Stats without Tears
Solutions to All Exercises

Nothing can eliminate sampling error, but you can reduce it by increasing your sample size. (Most nonsampling errors can be avoided by proper experimental design and technique.)

Although “numeric” or “quantitative” is correct, it’s not an adequate answer because it is not as specific as possible. Discrete and continuous data are treated differently in descriptive statistics, so it matters which type you have.

Students are sometimes fooled by the decimal. Always ask yourself what was the original question asked or the original measurement taken from each member of the sample.

Common mistake: Students sometimes answer “80” for population size, but this is not correct. You took data from 80 people, so those 80 people are your sample and 80 is your sample size.

What can be done to reduce response bias? Interviewers should be trained to be absolutely neutral in voice and facial expression, which is how the Kinsey team gathered data on sexual behavior. Or the question can be asked on a written questionnaire, so that the subject isn’t looking another person in the face when answering. The question can also be made less threatening: “Have you ever left an infant alone in the house, even for just a minute?”

• Random sample: get a list of the resident students. On your calculator, do randInt(1,2000) 50 times, not counting duplicates, and interview the students who came up in those positions.
• Systematic sample: You can’t station yourself in the cafeteria because that would exclude all students who don’t use it. Instead, station yourself at the main entrance to the dorm complex (or station yourself and confederates at the main entrance to each dorm) and interview every 20th person. Why k=20 and not 2000/50 = 40? Because whenever you’re there, you’re bound to miss a sizable proportion of students.

  To select the first person to survey, use randInt(1,20). Remember that a systematic survey begins with a randomly selected person from 1 to k, not 1 to 50 (sample size) or 1 to 2000 (population size).

  Notice that I didn’t suggest a time frame. What do you think would be a good time to do this?

  An alternative procedure might be to walk through the dorms (assuming you can get in) and interview the students in every 20th room. You may get better coverage that way than if you wait for them to come to you.

• Cluster sample: Randomly select 25 rooms, and interview both of the students in those rooms. (This is a single-stage cluster.)

Best balance? Probably the cluster sample. The true random sample is a lot of work for a sample of 50, because after selecting the names you have to track the students down. The systematic sample, no matter how you do it, is going to miss a lot of students, and you have that time-period problem. With the cluster sample, you can time it for when students are likely to be home, and you can go back to follow up on those you missed.

But nothing is perfect, in this life where we are born to trouble as the sparks fly upward. The cluster sample works if the students were randomly assigned to rooms. When students pick their own roommates, they tend to pick people with similar attitudes, interests, and activities. That means those two are more similar to each other than other students, and there’s no way you can treat that cluster sample as a random sample. The cluster would probably be safe for freshman, where the great majority would be randomly assigned, but less so for students in later years.

Students often answer questions like this with hand-waving arguments, either coming up with reasons why it’s a plausible conclusion or coming up with reasons why it isn’t. This is statistics, and we have to follow the facts. Whatever you may think about Fox News, the fact is that observational studies can’t prove causation.

Alternative: the more specific answer binomial data, which you may have heard in the lecture though it’s not in the book till Chapter 6.

(b) This is descriptive statistics because it’s reporting data actually measured: 42% of the sample. If it said “42% of Americans”, then it would be inferential because you know not every American was asked, so the investigators must have extrapolated from a sample to the population.

(c) It is a statistic because it is a number that summarizes data from a sample.

• The first people who present themselves are chosen. You should randomly select from among all volunteers. (Better still would be to randomly select from among all patients, and ask the selected individuals to volunteer.)
• Participants are not randomly assigned to control and experimental groups. This is always bad, but it’s especially bad when you accept a block of volunteers in order.
• The experiment is not double blind, only single blind. When doctors know who is getting a placebo and who is getting medicine, they may treat the two groups differently, consciously or unconsciously.

All of these are nonsampling errors.

To fix it, round to one decimal place: 1.9. (Don’t make the common mistake of “rounding” to 1.8.)

There’s no scale to interpret the quantities. And if one fruit in each row is supposed to represent a given quantity, then banana and apple have the same frequency, yet banana looks like its frequency is much greater.

90% of 15 is 13.5, 80% is 12, 70% is 10.5, and 60% is 9.

Alternatives: Instead of a title below the category axis, you could have a title above the graph. You could order the grades from worst to best (F through A) instead of alphabetically as I did here. And you could list the class boundaries as 13.5–15, 12–13.5, 10.5–12, and so on, with the understanding that a score of 12 goes into the 12–13.5 class, not the 10.5–12 class. (Data points “on the cusp” always go into the higher class.)

(a) The variable is discrete, “number of deaths in a corps in a given year”.
(b)
Alternatives: Some authors would draw a histogram (bars touching) or even a pie chart. Those are okay but not the best choice.

Relative frequency is f/n. f = 25, and n = 35+10+25+45+20 = 135. Dividing 25/135 gives 0.185185… ≈ 0.19 or 19%

skewed right

(a) See the histogram at right. Important features:

• The bars are labeled at their edges, not their centers, because this is a grouped histogram.
• Both axes are titled.
• The horizontal axis has a real-world title. (Sometimes you also need an overall title for the graph, but here the axis title says all that needs to be said.)

(b) 480.0−470.0 = 10.0 or just plain “10”.

Don’t make the common mistake of subtracting 479.9−470.0. Subtract consecutive lower bounds, always.

(c) skewed left

Alternative solution: In a normal distribution, the mean is half way between the given extremes: μ = (4.50+8.50)/2 = 6.50. Then the distance from the mean to 8.50 must be three SD: 8.50−6.50 = 2.00 = 3σ; σ = 0.67 ounces.

Caution! The midpoints are not midway between lower and upper bounds, such as (20+29)/2 = 24.5. They are midway between successive lower bounds, such as (20+30)/2 = 25.

1-VarStats L1,L2 (Check n first!)

x̅ = 63.85656971 → x̅ = 63.86

s = 15.43533244 → s = 15.44

n = 997

Common mistake: People tend to run 1-VarStats L1, leaving off the L2, which just gives statistics of the seven numbers 25, 35, …, 85. Always check n first. If you check n and see that n = 7, you realize that can’t possibly be right since the frequencies obviously add up to more than 7. You fix your mistake and all is well.

(b) You need the original data to make a boxplot, and here you have only the grouped data. A boxplot of a grouped distribution doesn’t show the shape of the data set accurately, because only class midpoints are taken into account. The class midpoints are good enough for approximating the mean and SD of the data, but not the five-number summary that is pictured in the boxplot.

1-VarStats L1,L2

n = 14 (This is the number of credits attempted. If you get 5, you forgot to include L2 in the command.)

x̅ = 2.93

Commute Distance, km
 0- 9     4
10-19    12
20-29     7
30-39     1
40-49     1
Total    25

(d) Data in a list such as L1. 1-VarStats L1 gives x̅ = 17.6 km, Median = 17, s = 9.0  km, n = 25

(e)

(f) Mean, because the data are nearly symmetric. Or, median, because there is an outlier.
Comment: The stemplot made the data look skewed, but that was just an artifact of the choice of classes. The boxplot shows that the data are nearly symmetric, except for that outlier. This is why the mean and median are close together. This is a good illustration that sometimes there is no uniquely correct answer. It’s why your justification or explanation is an important part of your answer.

(g) The five-number summary, from MATH200A part 2 [TRACE], is 1, 12, 17 22.5, 45. There is one outlier, 45.
(The five-number summary includes the actual min and max, whether they are outliers or not.)

Draw an auxiliary line at z = −2. You know that the area between z = −2 and z = +2 is 95%, so the area between z = 0 and z = 2 is half that, 47.5% or 0.475.

z_J = (2070−1500)/300 = 570/300 = 1.90

z_M = (129−100)/15 = 29/15 = about 1.93

Because she has the higher z-score, according to the tests Maria is more intelligent.

Remark: The difference is very slight. Quite possibly, on another day Jacinto might do slightly better and Maria slightly worse, reversing their ranking.

Put class midpoints in a list, such as L1, and frequencies go in another list, such as L2. (Either label the columns with the lists you use, as I did here, or state them explicitly: “class marks in L1, frequencies in L2”.)

1-VarStats L1,L2 (Always write down the command that you used.)

(a) n = 154

(b) x̅ = 499.81 (before rounding, 499.8051948)

(c) s = 12.74 (before rounding, 12.74284519)

Be careful with symbols. Use the correct one for symbol or population, whichever you have.

Common mistake: The SD is 12.74 (S_x), not 12.70 (σ), because this is a sample and not the population.

Common mistake: Don’t use any form of the word “correlation” in your answer. Your friend wouldn’t understand it, but it’s wrong anyway. Correlation is the interpretation of r, not R². Yes, r is related to R², but R² as such is not about correlation.

Common mistake: R² tells you how much of the variation in y is associated with variation in x, not the other way around. It’s not accurate to say 64% of variation in age is associated with variation in salary.

Common mistake: Don’t say “explained by” to non-technical people. The regression shows an association, but it does not show that growing older causes salary increases.

(b)  Yes

(c) The results of LinReg(ax+b) L1,L2,Y1 are shown at right. The correlation coefficient is r = 0.91

(d) ŷ = 0.1127x − 35.1786
Note: ŷ, not y. Note: −35.1786, not +−35.1786.

(e) The slope is 0.1127. An increase of 1000 power-boat registrations is associated with an increase of about 0.11 manatee deaths, on average.
It’s every 1000 boats, not every boat, because the original table is in thousands. Always be specific: “increase”, not just “change”.

Remark: Although this is mathematically accurate, people may not respond well to 0.11 as a number of deaths, which obviously is a discrete variable. You might multiply by 100 and say that 100,000 extra registrations are associated with 11 more manatee deaths on average; or multiply by 10 and round a bit to say that 10,000 extra registrations are associated with about one more manatee death on average.

(f) The y intercept is −35.1786. Mathematically, if there were no power boats there would be about minus 35 manatees killed by power boats. But this is not applicable because x=0 (no boats) is far outside the range of x in the data set.

(g) R² = 0.83. About 83% of variation in manatee deaths from power boats is associated with the variation in registrations of power boats.
It’s R², not r². And don’t use any form of the word “correlate” in your answer.

100% of manatee power-boat deaths come from power boats, so why isn’t the association 100%? The other 17% is lurking variables plus natural variability. For instance, maybe the weather was different in some years, so owners were more or less likely to use their boats. Maybe a campaign of awareness in some years caused some owners to lower their speeds in known manatee areas.

(h)  ŷ = 27.8

(i) y−ŷ = 34−27.8 = 6.2

(j) Remember that x is in thousands, so a million boats is x = 1000. But x=1000 is far outside the data range, so the regression can’t be used to make a prediction.

(b) The results of LinReg(ax+b) L3,L4,Y2 are shown at right. ŷ = −3.5175x+6.4561

(c) The slope is −3.5175. Increasing the dial setting by one unit decreases temperature by about 3.5°.
Again, state whether y increases or decreases with increasing x.

(d) The y intercept is 6.4561. A dial setting of 0 corresponds to about 6.5°.

(e) r = −0.99

(f) R² = 0.98. About 98% of variation in temperature is associated with variation in dial setting.

This seems almost too good to be true, as though the data were just made up. ☺ But it’s hard to think of many lurking variables. Maybe it happened that some measurements were taken just after the compressor shut off, and others were taken just before the compressor was ready to switch on again in response to a temperature rise.

(g)  ŷ = 2.9°

r, the linear correlation coefficient, would be roughly zero. Taking the plot as a whole, as x increases, y is about equally likely to increase or decrease. A straight line would be a terrible model for the data.

Clearly there is a strong correlation, but it is not a linear correlation. Probably a good model for this data set would be a quadratic regression, ŷ = ax²+bx+c. Though we study only linear regressions, your calculator can perform quadratic and many other types.

Remark: Don’t say “caused by” variation in family income. Correlation is not causation. You can think of some reasons why it might be plausible that wealthier families are more likely to produce smarter children, or at least children who do better on standardized tests, but you can’t be sure without a controlled experiment.

Remark: Though it’s an interesting fact, the correlation in twins’ IQ scores is not needed for this problem. In real life, an important part of solving problems and making decisions is focusing on just the relevant information and not getting distracted.

(b)

(c) Three events out of eight equally likely events: P(2H) = 3/8

Common mistake: Sometimes students write the sample space correctly but miss one of the combinations of 2 heads. I wish I could offer some “magic bullet” for counting correctly, but the only advice I have is just to be really careful.

(a) In a probability model, the probabilities must add to 1 (= 100%). The given probabilities add to 62.6%. What is the missing 37.4%? They’ve accounted for cell and landline, cell only, and nothing; the remaining possibility is landline only. The model is shown at right.

(b) P(Landline) = P(Landline only) + P(Landline and cell)

P(Landline) = 37.4% + 58.2% = 95.6%

Remark: “Landline” and “cell” are not disjoint events, because a given household could have both. But “landline only” and “landline and cell” are disjoint, because a given house can’t both have a cell phone with landline and have no cell phone with landline.

(b) That A and B are complementary means that one or the other must happen, but not both. Therefore P(B) = P(not A) → P(B) = 0.3

(c) Since the events are complementary, they can’t both happen: P(A and B) = 0

Common mistake: Many students get (c) wrong, giving an answer of 1. If events are complementary, they can’t both happen at the same time. That means P(A and B) must be 0, the probability of something impossible.

Maybe those students were thinking of P(A or B). If A and B are complementary, then one or the other must happen, so P(A or B) = P(A) + P(B) = 1. But part (c) was about probability and, not probability or.

This is the difference between theoretical and empirical probability. A truly impossible event has a theoretical probability of zero. But the 0 out of 412 figure is an empirical probability (based on past experience). Empirical probabilities are just estimates of the “real” theoretical probability. From the empirical 0/412, you can tell that the theoretical probability is very low, but not necessarily zero. In plain language, an unresolved complaint is unlikely, but just because it hasn’t happened yet doesn’t mean it can’t happen.

Common mistake: Students often try some sort of complicated calculation here. You would have to do that if conditions were stated on all five of those cards, but they weren’t. Think about it: any card has a 1/4 chance of being a spade.

(a) 0.0171 × 0.0171 = 0.0003

(b) The events are not independent. When a married couple are at home together or out together, any attack that involves one of them will involve the other also.

(b) About 10.38% of American adults in 2006 were divorced. If you randomly selected an American adult in 2006, there was a 0.1038 probability that he or she was divorced.

(c) Empirical or experimental

(d) P(divorced^C) = 1−P(divorced) = 1−22.8/219.7 ≈ 0.8962
About 89.62% of American adults in 2006 were not divorced (or, had a marital status other than divorced).

(e) P(man and married) = 63.6/219.7 ≈ 0.2895 (You can’t use a formula on this one.)

(f) Add up P(man) and P(not man but married):

P(man or married) = 106.2/219.7 + 64.1/219.7 ≈ 0.7751

Alternative solution: By formula:

P(man or married) = P(man) + P(married) − P(man and married)

P(man or married) = 106.2/219.7 + 127.7/219.7 − 63.6/219.7 = 0.7751

Remember, math “or” means one or the other or both.

(g) What proportion of males were never married? 30.3/106.2 = 28.53%.

(h) P(man | married) uses the sub-subgroup of men within the subgroup of married persons.

P(man | married) = 63.6/127.7 = 0.4980

49.80% of married persons were men.

Remark: You might be surprised that it’s under 50%. Isn’t polygamy illegal in the US? Yes, it is. But the table considers only resident adults. Women tend to marry slightly earlier than men, so fewer grooms than brides are under 18. Also, soldiers deployed abroad are more likely to be male.

(i) P(married | man) used the sub-subgroup of married persons within the subgroup of men.

P(married | men) = 63.6/106.2 = 0.5989

59.89% of men were married.

(a) 3 of 20 M&Ms are yellow, so 17 are not yellow. You want the probability of three non-yellows in a row:
(17/20)×(16/19)×(15/18) ≈ 0.5965

(b) The probability is zero, since there are only two reds to start with.

(a) Since the companies are independent, you can use the simple multiplication rule:

P(A bankrupt and W bankrupt) = P(A bankrupt) × P(W bankrupt)

P(A bankrupt and W bankrupt) = .9 × .8 = 0.72

At this point you could compute (b), but it’s little messy because you need the probability that A fails and W is okay, plus the probability that A is okay and W fails. (c) looks easier, so do that first.

(c) “Neither bankrupt” means both are okay. Again, the events are independent so you can use the simple multiplication rule.

P(neither bankrupt) = P(A okay and W okay)

P(A okay) = 1−.9 = 0.1; P(W okay) = 1−.8 = 0.2

P(neither bankrupt) = .1 × .2 = 0.02

(b) is now a piece of cake.

P(only one bankrupt) = 1 − P(both bankrupt) − P(none bankrupt)

P(only one bankrupt) = 1 − .72 − .02 = 0.26

Remark: If you have time, it’s always good to check your work and work out (b) the long way. You have only independent events (whether A is okay or fails, whether W is okay or fails) and disjoint events (A fails and W okay, A okay and W fails). The “okay” probabilities were computed in part (c).

P(only one bankrupt) = (A bankrupt and W okay) or (A okay and W bankrupt)

P(only one bankrupt) = (.9 × .2) + (.1 × .8) = 0.26

Common mistake: When working this out the long way, students often solve only half the problem. But when you have probability of exactly one out of two, you have to consider both A-and-not-W and W-and-not-A.

You can’t use the “or” formula here, even if you studied it. That computes the probability of one or the other or both, but you need the probability of one or the other but not both.

Remark: If you computed all three probabilities the long way, pause a moment to check your work by adding them to make sure you get 1. Whenever possible, check your work with a second type of computation.

(a) 15(You can assume independence because it’s a small sample from a large population.) P(red₁ and red₂ and red₃) = 0.13×0.13×0.13 = 0.0022

(b) P(red) = 0.13; P(red^C) = 1−0.13 = 0.87.
P(red₁^C and red₂^C and red₃^C) = 0.87×0.87×0.87 or 0.87³ = 0.6585

Common mistake: Students sometimes compute 1−.13³. But .13³ is the probability that all three are red, so 1−.13³ is the probability that fewer than three (0, 1, or 2) are red. You need the probability that zero are red, not the probability that 0, 1, or 2 are red. Think carefully about where your “not” condition must be applied!

(c) The complement is your friend with “at least” problems. The complement of “at least one is green” is “none of them is green”, which is the same as “every one is something other than green.”
P(green) = 0.16, P(non-green) = 1−0.16 = 0.84.
P(≥1 green of 3) = 1 − P(0 green of 3) = 1 − P(3 non-green of 3) = 1−0.84³ ≈ 0.4073

(d) (Sequences are the most practical way to solve this one.)
(A) G₁ and G₂^C and G₃^C; (B) G₁^C and G₂ and G₃^C; (C) G₁^C and G₂^C and G₃
.16×(1−.16)×(1−.16) + (1−.16)×.16×(1−.16) + (1−.16)×(1−.16)×.16 ≈ 0.3387

P(all 5 attended) = 0.45^5 = 0.0185

P(at least 1 had not attended) = 1 − 0.0185 = 0.9815

(cherry₁ and orange₂) or (orange₁ and cherry₂)

Common mistake: There are two ways to get one of each: cherry followed by orange and orange followed by cherry. You have to consider both probabilities.

There are 11+9 = 20 sourballs in all, and Grace is choosing the sourballs without replacement (one would hope!), so the probabilities are:

(11/20)×(9/19) + (9/20)×(11/19) = 99/190 or about 0.5211

P(win ≥1) = 1−P(win 0) = 1−P(lose 5).

P(lose) = 1−P(win) = 1−(1/500) = 499/500

P(lose 5) = [P(lose)]⁵ = (499/500)^5 = 0.9900

P(win ≥1) = 1−P(lose 5) = 1−0.9900 = 0.0100 or 1.00%

Common mistake: If you compute 1−(499/500)⁵ in one step and get 0.00996008, be careful with your rounding! 0.00996… rounds to 0.0100 or 1%, not 0.0010 or 0.1%.

Common mistake: 1/500 + 1/500 + … is wrong. You can add probabilities only when events are disjoint, and wins in the various years are not disjoint events. It is possible (however unlikely) to win more than once; otherwise it would make no sense for the problem to talk about winning “at least once”.

Common mistake: You can’t multiply 5 by anything. Take an analogy: the probability of heads in one coin flip is 50%. Does that mean that the probability of heads in four flips is 4×50% = 200%? Obviously not! Any process that leads to a probability >1 must be incorrect.

Common mistake: 1−(1/500)⁵ is wrong. (1/500)⁵ is the probability of winning five years in a row, so 1−(1/500)⁵ is the probability of winning 0 to 4 times. What the problem asks is the probability of winning 1 to 5 times.

(a) P(not first and not second) = P(not first) × P(not second) = (1−.7)×(1−.6) = 0.12

(c) P(first and second) = P(first) × P(second) = .7×.6 = 0.42

(b) 1−.12−.42 = 0.46

Alternative: You could compute (b) directly too, using sequences:

P(exactly one copy recorded) =

P(first and not second) + P(second and not first) =

P(first)×(1−P(second)) + P(second)×(1−P(first)) =

.7×(1−.6) + .6×(1−.7) = 0.46

A very common mistake on problems like this is writing down only one of the sequences. When you have exactly one success (or exactly any definite number), almost always there are multiple ways to get to that outcome.

You can’t use the “or” formula here, even if you studied it. That computes the probability of one or the other or both, but you need the probability of one or the other but not both.

(b) The probability of not getting a ticket on a given morning is 1−0.27 = 0.73. The probability of getting no tickets on five mornings in a row is therefore 0.73⁵ ≈ 0.2073 or about 21%.

P(man) = 106.2/219.7 ≈0.4834

P(man|divorced) = 9.7/22.8 ≈ 0.4254

Since P(man|divorced) ≠ P(man), the events are not independent.

Alternative solution:  You could equally well show that P(divorced|man) ≠ P(divorced):

P(divorced|man) = 9.7/106.2 ≈ 0.0913

P(divorced) = 22.8/219.7 ≈ 0.1038

(b) $ in L1, probabilities in L2. 1-VarStats L1,L2 yields μ = −2.70. The expected value of a ticket is −$2.70. This is a bad deal for you. (It’s a very good deal for the lottery company. They’ll make $2.70 per ticket, on average.)

Common mistakes: Students sometimes give hand-waving arguments such as the top prize being very unlikely, or the lottery company always getting to keep the ticket price, but these are not relevant. The only thing that determines whether it’s a good or bad deal for the player is the expected value μ.

μ = 1/p = 1/.066 ≈ 15.2

Over the course of her undead existence, taking each night’s hunt as a separate experience, the average of all nights has her first getting an O negative drink from her fifteenth victim.

(b) geometcdf(.066,10) = .4947936946 ≈ 0.4948. Velma has almost a 50% chance of getting O negative blood within her first ten victims.

(You could also do this as a binomial, n = 10, p = 0.066, x = 1 to 10.)

(c) This is a binomial model with n = 10, p = 0.066, and x = 2. Use MATH200A part 3 or binompdf(10,.066,2) = .1135207874 ≈ 0.1135. Velma has just over an 11% chance of getting exactly two O negative victims within her first ten.

(a) geometpdf(.08, 3) = .067712 → 0.0677

(b) geometcdf(.08, 3) = .221312 → 0.2213

(You could also do this as a binomial with n = 3, p = 0.08, x = 1 to 3.)

(a) This is a binomial distribution: each student passes or not, whether one student passes has nothing to do with whether anyone else passes, and there are a fixed seven trials.

μ = np = 7*0.8 ⇒ μ = 5.6 people

σ = √[npq] = √[7*0.8*(1−0.8)] = 1.058300524 ⇒ σ =  1.1 people

(b) Binomial again, n = 7, p = 0.8, x = 4 to 6. Use binompdf-sum or MATH200A part 3 to find P(4 ≤ x ≤ 6) = 0.7569.

(c) Geometric model: p = 0.8, x = 3. geometpdf(.8,3) = 0.0320

(d) geometcdf(.8,2) = 0.9600

Alternative solution: Binomial probability with n = 2, p = 0.8, x = 1 to 2 gives the same answer.

binomcdf(40,.49,13) or MATH200A Program part 3 with n=40, p=.49, x=0 to 13 gives .0259693307  →  0.0260, less than 5%, so you would be surprised though maybe not flabbergasted. ☺

(b) The company’s gross profit is $180.00−130.40 = $49.60, about 28%. But it could very well cost the company that much to sell the policy, pay the agent’s commission, and enter the policy in the computer. Also, all policies must bear part of the company’s general overhead costs. The price is not necessarily unfair in the plain English sense.

(a) x’s in L1, P’s in L2. 1-VarStats L1,L2 yields μ = 2 (exactly) and σ = 1.095353824 or σ ≈ 1.1. Interpretation: In the long run, on average you expect to get two heads per group of five flips. You expect most groups of five flips will yield between μ−σ = 1 head and μ+σ = 3 heads.

(b) (I wouldn’t use this part as a regular quiz question.) The long-term average is 2 heads out of 5 flips, which is p = 2/5 = 40%. Obviously coin flips are independent, so the probability of heads must be the same every time. Therefore you have a binomial model with n = 5 and p = 0.4.

(a) Binomial probability with n = 5, p = 0.7, x = 3 to 5. MATH200A part 3 5, .7, 3, 5 yields .83692 or P(x ≥ 3) = 0.8369. Or, binompdf(5,.7)→L6 and then sum(L6,4,6) to get the same answer. Or, use the complement: 1−binomcdf(5,.7,2).

(b) You need the mean of the binomial distribution:

μ = np = 10×0.7 = 7

(c) 5 is less than the expected number, so you compute P(x≤5):

MATH200A part 3 10, .7, 0, 5 yields 0.1503, or

binomcdf(10,.7.5) = 0.1503, not surprising

Common mistake: Don’t just compute P(x=5), which is 0.1029. When you want to know whether a result is unusual or surprising, you have to find the probability of that result or one even further from the expected value.

• On any given trip, there’s a 9% chance that Chantal’s commute will be less than 17 minutes.
• 9% of Chantal’s commutes are shorter than 17 minutes.

• The probability that a randomly selected man is 76.5″ or taller is 0.0068 or 0.68%.
• Only 0.68% of men are 76.5″ tall or taller.

P(64 ≤ x ≤ 67) = normalcdf(64, 67, 64.1, 2.75) = 0.3686871988 → 0.3687

36.87% of women are 64″ to 67″ tall.

x₁ = invNorm(.025, 69.3, 2.92) = 63.57690516

x₂ = invNorm(1−.025, 69.3, 2.92) = 75.02309484

Heights under 63.6″ or over 75.0″ would be considered unusual.

P15 = invNorm(.15, 69.3, 2.92) = 66.27361453

You must be at least 66″ or 5′6″ tall. Also acceptable: at least 66¼ inches, or at least 66.3 inches.

P25 = invNorm(.25, 64.1, 2.75) = 62.24515319 → P25 = 62.2″

P75 = invNorm(.75, 64.1, 2.75) = 65.95484681 → P75 = 66.0″

(b) Q3 is P75 and Q1 is P25, so the IQR is P75−P25 = 65.95484681−62.24515319 = 3.70969362 → IQR = 3.7″.

(c) 1.35σ = 1.35×2.75 = 3.7125 → 3.7″, matching the IQR as expected. (The match isn’t perfect, because 1.35 is a rounded number.)

P(x ≤ 735) = normalcdf(−10^99, 735, 500, 100) = 0.9906.

A score of 735 is at the 99th percentile.

x₁ = invNorm(1−.02, 1500, 300) = 2116.124673

You must score at least 2117. (If you round to 2116, you get a number that is a bit less than the computed minimum. While rounding usually makes sense, there are situations where you have to round up, or round down, instead of following the usual rule.)

Common mistake: The probability is not 7.24! That’s not just wrong, it’s very wrong — probabilities are never greater than 1. “E−4” on your calculator comes at the end of the number, but it’s critical info. It means “times 10 to the minus 4th power”, so the probability is 7×10⁻⁴ or 0.0007.

• The probability that a randomly selected man is under 60″ tall is 0.0007 or 0.07%.
• 0.07% of men are under 60″ tall.

x_m1 = invNorm(.05, 69.3, 2.92) = 64.49702741

x_m2 = invNorm(1−.05, 69.3, 2.92) = 74.10297259

x_f1 = invNorm(.05, 64.1, 2.75) = 59.57665253

x_f2 = invNorm(1−.05, 64.1, 2.75) = 68.62334747

Men must be 64.5 to 74.1 inches tall; women must be 59.6 to 68.6 inches tall.

• Center: The mean of the sampling distribution is the same as the mean of the population, 800 hours.
• Spread: The standard error of the mean is σ_x̅ = σ/√n = 50/√100 = 5 hours.
• Shape: You have a random sample, 10n = 10×100 = 1000 is certainly less than the total number of light bulbs, and your sample size is comfortably larger than 30. Therefore you can use the normal model for the sampling distribution.

Sample means are ND with mean 800 hours and SD 5 hours. The sketch is at right.

Common mistake: The correct standard deviation is 5 hours, not 50. You’re not sketching the population of light bulbs. Rather, you’re now interested in the distribution of average lifetimes in samples of 100 bulbs. (The axis is the x̅ axis, not the x axis.)

780 hours, the sample mean that the problem asks about, is 20 hours below the population mean of 800. 20/5 = 4 standard errors, so you should have marked 780 hours at four standard deviations below the mean.

A sample mean of 780 is less than the population mean of 800 hours. Therefore you compute the probability of a sample mean of 780 hours or less. It will be surprising (unusual, unexpected) if the probability is under 5%.

P(x̅ ≤ 780) = normalcdf(−10^99, 780, 800, 50/√(100)) = 3.1686E-5 → P(x̅ ≤ 780) = 0.00003

(You can also give the probability as <0.0001.) Yes, this is surprising.

Common mistake: Don’t give the probability as 3.1686. Probabilities are never greater than 1.

(b) If the manufacturer’s claim is true, there are only three chances in a hundred thousand of getting a sample mean this low. It’s very unlikely that the manufacturer’s claim is true.

• μ_p̂ = p = 0.72
• σ_p̂ or SEP = √pq/n = √.72×(1−.72)/500 = 0.0200798406
• Expected “yes” per sample: np = 500×.72 = 360; expected “no” = 500−360 = 140; both are well above 10. You have a random sample, and 10×500 = 5000 is far less than the American population. Therefore the normal approximation is valid.

Answer: normally distributed with mean = 0.72, standard deviation (standard error) = 0.020

Common mistake: Don’t write n≥30 when testing the normal approximation. The n≥30 test applies to numeric data, but in this problem you have binomial data.

(b) 350/500 = 0.70 exactly, and 370/500 = 0.74 exactly. In a sample of 500, finding 350 to 374 successes is the same as finding 70% to 74% successes.

If you stored the computed SEP in part (a), then your screen will look like the one on the left. Otherwise, it will look like the one on the right:

  or  

Answer: P(70% ≤ p̂ ≤ 74%) = 0.6808.

Remark: Always check for reasonableness. 70% and 74% are one standard error below and above the mean, so you know from the Empirical Rule that about 68% of the data should be within that region.

Remark: The problem wanted you to use the normal approximation, but it’s always good to check answers by a different method if possible. 70%×500 = 350; 74%×500 = 370. MATH200A part 3 with n=500, p=.72, from 350 to 370, gives a probability of 0.7044, pretty good agreement.

P(x̅ ≤ $31,000) = normalcdf(−10^99, 31000, 32400, 19000/√(1000)) = 0.0099, almost exactly 1%. That would be pretty unlikely if the population mean was still $32,400, so the city manager is most likely correct.

Remark: This problem was adapted from Freedman, Pisani, Purves (2007, 415) [see “Sources Used” at end of book].

(a) The model is at right. You could list green and black separately, but since they have the same outcome there’s no need to do that. It’s important to have the probabilities as exact fractions, not approximate decimals.

(b) x’s in L1, P’s in L2. 1-VarStats L1,L2 gives μ = −$0.53, σ = $9.99. Interpretation: In the long run, a player who bets $10 on red will lose an average of 53¢ per bet.

Remark: Notice that the SD is about 20 times the mean. This is why gambling is so exciting for the player: there’s a lot of variability from one bet to the next.

• μ_x̅ = μ = 5.00 lb
• The SEM is σ_x̅ = 0.05/√15 ≈ 0.013 lb.
• The sample means are normally distributed, even for this small sample, because the original population is normally distributed.

P(∑x > 75.6) = P(x̅ > 5.04) = normalcdf(75.6/15, 10^99, 5.00, 0.05/√(15)) = 9.7295E-4 ≈ 0.0010, about one chance in a thousand.

Answer: P(x > 43.0) = 0.1634

(b) The sampling distribution of x̅ is ND, even for this small sample, because the population is ND. The standard error is σ_x̅ = 5.1/√14 ≈ 1.4.

P(x̅ > 43.0) = normalcdf(43, 10^99, 38, 5.1/√(14))  1.2212E-4 → P(x̅>43.0) = 0.0001 or 0.01%

Remark: This sketch is not very well proportioned, because it makes the probability look much larger than it actually is.

The standard error of the mean is σ_x̅ = 3.5/√1000, about 0.11. The sampling distribution of the mean is normal because data are numeric and n =1000, greater than 30. (Treat the sample as random because it’s a “typical neighborhood”. And a thousand households is less than 10% of all the households that there are.)

P(x̅ > 12.778) = normalcdf(12.778, 10^99, 12.5, 3.5/√(1000) = 0.0060

• The doctors were randomized between treatment and placebo groups.
• 10×11,037 = 110,370. There are more adult males than that.
• np = 11037×.0171 = about 189; nq = 11037−189 = 10848. Both are well above 10.

Therefore the sampling distribution can be approximated by a normal distribution.

The standard error of the proportion or SEP is σ_p̂ = √pq/n = √.0171(1−.0171)/11037 ≈ 0.0012

If you use my shortcut, your screen will look like the one at the left; if not, it will look like the one at the right.

 or 

Either way, the probability is 2.2013×10^-10, or 0.000 000 000 2. There are only two chances in ten billion of getting a sample proportion of 0.94% or less with sample size 11,037, if the true population proportion is 1.71%. That’s pretty darn unlikely, so based on this experiment you can rule out coincidence and decide that aspirin does reduce the chance of a heart attack among adult males.

The standard error of the mean or SEM is σ_x̅ = σ/√n = 2.92/√16 = 0.73″.

μ_x̅ ± 2σ_x̅ = 69.3 ± 2×.73 = 67.84 to 70.76.

Sample means between those values would not be surprising, and therefore a sample mean would be surprising if it is under 67.84″ or over 70.76″.

Alternative solution: That back-of-the-envelope calculation is good enough, but you could also get a more precise answer:

L = invNorm(0.025, 69.3, 2.92/√(16)) = 67.87

H = invNorm(1−0.025, 69.3, 2.92/√(16)) = 70.73

What does the sampling distribution look like? The center is μ_p̂ = p = 0.45. The standard error is σ_p̂ = √0.45×(1−.45)/1504 ≈ 0.013. Check requirements to make sure that a normal model can be used for the sampling distribution:

• Random sample? Yes, given.
• Sample less than 10% of population? 10×1504 = 15,040, compared to millions of American adults, OK.
• Sample large enough? Yes, 0.45×1504 ≈ 677 successes and 1504−677 ≈ 827 failures expected, both above 10.

P(x ≥ 737) = P(p̂ ≥ 49%) = normalcdf(737/1504, 10^99, .45, √(.45*(1−.45)/1504)) ≈ 9E-4 or 0.0009.

Can you draw a conclusion? Yes, you can. In a population with 45% unfavorable rating of the Tea Party, there are only 9 chances in 10,000 of getting a sample as unfavorable as this one (or more unfavorable). That’s pretty unlikely, so you conclude that the true unfavorable rating in October was most likely more than 45% of all Americans. (In Chapter 9, you’ll learn how to estimate that proportion from a sample.)

(Statisticians would say, “the population mean or proportion is not a random variable.” By that, they mean just what I said in less technical language.)

Answer: A confidence interval for numeric data is an estimate of the average, and tells you nothing about individuals. Correct his conclusion to I’m 90% confident that the average food expense for all TC3 students is between $45.20 and $60.14 per week..

Remark: Use all or a similar word to show that you’re estimating the mean for the population, not just the sample of 40 students. There’s no need to estimate the mean of the sample, because you know the exact sample mean x̅ for your sample.

Remark: Be clear in your mind that you’re estimating the average spending per student at $45–60 a week. Some individual students will quite likely spend outside that range, so your interpretation shouldn’t say anything about individual student spending.

Answer: It’s the use of the word average. When you collect data points that are all yes/no or success/failure, you have a sample proportion p̂, equal to the number of successes divided by sample size, and you can estimate a population proportion. There is no “average” with non-numeric data.

Your 90% confidence estimate is simply that 27% to 40% usually or always prepare their own food.

This is a confidence interval about a mean, Case 1 in Inferential Statistics: Basic Cases.

Requirements: random sample, OK. 10n = 10×40 = 400 is less than total number of batteries made; OK. n = 40 >30, OK.

TInterval 1756, 142, 40, .95

(1710.6, 1801.4)

Neveready is 95% confident that the average Neveready A cell, operating a wireless mouse, lasts 1711 to 1801 minutes (28½ to 30 hours).

Common mistake: Don’t make any statement about 95% of the batteries! Your CI is about your estimate of one number, the average life of all batteries. Your CI has a margin of error of ±15 minutes; the 95% range for all batteries would be about 4 to 5 hours.

Don’t make the term “point estimate” harder than it is! The point estimate for the population mean (or proportion, standard deviation, etc.) is just the sample mean (or proportion, standard deviation, etc.).

(b) The sample is his actual data, the 10,000 flips. Therefore the sample size is n = 10,000. The population is what he wants to know about, all possible flips. The population size is infinite or “indefinitely large”.

This is sample size for a confidence interval about a proportion, Case 2 in Inferential Statistics: Basic Cases. Since you have no prior estimate, use 0.5 for p̂.

With the MATH200A program (recommended):	If you’re not using the program:
MATH200a/sample size/binomial, p̂ = .5, E = .035, C-Level = .95, sample size is at least 784	The formula is . 1−α = .95 ⇒ α/2 = 0.025. z0.025 = invNorm(1−.025, 0, 1) Divide by .035, square the result, and multiply by .5*(1−.5). Answer: at least 784. Remember — you’re not rounding, you’re going up to a whole number.

This is a confidence interval about a proportion, Case 2 in Inferential Statistics: Basic Cases.

Requirements:

• Random sample, OK.
• 10n = 10×100 = 1000 < 68,917, OK.
• 40 successes, 100−40 = 60 failures, both > 10, OK.

Common mistake: Don’t say “n > 30” or “n ≥ 30”. That’s true, but it doesn’t help you with binomial data. For computing a confidence interval about a proportion from binomial data, the “sample size large enough” condition is at least 10 successes and at least 10 failures, not sample size at least 30.

1-PropZInt 40, 100, .9 → (.31942, .48058), p̂ = .4

31.9% to 48.1% of all claims at that office have been open for more than a year (90% confidence).

This is a confidence interval about a mean, Case 1 in Inferential Statistics: Basic Cases.

Requirements check:

• Random sample, OK.
• 10×40 = 400, less than the number of times she could commute (past, present, and future), OK.
• Sample size 40 > 30, OK.

TInterval 17.7, 1.8, 40, .95 → (17.124, 18.276)
She’s 95% confident that the average of all her commutes is 17.1 to 18.3 minutes.

This is a confidence interval about a mean, Case 1 in Inferential Statistics: Basic Cases.

Requirements check:

• Random sample, OK.
• 10×15 = 150 is less than total number of women in their 20s.
• MATH200A/Normality: r=.9667, CRIT=.9383, r>CRIT, OK.
• MATH200A/Box-whisker: no outliers, OK.

TInterval L6, 1, .95 → (62.918, 65.016), x̅=63.96666667, s=1.894226818, n=15

The average height of women aged 20–29 is 62.9 to 65.0 inches (95% confidence).

Remark: Since adult women’s heights are known to be normally distributed, you could get away without checking for normality and outliers in this sample. But it does no harm to check every time.

This is a confidence interval about a mean, Case 1 in Inferential Statistics: Basic Cases.

Requirements check:

• Random sample, OK.
• 10×18 = 180. There are far more than 180 male students; OK.
• MATH200A/box-whisker: no outliers, OK.
• MATH200A/Normality check: r=.9787, CRIT=.9461, r>CRIT, OK.

TInterval L5, 1, .9 → (97.757, 98.343), x̅ = 98.05, s =.7155828558 → 0.72, n = 18.
(a) Fred is 90% confident that the average body temperature of healthy male students is 97.8 to 98.3 °F.

(b) He’s 90% confident that the average body temperature is not more than 98.3°, so 98.6° as normal (average) temperature is inconsistent with his data.

(c) E = 98.343−98.05 = 0.3°, or E = 98.05−97.757 = 0.3°, or (98.343−97.757)/2 = 0.3°.

With the MATH200A program (recommended):	If you’re not using the program:
(d) MATH200A/Sample size/Num unknown σ: s=.7155828558, E=.1, C-Level=.95, n≥202. He will need at least 202 in his sample.	(d) Confidence level = 1−α = 0.95 ⇒ α = 0.05 ⇒ α/2 = 0.025. z0.025 = invNorm(1−.025) Multiply by s, divide by E, and square the result. This gives 197. But the t distribution is more spread out than the normal (z) distribution, so you probably want to bump that number up a bit, say to 200 or so.

This problem is about a confidence interval about a proportion, Case 2 in Inferential Statistics: Basic Cases.

(a) Requirements check:

• Random sample, OK.
• 10×500 = 5000. A city of 6.4 million must have more than 5000 in that age range. OK.
• 219 successes, 500−219 = 281 failures. Both > 10, OK.

1-PropZInt, 219, 500, .9 → (.4015, .4745), p̂ = .438

You’re 90% confident that 40.2% to 47.5% of Metropolis adults aged 50–75 have had a colonoscopy in the past ten years.

(b) MATH200A/sample size/binomial, p̂ = .438, E = .02, C-Level = .9 → at least 1665

This is a confidence interval about a mean, Case 1 in Inferential Statistics: Basic Cases.

Requirements check:

• Random sample, OK.
• 10×20 = 200, less than the total number of cash deposits, OK.
• MATH200A/normality check, r=.9864, CRIT=.9503, r>CRIT, OK.
• MATH200a/box-whisker, no outliers, OK.

TInterval L4, 1, .95 → (179.86, 198.93), x̅ = 189.40, s = 20.37, n = 20

You’re 95% confident that the average of all cash deposits is between $179.86 and $198.93.

Common mistake: Don’t say that 95% of deposits are between those values — if you look at the sample you’ll see that’s pretty unlikely. You’re estimating the average, not the individual deposits in the population.

This is a confidence interval about a proportion, Case 2 in Inferential Statistics: Basic Cases.

Requirements check:

• Systematic sample, OK.
• Sample 10n = 10×1000 = 10,000, less than the number of voters; OK.
• 520 successes and 1000−520 = 480 failures, OK.

1-PropZInt 520, 1000, .95 → (.48904, .55096), p̂ = .52

With 95% confidence, 48.9% to 55.1% of voters voted Snake. At the 95% confidence level, we can’t tell whether more or less than 50% of voters voted for Abe Snake.

1. Hypotheses.   2. Significance level   RC. Requirements check   3–4. Test statistic and p-value   5. Decision rule (or, conclusion in statistics language)   6. Conclusion (in English)

It keeps you honest. If you could select a significance level after computing the value, you could always get the result you want, regardless of evidence.

Answers will vary here. But you should get in the key idea that If H₀ is true, the p-value is the chance of getting the sample you got, or a sample even further from H₀, purely by random chance. For more correct statements, and common incorrect statements, see What Does the p-Value Mean?

(a) It’s too wishy-washy. When p<α, you can reach a conclusion. Correction: The accelerant makes a difference, at the 0.05 significance level.

(b)You can never prove the null hypothesis of “no difference”. You can’t even say “The accelerant may make no difference,” because that’s only part of the truth: it equally well may make a difference. You must say something like, “At the 0.05 significance level it’s impossible to say whether the accelerant makes a difference or not.”

(a) A Type I error is rejecting the null hypothesis when it’s actually true. In this case, a Type I error would be concluding “the accelerant makes paint dry faster” when actually it makes no difference. This would lead you to launch the product and expose yourself to a lot of warranty claims.

(b) A Type II error is failing to reject the null hypothesis when it’s actually false. In this case, a Type II error would be concluding “the accelerant doesn’t makes paint dry faster” when actually it does. This would lead you to keep the product off the market even though it could add to your sales and would perform as promised.

They are not necessarily mistakes. Type I and II errors are an unavoidable part of sample variability. Nothing can prevent them entirely. The only way to make them both less likely at the same time is to use a larger sample size.

That said, if you make mistakes in data collection or analysis you definitely make Type I or Type II errors (or both of them) more likely.

Make your significance level α smaller. The side effect is making a Type II error more likely.

Your own words will vary from mine, but the main difference is that when p > α you can’t reach a conclusion. Accepting H₀ is wrong because it reaches the conclusion that H₀ is true. Failing to reject H₀ is correct because it leaves both possibilities open.

It’s like a jury verdict of “not guilty beyond a reasonable doubt. The jury is not saying the defendant didn’t do it. They are saying that either he didn’t do it or he did it but the prosecution didn’t present enough evidence to convince them.

A hypothesis test can end up rejecting H₀ or failing to reject it, but the result can never be to accept H₀.

H₀: μ = 500
H₁: μ ≠ 500

Remark: It must be ≠, not > or <, because the claim is that the mean is 500 minutes, and a difference in either direction would destroy the claim.

(a) p > α; fail to reject H₀. At the 0.01 significance level, we can’t determine whether the directors are stealing from the company or not.

(b) p < α; reject H₀ and accept H₁. At the 0.01 level of significance, we find that the directors are stealing from the company.

α is the probability of a Type I error that you can tolerate. A Type I error in this case is determining that the defendant is guilty (calling H₀ false) when actually he’s innocent (H₀ is really true), and the consequence would be putting an innocent man to death. You specify a low α to make it less likely this will happen. Of the given choices, 0.001 is best.

This is binomial data, a Case 2 test of proportion in Inferential Statistics: Basic Cases.

This is binomial data (against or not against): a Case 2 test of population proportion in Inferential Statistics: Basic Cases.

Requirements check: Random sample? NO, this is a self-selected sample, consisting only of those who returned the poll. (That could be overcome by following up on those who did not return the poll, but nobody did that.)

The 10n≤N requirement also fails. 10n = 10×380 = 3800, much larger than the 1366 population size.

Answer: No, you cannot do any inferential procedure because the requirements are not met.

(b) The population is all persons who do the primary grocery shopping in their households. We don’t know the precise number, but it is surely in the millions since there are millions of households. We can say that it is indefinitely large.

(c) The number 182 is x, the number of successes in the sample.

(d) She wanted to know whether the true proportion is greater than 40%, so her alternative hypothesis is H₁: p > 0.4 and p_o is 0.4.

(e) No. The researcher is interested in the habits of the primary grocery shoppers in households; therefore she must sample only people who are primary grocery shoppers in their households. If you even thought about saying Yes, please go back to Chapter 1 and review what bias actually means.

(a) This is inference about the proportion in one population, Case 2 in Inferential Statistics: Basic Cases.

(b) Requirements have already been checked.
1-PropZInt 18, 30, .95. Results: (.4247, .7753), p̂ = .6.
We’re 95% confident that the true probability of winning if you switch doors is between 42.5% and 77.5%.

(c) It’s possible that the true probability of winning if you switch doors is 1/3 (33.3%) or even worse, but it’s very unlikely. Why? You’re 95% confident that it’s at least 42.5%. Therefore you’re better than 95% confident that the true probability if you switch is better than the 1/3 probability if you don’t switch doors. Switching is extremely likely to be the good strategy.

(a) A Type I error is rejecting the null hypothesis when it’s actually true. Here, a Type I error means deciding a piece of mail is spam when it’s actually not, so if Heather’s spam filter makes a Type I error then it will delete a piece of real mail. A Type II error is failing to reject H₀ when it’s actually false, treating a piece of spam as real mail, so a Type II error would let a piece of spam mail into Heather’s in-box..

(b) Most people would rather see a piece of spam (Type II) than miss a piece of real mail (Type I), so a Type I error is more serious in this situation. Lower significance levels make Type I errors less likely (and Type II errors more likely), so a lower α is appropriate here.

(a)The population parameter is missing. It should be either μ or p, but since a proportion can’t be greater than 1 it must be μ.
Correction: H₀: μ = 14.2;  H₁: μ > 14.2

(b) H₀ must have an = sign. Correction: H₀: μ = 25;  H₁: μ > 25

(c) You used sample data in your hypotheses. Correction: H₀:μ=750; H₁:μ>750

(d) You were supposed to test “makes a difference”, not “is faster than”. Never do a one-tailed test (> or <) unless the other direction is impossible or of no interest at all. It’s possible that your “accelerant” could actually increase drying time, and if it does you’d definitely want to know.
Correction: H₀: μ = 4.3 hr;  H₁: μ ≠ 4.3 hr

This is numeric data, and you don’t know the standard deviation (SD) of the population. In Inferential Statistics: Basic Cases this is Case 1, a test of population mean.

This is numeric data with unknown SD of the population, Case 1 (test of population mean) in Inferential Statistics: Basic Cases.

This is numeric data with unknown SD of population. You’re testing a population mean, Case 1 in Inferential Statistics: Basic Cases.

