stands for cumulative, because the cdf functions accumulate the probabilities for a range of outcomes.)

They’re both in the [2nd VARS makes DISTR] menu.

(If you have a calculator in the TI-89 family, use the [F5] Distr menu. Select Geometric Pdf and Geometric Cdf.)

Let’s use the calculator to find the answers for Example 13. Here p, the probability of success in any given household, is 30.4% or 0.304.

Part (a) wants the probability of four failures followed by a success on the fifth try. For that you use geometpdf. Press [2nd VARS makes DISTR] [▲] [▲] to get to geometpdf, and press [ENTER].

There’s about a 7% chance you won’t find any cat owners in the first four households but you will in the fifth household.

(You could calculate this the long way. The probability of four failures followed by a success is (1−.304)⁴×.304. But the geometric model is easier. That’s the point of a model: one general rule works well enough for all cases, so you don’t have to treat each situation as a special case with its own unique methods.)

Part (b) wants the probability of a success occurring anywhere in the first five trials. This is a geometcdf problem. Press [2nd VARS makes DISTR] [▲] to get to geometcdf, and press [ENTER].

There’s almost an 84% chance you will find at least one cat-owning household among the first five.

(Doing this the long way, you would use the complement. The complement of “at least one cat-owning household in the first five” is “no cat-owning households in the first five”. The probability that a given household doesn’t own a cat is q = 1−.304 = 0.696, and the probability that five in a row don’t own cats is 0.696⁵. Therefore the original probability you wanted is 1−(.696⁵) = 0.8367.)

You don’t actually need formulas for the geometric model, but if you’re curious about what your calculator is doing, here they are:
       geometpdf(p,x) = q^x−1p     geometcdf(p,x) = 1−q^x
where q = 1−p as usual. You can see that the two “long way” paragraphs above actually used those formulas.

6D2.  Mean and Standard Deviation of a Geometric Distribution

The geometric distribution is completely specified by p, so you can compute the mean and standard deviation quite easily:

μ = 1/p          σ = μ √q  or  (1/p) √(1−p)

Example 14: 30.4% of US households own cats. How many households do you expect you’ll need to visit to find a cat-owning household?

Solution: The expected value of a distribution is the mean. μ = 1/p = 1/.304 = 3.289473684. μ = 3.3. Interpretation: On average, you expect to have to visit between 3 and 4 households to find the first cat owners.

Caution! The expected value (mean) is not the most likely value (mode). Take a look back at the histogram, and you’ll see that the most likely value is 1: you’re more likely to get lucky on the first trial than on any other specific trial. But the distribution is highly skewed right, so the average gets pulled toward the higher numbers.

To compute the SD, just multiply the mean by √q. A handy technique is called chaining calculations. After first calculating the mean, press the [×] key, and the calculator knows you are multiplying the previous answer by something. Here you see that σ = 2.7.

Interpreting σ is a bit harder. The geometric distribution is a type of discrete probability distribution, so you interpret its standard deviation the same way as for any other DPD. In this particular example, σ is almost as large as μ, so you expect a lot of variability. If you and a lot of co-workers go out independently looking for households with cats, the group average number of visits will be 3.3 households, but there will be a lot of variability between different workers’ experience. You can’t use the Empirical Rule here because the geometric model is not a bell curve, but you can at least say you won’t be surprised to find workers who get lucky on the first house (μ−σ ≈ 0.5), and workers who have to visit six houses or more (μ+σ ≈ 6.0).

6D3.  Making a Decision

Some people find it very hard to make choices because they feel they must consider all the pros and cons of every possibility. Others look at possibilities one at a time and take the first one that’s acceptable. Studies such as The Tyranny of Choice (Roets, Schwartz, Guan 2012 [see “Sources Used” at end of book]) show that the first group may make better choices objectively, but the second group is happier with the items they choose.

Example 15: You have to buy a new sofa. You’d be content with 55% of the sofas out there. Let’s assume that your Web search presents sofas in an order that has nothing to do with your preferences. There are hundreds to choose from, so you decide to adopt the “first one that’s acceptable” strategy. How likely is it that you’d order the third sofa you’d see?

Solution: This is a geometric model, with two failures followed by one success. p = 55%. geometpdf(.55,3) = .111375. There’s about an 11% chance you’d order the third sofa.

6D4.  Baseball

Example 16: Larry’s batting average is .260. During which time at bat would he expect to get his first hit of the game? How likely is he to get his first hit within his first four times at bat?

Solution: This is a geometric model with p = 0.260. The mean or expected value is 1/p = 1/.26 = 3.85, about 4. On average, his first hit each game will come on his fourth time at bat. For the second question, geometcdf(.26,4) = .70013424; there’s about a 70% chance he’ll get his first hit within his first four times at bat.

6E.  The Binomial Model

In the previous section, we looked at the geometric model, where you just keep trying until you get a success. In this section, we’ll look at the binomial model, where you have a fixed number of trials and a varying number of successes.

Example 17: Cats again! 30.4% of US households own one or more cats. You visit five households, selected randomly.
(a) What’s the chance that no more than two have cats?
(b) What’s the chance that exactly two have cats?
(c) What’s the chance that at least two have cats?
(d) What’s the chance that two to four have cats?