This is binomial data (each person either has a bachelor’s or doesn’t) for a one-population test of proportion: Case 2 in Inferential Statistics: Basic Cases

(a) Requirements:

• Random sample, yes.
• 10n = 10×120 = 1200, and there are many more than 1200 residents of Tompkins County aged 25+.
• Sample has 52 successes and 120−52 = 68 failures, both ≥ 10, check.

1-PropZInt: x=52, n=120, C-Level=.95
Results: (.34467, .52199); p̂=.4333333333 → p̂ = .4333

We’re 95% confident that 34.5 to 52.2% of Tompkins County residents aged 25+ have at least a bachelor’s degree.

(b) Requirements have already been checked. A two-tailed test at the 0.05 level is equivalent to a confidence interval at the 95% level. The statewide proportion of 32.8% is outside the 95% CI for Tompkins County, and therefore at the 0.05 significance level, the proportion of bachelor’s degrees among Tompkins County residents aged 25+ is different from the statewide proportion of 32.8%. In fact, Tompkins County’s proportion is higher.

This is numeric data, with population SD unknown: test of population mean, Case 1 in Inferential Statistics: Basic Cases.

This is numeric data, with σ unknown: test of population mean, Case 1 in Inferential Statistics: Basic Cases.

Binomial data (head or tail) implies Case 2, test of population proportion on Inferential Statistics: Basic Cases. A fair coin has heads 50% likely, or p = 0.5.

You have numeric data, and you don’t know the SD of the population, so this is a Case 1 test of population mean in Inferential Statistics: Basic Cases.

(a) Check requirements: random sample, n = 45 > 30, and there are more than 10×45 = 450 people with headaches.

TInterval: x̅=18, s=8, n=45, C-Level=.95

Results: (15.597, 20.403)

We’re 95% confident that the average time to relief for all headache sufferers using PainX is 15.6 to 20.4 minutes.

(b) Requirements have already been checked. A two-tailed test (a test for “different”) at the 0.05 level is equivalent to a confidence interval at the 1−0.05 = .95 = 95% confidence level. Since the 95% CI includes 20, the mean time for aspirin, we cannot determine, at the 0.05 significance level, whether PainX offers headache relief to the average person in a different time than aspirin or not.

(a) Use MATH200A part 5 and select 2-pop binomial. You have no prior estimates, so enter 0.5 for p̂₁ and p̂₂. E is 0.03, and C-Level is 0.95. Answer: you need at least 2135 per sample, 2135 people under 30 and 2135 people aged 30 and older. Here’s what it looks like, using MATH200A part 5:

Caution! Even if you don’t identify the groups, at least you must say “per sample”. Plain “2135” makes it look like you need only that many people in the two groups combined, or around 1068 per group, and that is very wrong.

Caution! You must compute this as a two-population case. If you compute a sample size for just one group or the other, you get 1068, which is just about half of the correct value.

If you don’t have the program, you have to use the formula: [p̂₁(1−p̂₁)+p̂₂(1−p̂₂)]·(z_α/2/E)². You don’t have any prior estimates, so p̂₁ and p̂₂ are both equal to 0.5. Multiply out p̂₁ × (1−p̂₁) × p̂₂ × (1−p̂₂) to get .5.

Next, 1−α = 0.95, so α = 0.05 and α/2 = 0.025. z_α/2 = z_0.025 = invNorm(1−0.025). Divide that by E (.03), square, and multiply by the result of the computation with the p̂’s.

(b) Using MATH200A Program part 5 with .3, .45, .03, .95 gives 1953 per sample.

Alternative solution: Using the formula, .3(1−.3)+.45(1−.45) = .4575. Multiply by (invNorm(1−.05/2)/.03)² as before to get 1952.74157 → 1953 per sample.

Again, you must do this as two-population binomial. If you do the under-30 group and the 30+ group separately, you get sample sizes of 897 and 1057, which are way too small. If your samples are that size, the margins of error for under-30 and 30+ will each be 3%, but the margin of error for the difference, which is what you care about, will be around 4.2%, and that’s greater than the desired 3%.

(a) You have numeric data in two independent samples. You’re testing the difference between the means of two populations, Case 4 in Inferential Statistics: Basic Cases. (The data aren’t paired because you have no reason to associate any particular Englishman with any particular Scot.)

(b) Requirements are already covered.
2-SampTInt, C-Level=.90
Results: (−.2025, 3.5775)
We’re 90% confident that, on a scale from 1=hate to 10=love, the average Englishman likes soccer between 0.2 points less and 3.6 points more than the average Scot.

(a) This is the difference of proportions in two populations, Case 5 in Inferential Statistics: Basic Cases.

(b) Requirements already checked.

2-PropZInt with C-Level = .95 → (−.1919, −.0081)

That’s the estimate for p₁−p₂, English minus Scots. Since that’s negative, English like soccer less than Scots do. With 95% confidence, Scots are more likely than English to be soccer fans, by 0.8 to 19.2 percentage points.

(c) [(−.0081) − (−.1919)] / 2 = 0.0919, a little over 9 percentage points.

(d) MATH200A part 5, 2-pop binomial, p̂₁=.7, p̂₂=.8, E=.04, C-Level .95 gives 889 per sample

By formula, z_α/2 = z_0.025 = invNorm(1−0.025) = 1.96.
n₁ = n₂ = [.7(1−.7)+.8(1−.8)]×(1/96/.04)² = 888.37 → 889 per sample

(a) This is before-and-after paired data, Case 3 in Inferential Statistics: Basic Cases. You’re testing the mean difference.

(b) TInterval with C-Level .9 gives (1.3951, 7.8049).

Interpretation: You are 90% confident that running an average of four miles a day for six months will raise HDL by 1.4 to 7.8 points for the average woman.

Caution! Don’t write something like “I’m 90% confident that HDL will be 1.4 to 7.8”. The confidence interval is not about the HLD level, it’s about the change in HDL level.

Remark: Notice the correspondence between hypothesis test and confidence interval. The one-tailed HT at α = 0.05 is equivalent to a two-tailed HT at α = 0.10, and the complement of that is a CI at 1−α = 0.90 or a 90% confidence level. Since the HT did find a statistically significant effect, you know that the CI will not include 0. If the HT had failed to find a significant effect, then the CI would have included 0. See Confidence Interval and Hypothesis Test.

(a) Each participant either had a heart attack or didn’t, and the doctors were all independent in that respect. This is binomial data. You’re testing the difference in proportions between two populations, Case 5 in Inferential Statistics: Basic Cases.

Remark The study was conducted from 1982 to 1988 and was stopped early because the results were so dramatic. For a non-technical summary, see Physicians’ Health Study (2009) [see “Sources Used” at end of book]. More details are in the original article from the New England Journal of Medicine (Steering Committee 1989 [see “Sources Used” at end of book]).

(b) 2-PropZInt with C-Level .95 gives (−.0125, −.0056). We’re 95% confident that 325 mg of aspirin every other day reduces the chance of heart attack by 0.56 to 1.25 percentage points.

Caution! You’re estimating the change in heart-attack risk, not the risk of heart attack. Saying something like “with aspirin, the risk of heart attack is 0.56 to 1.25%” would be very wrong.

(a) You’re estimating the difference in means between two populations. This is Case 4 in Inferential Statistics: Basic Cases. Requirements:

• Random samples (given).
• Sample sizes both >30.
• 10×30 = 300 and 10×32 = 320 are less than the numbers of houses in the two counties.

Population 1 = Cortland County houses, population 2 = Broome County houses.
2-SampTInt, 134296, 44800, 30, 127139, 61200, 32, .95, No

results: (−20004, 34318)

June is 95% confident that the average house in Cortland County costs $20,004 less to $34,318 more than the average house in Broome County.

(b) A 95% confidence interval is the complement of a significance test for ≠ at α = 0.05. Since 0 is in the interval, you know the p-value would be >0.05 and therefore June can’t tell, at the 0.05 significance level, whether there is any difference in average house price in the two counties or not.

If both ends of the interval were positive, that would indicate a difference in averages at the 0.05 level, and you could say Cortland’s average is higher than Broome’s. Similarly, if both ends were negative you could say Cortland’s average is lower than Broome’s. But as it is, nada.

Remark: Obviously Broome County is cheaper in the sample. But the difference is not great enough to be statistically significant. Maybe the true mean in Broome really is less than in Cortland; maybe they’re equal; maybe Broome is more expensive. You simply can’t tell from these samples.

The immediate answer is that those are proportions in the sample, not the proportions among all voters. This is two-population binomial data, Case 5 in Inferential Statistics: Basic Cases.
Requirements check:

• Random samples, OK.
• Each sample 10n = 10×1000 = 10,000. There are far more than 10,000 voters nationally; OK.
• The two samples were independent, OK.
• Red: 520 successes and 1000−520 = 480 failures, OK.
  Blue: 480 successes and 1000−480 = 520 failures, OK.

Population 1 = Red voters, population 2 = Blue voters.
2-PropZInt 520, 1000, 480, 1000, .95
Results: (−.0038, .08379), p̂₁=.48, p̂₂=.52

With 95% confidence, the Red candidate is somewhere between 0.4 percentage points behind Blue and 8.4 ahead of Blue. The confidence interval contains 0, and so it’s impossible to say whether either one is leading.

Remark: Newspapers often report the sample proportions p̂₁ and p̂₂ as though they were population proportions, but now you know that they aren’t. A different poll might have similar results, or it might have samples going the other way and showing Blue ahead of Red.

(b) For a hypothesis test, we often use “at least 10 successes and 10 failures in each sample” as a shortcut requirements test, but the real requirement is at least 10 successes and 10 failures expected in each sample, using the blended proportion p̂. If the shortcut procedure fails, you must check the real requirement. In this problem, the blended proportion is

p̂ = (x₁+x₂)/(n₁+n₂) = (7+18)/(28+32) =25/60, about 42%.

For sample 1, with n₁ = 28, you would expect 28×25/60 ≈ 11.7 successes and 28−11.7 = 16.3 failures. For sample 2, with n₂ = 32, you would expect 32×25/60 ≈ 13.3 successes and 32−13.3 = 18.7 failures. Because all four of these expected numbers are at least 10, it’s valid to compute a p-value using 2-PropZTest.

Solution: Use Case 7, 2-way table, in Inferential Statistics: Basic Cases.

(1)	H0: The tested treatments with Echinacea make no difference to the proportion who catch cold. H1: The treatments do make a difference. …
(2)	α = 0.01
(3/4)	There were seven treatments and two outcomes, so enter your 7×2 matrix and run a χ²-Test or MATH200A part 7. Results: χ² = 4.74, df = 6, p-value = 0.5769 Common mistake: Never enter the totals in a two-way test.
(RC)	Random sample? Yes, randomized experimental design. ✔ Sample less than 10% of population? Yes, the population of people exposed to the common cold is indefinitely large. ✔ All expected values ≥5? Yes, matrix B shows all values at least 5.6. ✔
(5)	p > α. Fail to reject H0.
(6)	At the 0.01 significance level, we can’t determine whether Echinacea is effective against the common cold or not. Or, We can’t determine whether Echinacea is effective against the common cold or not (p = 0.5769). Remark: Researchers might write something like “Echinacea made no significant difference to infection rates in our study” with the p-value or significance level. It’s understood that this does not prove Echinacea ineffective — this particular study fails to reach a conclusion. But as additional studies continue to find p > α, our confidence in the null hypothesis increases.

Remark: If you used MATH200A part 7, there’s some interesting information in matrix C. The top left 7 rows and 2 columns are the χ² contributions for each of the seven treatments and two outcomes. All are all quite low, in light of the rule of thumb that only numbers above 4 or so are significant, even at the less stringent 0.05 level.

The last two rows are the total numbers and percentages of people who did and didn’t catch cold: 349 (87.5%) and 50 (12.5%). If Echinacea is ineffective, you’d expect to see about that same infection rate for each of the seven treatments. Sure enough, compute the rates from the rows of the data table, and you’ll find that they vary between 81% and 92%.

The third column is the total subjects in each of the seven treatments, and the overall total. Of course you were given those in the data table, but it’s always a good idea to use this information to check your data entry.

The fourth column is the percentage of subjects who were assigned to each of the seven treatments, totaling 100% of course.

Common mistake: Binomial is a subtype of qualitative data so it’s not really a synonym. Discrete and continuous are subtypes of numeric data.

The Gambler’s Fallacy is believing that the die is somehow “due for a 6”. The Law of Large Numbers says that in the long run the proportion of 6’s will tend toward 1/6, but it doesn’t tell us anything at all about any particular roll.

Common mistake: You must specify which is population 1 and which is population 2.

Common mistake: The data type is binomial: a student is in trouble, or not. There are no means, so μ is incorrect in the hypotheses.

• Are there a fixed number of trials? Yes, you are rolling five dice, n = 5.
• Are there only two outcomes, success and failure? Yes, each die is either a 3 or not.
• Is the probability of success the same from trial to trial; are the trials independent? Yes, p = 1/6 for each die, and the dice are independent.

This is a binomial PD.

Remark: The significance level α is the level of risk of a Type I error that you can live with. If you can live with more risk, you can reach more conclusions.

Complementary events can’t happen at the same time and one or the other must happen. Example: rolling a die and getting an odd or an even. Complementary events are a subtype of disjoint events.

For a small- to moderate-sized set of numeric data, you might prefer a stemplot.

Remark: C is wrong because “model good” is H₀. D is also wrong: every hypothesis test, without exception, compares a p-value to α. For E, df is number of cells minus 1. F is backward: in every hypothesis test you reject H₀ when your sample is very unlikely to have occurred by random chance.

Remark: As stated, what you can prove depends partly on your H₁. There are three things it could be:

• If H₁: p > 0.01 (machine producing too many defectives) and you calculate p-value<α, your conclusion is C.
• If H₁: p ≠ 0.01 (machine not operating as specified) and you calculate p-value<α, your conclusion depends on the sample data. Your conclusion is either C above or the unlisted conclusion “the machine is producing fewer defectives than allowed”, in other words performing better than specified. See p < α in Two-Tailed Test: What Does It Tell You? for reminders about interpreting a two-tailed test when p-value<α.
• There is little reason to choose H₁: p < 0.01 (the machine is performing better than specified), but if that is H₁ and p-value<α then your conclusion is that the machine is performing better than specified; that conclusion is not listed above.

Regardless of H₁, if p-value>α your conclusion will be D or similar to it.

Common mistake: Conclusion A is impossible because it’s the null hypothesis and you never accept the null hypothesis.

Conclusion B is also impossible. Why? because “no more than” translates to ≤. But you can’t have ≤ in H₁, and H₁ is the only hypothesis that can be accepted (“proved”) in a hypothesis test.

Remark: A Type I error is a wrong result, but it is not necessarily the result of a mistake by the experimenter or statistician.

(b) The size is unknown, but certainly in the millions. You also could call it infinite, or uncountable. Common mistake: Don’t confuse size of population with size of sample. The population size is not the 487 from whom you got surveys, and it’s not the 321 churchgoers in your sample.

(c) The sample size n is the 321 churchgoers from whom you collected surveys. Yes, you collected 487 surveys in all, but you have to disregard the 166 that didn’t come from churchgoers, because they are not your target group. Common mistake: 227 isn’t the sample size either. It’s x, the number of successes within the sample.

(d) No. You want to know the attitudes of churchgoers, so it is correct sampling technique to include only churchgoers in your sample.

If you wanted to know about Americans in general, then it would be selection bias to include only churchgoers, since they are more likely than non-churchgoers to oppose teaching evolution in public schools.

Common mistake: Your answer will probably be worded differently from that, but be careful that it is a conditional probability: If H₀ is true, then there’s a p-value chance of getting a sample this extreme or more so. The p-value isn’t the chance that H₀ is true.

Remark: If you are at all shaky about this, review What Does the p-Value Mean?

(b) This is a binomial distribution, for exactly the same reasons.
MATH200A part 3, or binompdf(100,.08,5)

(c) The probability of success is p = 0.08 on every trial, but you don’t have a fixed number of trials. This is a geometric distribution.
geometpdf(.08,5)

Remark: There is no specific claim, so this is not a hypothesis test.

Caution: The percentages must add to 100%. Therefore you must have complete data on all categories to display a pie chart. Also, if multiple responses from one subject are allowed, then a pie chart isn’t suitable, and you should use some other presentation, such as a bar graph.

Remark: This problem tests for several very common mistakes by students. Always make sure that

• Your hypotheses include a population parameter (rules out A, E, and I because they have no symbol; rules out B, D, F, H, J, L because their symbols aren’t a population parameter)
• The “=” sign is in H₀ (rules out A–D)

This leaves you with G and K as possibilities. Either can be correct, depending on your textbook. The most common practice is always to put a plain = sign in H₀ regardless of H₁, which makes G the correct answer. But some textbooks or profs prefer ≤ or ≥ in H₀ for one-tailed tests, whch makes K the correct answer.

Remark: The Z-Test is wrong because you don’t know the SD of the selling price of all 2006 Honda Civics in the US. The 1-PropZTest and χ²-test are for non-numeric data. There is no such thing as a 1-PropTTest.

Example: “812 of 1000 Americans surveyed said they believe in ghosts” is an example of descriptive statistics: the numbers of yeses and noes in the sample were counted. “78.8% to 83.6% of Americans believe in ghosts (95% confidence)” is an example of inferential statistics: sample data were used to make an estimate about the population. “More than 60% of Americans believe in ghosts” is another example of inferential statistics: sample data were used to test a claim and make a statement about a population.

Remark: Remember that the confidence interval derives from the central 95% or 90% of the normal distribution. The central 90% is obviously less wide than the central 95%, so the interval will be less wide.

Example: You want to know the average amount of money a full-time TC3 student spends on books in a semester. The population is all full-time TC3 students. You randomly select a group of students and ask each one how much s/he spent on books this semester. That group is your sample.

Remark: This is unpaired numeric data, Case 4.

(b) For binomial data, requirements are slightly different between CI and HT. Here you are doing a hypothesis test.

• Random sample? ✔
• At least 10 successes and 10 failures expected? np_o = 500×0.25 = 125, and 500−125 = 375, both >10. ✔

  Common mistake: For hypothesis test, you need expected successes and failures. It’s incorrect to use actual successes (150) and failures (350).

• Check that the sample is not too large: 10n = 10×500 = 5000, and far more than 5000 students graduate from US high schools each year. ✔

Common mistake: Some students answer this question with “n > 30”. That’s true, but not relevant here. Sample size 30 is important for numeric data, not binomial data.

Common mistake: You cannot do a 2-SampZTest because you do not know the standard deviations of the two populations.

Some instructors have you do a preliminary F-test. It gives p=0.9089>0.05, so after that test you would use Pooled:Yes in the 2-SampTTest and get p=0.1553.

Solution: This is one-population numeric data, and you don’t know the standard deviation of the population: Case 1. Put the data in L1, and 1-VarStats L1 tells that x̅ = 4.56, s = 1.34, n = 8.

(b) TInterval: C-Level=.95
Results: (3.4418, 5.6832)

(There’s no need to repeat the requirements check or to write down all the sample statistics again.)

With 95% confidence, the true mean decrease in drying time is between 3.4% and 5.7%.

n = 5, p = 0.28, from = 0, to = 0. Answer: 0.1935

Alternative solution: If you don’t have the program, you can compute the probability that one rabbit has short hair (1−.28 = 0.72), then that all the rabbits have short hair (0.72^5 = 0.1935), which is the same as the probability that none of the rabbits have long hair.

(b) The complement of “one or more” is none, so you can use the previous answer.

P(one or more) = 1−P(none) = 1−0.1935 = 0.8065

Alternative solution: MATH200A part 3 with n=5, p=.28, from=1, to=5; probability = 0.8065

(c) Again, use MATH200A part 3 to compute binomial probability: n = 5, p = 0.28, from = 4, to = 5. Answer: 0.0238

Alternative solution: If you don’t have the program, do binompdf(5, .28) and store into L3, then sum(L3,5,6) or L3(5)+L3(6) = 0.0238.  Avoid the dreaded off-by-one error! For x=4 and x=5 you want L3(5) and L3(6), not L3(4) and L3(5).

For n=5, P(x≥4) = 1−P(x≤3). So you can also compute the probability as 1−binomcdf(5, .28, 3) = 0.0238.

(d) For this problem you must know the formula:

μ = np = 5×0.28 = 1.4 per litter of 5, on average

Common mistake: It might be tempting to do this problem as a goodness-of-fit, Case 6, taking the Others row as the model and the doctors’ choices as the observed values. But that would be wrong. Both the Doctors row and the Others row are experimental data, and both have some sampling error around the true proportions. If you take the Others row as the model, you’re saying that the true proportions for all non-doctors are precisely the same as the proportions in this sample. That’s rather unlikely.

(b) 70−67.6 = 2.4″, and therefore z = −1. By the Empirical Rule, 68% of data lie between z = ±1. Therefore 100−68 = 32% lie outside z = ±1 and 32%/2 = 16% lie below z = −1. Therefore 67.6″ is the 16th percentile.

Alternative solution: Use the big chart to add up the proportion of men below 67.6″ or below z = −1. That is 0.15+2.35+13.5 = 16%.

(c) z = (74.8−70)/2.4 = +2. By the Empirical Rule, 95% of men fall between z = −2 and z = +2, so 5% fall below z = −2 or above z = +2. Half of those, 2.5%, fall above z = +2, so 100−2.5 = 97.5% fall below z = +2. 97.5% of men are shorter than 74.8″.

Alternative solution: You could also use the big chart to find that P(z > 2) = 2.35+0.15 = 2.5%, and then P(z < 2) = 100−2.5 = 97.5%.

(b) Compute the class marks or midpoints: 575, 725, and so on. Put them in L1 and the frequencies in L2. Use 1-VarStats L1,L2 and get n = 219.
See Summary Numbers on the TI-83.

(c) Further data from 1-VarStats L1,L2: x̅ = 990.1 and s = 167.3

Common mistake: If you answered x̅ = 950 you probably did 1-VarStats L1 instead of 1-VarStats L1,L2. Your calculator depends on you to supply one list when you have a simple list of numbers and two lists when you have a frequency distribution.

(d) f/n = 29/219 ≈ 0.13 or 13%

invNorm(0.85, 57.6, 5.2) = 62.98945357 → 63.0 mph

MATH200A/sample size/binomial: p̂ = .2, E = 0.04, C-Level = 0.90

answer: 271.

Common mistake: The margin of error is E = 4% = 0.04, not 0.4.

Alternative solution: See Sample Size by Formula and use the formula at right. With the estimated population proportion p̂ = 0.2 in the formula, you get z_α/2 = z_0.05 = invNorm(1−0.05) = 1.6449, and n = 270.5543 → 271

(b) If you have no prior estimate, use p̂ = 0.5. The other inputs are the same, and the answer is 423

You expect positive correlation because points trend upward to the right (or, because y tends to increase as x increases). Even before plotting, you could probably predict a positive correlation because you assume higher calories come from fat; but you can’t just assume that without running the numbers.

(b) See Step 2 of Scatterplot, Correlation, and Regression on TI-83/84.
r = .8863314629 → r = 0.8862
a = .0586751909 → a = 0.0587
b = −3.440073602 → b = −3.4401

ŷ = 0.0587x − 3.4401

Common mistake: The symbol is ŷ, not y.

(c) The y intercept is −3.4401. It is the number of grams of fat you expect in the average zero-calorie serving of fast food. Clearly this is not a meaningful concept.

Remark: Remember that you can’t trust the regression outside the neighborhood of the data points. Here x varies from 130 to 640. The y intercept occurs at x = 0. That is pretty far outside the neighborhood of the data points, so it’s not surprising that its value is absurd.

(d) See How to Find ŷ from a Regression on TI-83/84. Trace at x = 310 and read off ŷ = 14.749… ≈ 14.7 grams fat. This is different from the actual data point (x=310, y=25) because ŷ is based on a trend reflecting all the data. It predicts the average fat content for all 310-calorie fast-food items.

Alternative solution: ŷ = .0586751909(310) − 3.440073602 = 14.749 ≈ 14.7.

(e) The residual at any (x,y) is y−ŷ. At x = 310, y = 25 and ŷ = 14.7 from the previous part. The residual is y−ŷ = 10.3

Remark: If there were multiple data points at x = 310, you would calculate one residual for each point.

(f) From the LinReg(ax+b) output, R² = 0.7855834621 → R² = 0.7856 About 79% of the variation in fat content is associated with variation in calorie content. The other 21% comes from lurking variables such as protein and carbohydrate count and from sampling error.

(g) See Decision Points for Correlation Coefficient. Since 0.8862 is positive and 0.8862 > 0.602, you can say that there is some positive correlation in the population, and higher-calorie fast foods do tend to be higher in fat.

(b) normalcdf(-10^99, 24, 27, 4/√5) = .0467662315 → 0.0468 or about a 5% chance

(b) For two-population numeric data, paired data do a good job of controlling for lurking variables. You would test each person’s blood sugar, then enroll all thirty patients in the course and test their blood sugar six months after the end of the course. Your variable d is blood sugar after the course minus blood sugar before, and your H₁ is μ_d < 0.

One potential problem is that all 30 patients receive a heightened level of attention, so you have to worry about the placebo effect. (With the original experiment, the control group did not receive the extra attention of being in the course, so any difference from the attention is accounted for in the different results between control group and treatment group.)

It seems unlikely that the placebo effect would linger for six months after the end of a short course, but you can’t rule out the possibility. There are two answers to that. You could re-test the patients after a year, or two years. Or, you could ask whether it really matters why patients do better. If they do better because of the course itself, or because of the attention, either way they’re doing better. A short course is relatively inexpensive. If it works, why look a gift horse in the mouth? In fact, medicine is beginning to take advantage of the placebo effect in some treatments.

(b) 2-PropZInt with the above inputs and C-Level=.95 gives (−.0268, .14682). The English sentence needs to state both magnitude and direction, something like this: Regarding men and women who refused promotion for family reasons, we’re 95% confident that men were between 2.7 percentage points less likely than women, and 14.7 percentage points more likely.

Common mistake: With two-population confidence intervals, you must state the direction of the difference, not just the size of the difference.

If the middle 95% runs from 70 to 130, then the mean must be μ = (70+130)÷2 → μ = 100

95% of any population are within 2 standard deviations of the mean. The range 70 to 100 (or 100 to 130) is therefore two SD. 2σ = 100−70 = 30 → σ = 15

P(Brand A and mislabeled) = P(Brand A) × P(mislabeled | Brand A)

and similarly for brand B.

P(mislabeled) = 0.40 × 0.025 + 0.60 × 0.015 = 0.019 or just under 2%

Alternative solution: The formulas can be confusing, and often there’s a way to do without them. You could also do this as a matter of proportions:

Out of 1000 shoes, 400 are Brand A and 600 are Brand B.

Out of 400 Brand A shoes, 2.5% are mislabeled. 0.025×400 = 10 brand A shoes mislabeled.

Out of 600 Brand B shoes, 1.5% are mislabeled. 0.015×600 = 9 brand B shoes mislabeled.

Out of 1000 shoes, 10 + 9 = 19 are mislabeled. 19/1000 is 1.9% or 0.019.

This is even easier to do if you set up a two-way table, as shown below. The values in bold face are given in the problem, and those in light face are derived from them.

Solution: This is paired numeric data, Case 3.

Common mistake: You must do this as paired data. Doing it as unpaired data will not give the correct p-value.

Solution: (a) Use MATH200A part 3 with n=2, p=0.9, from=1, to=1. Answer: 0.18

You could also use binompdf(2, .9, 1) = 0.18.

Alternative solution: The probability that exactly one is tainted is sum of two probabilities: (i) that the first is tainted and the second is not, and (ii) that the first is not tainted and the second is. Symbolically,

P(exactly one) = P(first and second^C) + P(first^C and second)

P(exactly one) = 0.9×0.1 + 0.1×0.9

P(exactly one) = 0.09 + 0.09 = 0.18

Solution: (b) When sampling without replacement, the probabilities change. You have the same two scenarios — first but not second, and not first but second — but the numbers are different.

P(exactly one) = P(first and second^C) + P(first^C and second)

P(exactly one) = (9/10)×(1/9) + (1/10)×(9/9)

P(exactly one) = 1/10 + 1/10 = 2/10 = 0.2

Common mistake: Many, many students forget that both possible orders have to be considered: first but not second, and second but not first.

Common mistake: You can’t use binomial distribution in part (b), because when sampling without replacement the probability changes from one trial to the next.

For example, if the first card is an ace then the probability the second card is also an ace is 3/51, but if the first card is not an ace then the probability that the second card is an ace is 4/51. Symbolically, P(A₂|A₁) = 3/51 but P(A₂| not A₁) = 4/51.

(a) p̂_T = 128/300 = 0.4267. p̂_C = 135/400 = 0.3375. p̂_T−p̂_C = 0.0892 or about 8.9%

Remark: The point estimate is descriptive statistics, and requirements don’t enter into it. But the confidence interval is inferential statistics, so you must verify that each sample is random, each sample has at least 10 successes and 10 failures, and each sample is less than 10% of the population it came from.

The problem states that the samples were random, which takes care of the first requirement. There were 128 successes and 300−128 = 172 failures in Tompkins, 135 successes and 400−135 = 265 failures in Cortland, so the second reqirement is met.

What about the third requirement? You don’t know the populations of the counties, but remember that you can work it backwards. 10×300 = 3000 (Tompkins) and 10×400 = 4000 (Cortland), and surely the two counties must have populations greater than 3000 and 4000, so the third requirement must be met.

(b) 2-PropZInt: The 98% confidence interval is 0.0029 to 0.1754 (about 0.3% to 17.5%), meaning that with 98% confidence Tompkins viewers are more likely than Cortland viewers, by 0.3 to 17.5 percentage points, to prefer a movie over TV.

(c) E = 0.1754−0.0892 = 0.0862 or about 8.6%

You could also compute it as 0.0892−0.0029 = 0.0863 or (0.1754−0.0029)/2 = 0.0853. All three methods get the same answer except for a rounding difference.

Alternative solution: You could also do this as a test of homogeneity, Case 7. The χ²-Test gives χ² = 4.98, df = 1, p=0.0256