This problem fits the binomial model: n = 5 trials, each household does or does not have cats, and the probability p = 30.4% is the same for each household.

A picture of this binomial distribution is shown at right, and you can see some differences from the picture of the geometric distribution:

• The geometric distribution extends from one trial to ∞, but the binomial distribution can have only 0 to n successes in n trials.
• The binomial distribution is less strongly skewed than the geometric distribution, and 1 is not necessarily the most likely value of X (though it happens that way in this particular distribution).

How do you read the picture? There’s about a 17% probability that none of the five households will have cats, about 36% that one of the five will have cats, and so on. (Why 36% and not 30.4%? Because there’s a greater chance of “winning” one out of five than one out of one.)

In this book we’re more concerned with computing probabilities, but it can be nice to get an overall picture of a distribution. I made this particular graph by using @RISK from Palisade Corporation, but you can also make histograms of binomial distributions by using MATH200A Program part 1(5).

6E1.  Computing Probabilities

Here you have a choice. Your TI-83/84 calculator comes with two menu selections for the binomial model, but the MATH200A program gives you a simpler interface. Here’s a quick overview of both, before we start on computations:

Got a TI-89 family calculator? Use the [F5] Distr menu. Select Binomial Pdf or Binomial Cdf. The Cdf function can handle any range of successes, not just 0 to x. See Binomial Probability Distribution on TI-89 for full instructions.)

Now let’s use your TI-83/84 to answer the questions in Example 17. You have five trials, so n = 5. The probability of success on any given household is 30.4%, so p = 0.304.

(a) What’s the probability that no more than two of the five randomly selected households have cats?

With the MATH200A program (recommended):

If you’re not using the program:

Press [PRGM], select MATH200A, and press [3] in the MATH200A menu. Enter n and p. “No more than two cats” is from 0 to 2 cats, so enter those values when prompted. The program echoes back your inputs and shows the computed probability. To show your work, write down the screen name, the inputs, and the result. Conclusion: P(x ≤ 2) or P(0 ≤ x ≤ 2) = 0.8316.

The probability that no more than two of your five households have cats (in other words, the probability that 0 to 2 have cats) is binomcdf(5,.304,2). Press [2nd VARS makes DISTR] and scroll up to binomcdf. If you don’t have the “wizard” interface, or you have it turned off, binomcdf( will appear on your screen, Enter n, p, and the desired maximum number of successes, in that order, then the closing paren and [ENTER]. If you have the “wizard” interface, you get a menu screen, but you enter the same information. Press [ENTER] once on Paste and then again when the command is pasted to your home screen. Either way, write down the binomcdf command and the argument numbers to show your work. Conclusion: P(x ≤ 2) or P(0 ≤ x ≤ 2) = 0.8316.

(b) What’s the probability that exactly two of five randomly selected households are cat owners?

With the MATH200A program (recommended):	If you’re not using the program:
You need a specific number of successes, instead of a range. It’s almost exactly the same deal: you just enter the same number for from and to. In this example, to get the probability of exactly two successes, enter number of successes from 2 to 2. Conclusion: P(x = 2) or P(2) = 0.3116.	(a) The probability of exactly two cat-owner households in five is binompdf(5,.304,2). Press [2nd VARS makes DISTR] and then press [▲] several times to get to binompdf. (Caution! pdf, not cdf.) Press [ENTER], type in the numbers, and press [)] [ENTER]. (The “wizard” interface screen is the same as it was for binomcdf.) Conclusion: P(x = 2) or P(2) = 0.3116.

(c) What’s the probability that at least two of the five randomly selected households have cats?

(d) What’s the chance for two to four cat-owning households in your random sample of five households?

You don’t actually need a formula for the binomial model, but if you’re curious about what your calculator is doing, here it is:
       binompdf(n,p,x) = _nC_x · p^x q^n−x
Why? p^x is the probability of getting successes on all of the first x trials. q is the probability of failure on one trial, and therefore q^n−x is the probability of failure on the remaining trials, after the x out of n successes. But in a binomial probability model, you care how many successes and failures there are, not in what order they occur. To account for the fact that order doesn’t matter, the formula has to multiply by _nC_x, “the number of ways to choose x objects out of n”. (If you want to know more about _nC_x, search “combinations” at your favorite math site.)

Unlike the geometric case, there’s no simple formula for binomcdf. Your calculator just has to compute probabilities for x = 0, 1, and so on and add them up.

6E2.  Baseball Again!

Example 18: Larry’s batting average is .260. How likely is it that he’ll get more than one hit in four times at bat?

Solution: This is a binomial model with n = 4, p = 0.26, x = 2 to 4. You can use MATH200A part 3 or the binompdf-sum technique to get .27870128. P(x > 1) = 0.2787 or about 28%. (The program is completely straightforward, so I’m showing only the tricky binompdf-sum sequence here.)

Alternative solution: If you don’t have the program, can you see how to use the complement to solve this problem more easily? Check your answer against mine to be sure that your method is correct.

6E3.  Mean and Standard Deviation of a Binomial Distribution

The binomial distribution depends on the proportion in the population (p) and your sample size (n). You can compute the mean and SD quite easily: