Stats without Tears
9. Estimating Population Parameters

9A.  Estimating Population Proportion p

In the Physicians’ Health Study [see “Sources Used” at end of book], about 22,000 male doctors were randomly assigned to take aspirin or a placebo every night. Of 11,037 in the treatment group, 104 had heart attacks and 10,933 did not. Can you say how likely it is for people in general (or, at least, male doctors in general) to have a heart attack if they take aspirin nightly?

As always, probability of one equals proportion of all. So you could just as well ask, what proportion of people who take aspirin would be expected to have heart attacks?

Before statistics class, you would divide 104/11037 = 0.0094 and say that 0.94% of people taking nightly aspirin would be expected to have heart attacks. This is known as a point estimate.

But you are in statistics class. You know that a sample can’t perfectly represent the population, and therefore all you can say is that the true proportion of heart attacks in the population of aspirin takers is around 0.94%. Can you be more specific?

Yes, you can. You can compute a confidence interval for the proportion of heart attacks to be expected among aspirin takers, based on your sample, and that’s the subject of this chapter. We’ll get back to the doctors and their aspirin later, but first, let’s do an example with M&Ms.

9A1.  Confidence Interval for p (Binomial Data)

Example 1: You take a random sample of 605 plain M&Ms, and 87 of them are red. What can you say about the proportion of reds in all plain M&Ms?

For all the cases you’ll study in this course, the point estimate — the mean or proportion of your sample — is at the middle of the confidence interval. But that’s not true for some other cases, such as estimating the standard deviation of a population. For those cases, computing the margin of error is uglier.

Computing a Confidence Interval

As you might expect, your TI-83/84 and lots of statistical packages can compute confidence intervals for you. But before doing it the easy way, let’s take a minute to understand what’s behind computing a confidence interval.

You can compute an interval to any level of confidence you desire, but 95% is most common by far, so let’s start there. How do you use those 87 reds in a sample of 605 M&Ms to estimate the proportion of reds in the population, and have 95% confidence in your answer?

In Chapter 8, you learned how to find the sampling distribution of p̂. Given the true proportion p in the population, you could then determine how likely it is to get a sample proportion p̂ within various intervals. To find a confidence interval, you simply run that backward.

You don’t know the proportion of reds in all plain M&Ms, so call it p. You know that, if the sample size is large enough, sample proportions are ND and there’s a 95% chance that any given sample proportion will be within 2 standard errors on either side of p, whatever p is.

The standard error of the proportion is σ_p̂ = √pq/n. You don’t know p — that’s what you’re trying to find. Are you stuck? No, you have an estimate for p. Your point estimate for the population proportion p is the sample proportion p̂ = 87/605. You can estimate the standard error of the proportion (the SEP) by using the statistics of your sample:

σ_p̂ ≈  √(87/605)(1−87/605)/605 = 0.0142656 or about 1.4%

Two standard errors is 0.0285312 → 0.029 or 2.9%.

How good is this estimate? For decent-sized samples, it’s quite good. For example, suppose the true population proportion p is 50% or 0.5. For a sample of n = 625, the SEP is √.5(1−.5)/625 = 0.0200 or 2.00%. Your sample proportion is very, very, very unlikely to be as far away as 40% or 0.4, but even if it is then you would estimate the SEP as √.4(1−.4)/625 = 0.0196 or 1.96%, which is extremely close.

Different authors use the term “standard error” slightly differently. Some use it only for the standard deviation of the sampling distribution, which you never know exactly because you never know the population parameters exactly. Others use it only for the estimate based on sample statistics, which I computed just above. Still others use it for either computation. In practice it doesn’t make a lot of difference. I don’t see much point to getting too fussy about the terminology, given that only one of them can be computed anyway.

Any given sample proportion is 95% likely to be within two standard errors or 2.9% of the population proportion:

p−0.029 ≤ p̂ ≤ p+0.029 (probability = 95%)

Now the magic reverso: Given a sample proportion, you’re 95% confident that the population proportion is within 2.9% of that sample proportion:

p̂−0.029 ≤ p ≤ p̂+0.029 (95% confidence)

In this case, your sample proportion is 87/605 ≈ 0.144:

0.144−0.029 ≤ p ≤ 0.144+0.029 (95% confidence)

0.115 ≤ p ≤ 0.173 (95% confidence)

So your 95% confidence interval is 0.115 to 0.173, or 11.5% to 17.3%.

You might have noticed that I changed from 95% probability to 95% confidence. What’s up with that? Well, the sample proportion is a random variable — different samples will have different sample proportions p̂, and you can compute the probability of getting p̂ in any particular range.

But the population proportion p is not a random variable. It has one definite value, even though you don’t know what that definite value is. Probability statements about a definite number make about as much sense as discussing the probability of precipitation for yesterday. The population proportion is what it is, and you have some level of confidence that your estimated range includes that true value.

What does “95% confident” mean, then? Simply this: In the long run, when you do everything right, 95% of your 95% intervals will actually include the population proportion, and the other 5% won’t. 5% is 5/100 = 1/20, so in the long run about one in 20 of your 95% confidence intervals will be wrong, just because of sample variability.

Probability of one = proportion of all, so there’s one chance in twenty that this interval is wrong, meaning that it doesn’t contain the true population proportion, even if you did everything right. If that makes you too nervous, you can use a higher confidence level, but you can never reach 100% confidence.

There’s one more wrinkle. That margin of error of 0.029 was 2σ_p̂, two standard errors. The figure of 2 standard errors for the middle 95% of a ND comes from the Empirical Rule or 68–95–99.7 Rule, so it’s only approximately right.

But you can be a little more precise. In Chapter 7 you learned to find the middle any percent, and that lets you generalize to any confidence level:

	This Example	General Case
Confidence level (middle area of the ND)	95%	1−α
Area in the two tails combined	100%−95% = 5% or 0.05	1−(1−α) = α
Area in each tail	0.05/2 = 0.025	α/2
The boundaries are	±z0.025 = invNorm(1−0.025) = 1.9600	±zα/2
The margin of error is	E = 1.96σp̂	E = zα/2 σp̂
And you compute it as	E = 1.96	E = zα/2

The margin of error on a 1−α confidence interval is z_α/2 standard errors. (This will be important when you determine necessary sample size, below.)

The margin of error on a 95% confidence interval is close to 2σ_p̂, but more accurately it’s 1.96σ_p̂. For the proportion of red M&Ms, where the SEP was σ_p̂ = 0.0142656, the margin of error is 1.96σ_p̂ = 0.0279606 → 0.028 or 2.8%. Since the point estimate was 14.4%, you’re 95% confident that the proportion of reds in plain M&Ms is within 14.4%±2.8%, or 11.6% to 17.2%.

Interpreting a Confidence Interval

You’ve seen that there are two ways to state a confidence interval: from ____ to ____ with ____% confidence, or ____ ± ____ with _____% confidence. Mathematically these are equivalent, but psychologically they’re very different. The first form is better than the second.

What’s wrong with the ____ ± ____ form? It’s easy to misinterpret.

If you say “I’m 95% confident that the proportion of reds in plain M&Ms is within 14.4%±2.8%”, some people will read 14.4% and stop — they’ll think that the population proportion is 14.4%. And even people who get past that will probably think that there’s something special about 14.4%, that somehow it’s more likely to be the true proportion of reds among all plain M&Ms. But 14.4% is just a value of a random variable, namely the proportion of reds in this sample. Another sample would almost certainly have a different p̂ and therefore a different midpoint for the interval.

It’s much better to use the endpoint form, because the endpoint form is harder to misinterpret. When you say “I’m 95% confident that the proportion of reds in plain M&Ms is 11.6% to 17.2%”, you lead the reader, even the non-technical reader, to understand that the proportion could be anything in that range, and even that there’s a slight chance that it’s outside that range.

Requirements check (RC): This is an essential step — do it before you compute the confidence interval. Computing the CI assumes that the sampling distribution of p̂ is a ND, but “assumes” in statistics means you don’t assume, you check it.

The requirements are stated in Chapter 8 as simple random sample (or equivalent), np and nq both ≥ about 10, 10n ≤ N. You don’t know p, but for binomial data it’s okay to use p̂ as an estimate. But np̂ is just the number of yeses or successes in your sample, and nq̂ is just the number of noes or failures in your sample, so you really don’t need to do any multiplications.

Here’s how you check the requirements:

• Random sample: stated at start of section, OK.
• Successes in sample: 87; failures in sample: 605−87 = 518; both ≥ 10, OK.
• 10n = 10×605 = 6050. We don’t know how many plain M&Ms there are in the world, but surely M&M Mars makes far more than that every second, so this is also OK.

Easy CIs with TI-83/84

Your TI-83 or TI-84 can easily compute confidence intervals for a population proportion. With binomial data, this is Case 2 in Inferential Statistics: Basic Cases. (Excel can do it too, but it’s significantly harder in Excel.)

Example 2: Let’s do the red M&Ms, since you already know the answer. See the requirements check above. Press [STAT] [◄] to get to the STAT TESTS menu, and scroll up or down to find 1-PropZInt. (Caution: you don’t want 1-PropZTest. That’s reserved for Chapter 10.) Enter the number of successes in the sample, the sample size, and the confidence level — easy-peasy! Write down the screen name and your inputs, then proceed to the output screen and write down just the new stuff:

Here’s how you show your work:

1-PropZInt 87, 605, .95 (not PropZInt, please!)

(.11584, .17176), p̂ = .1438016529

There’s no need to write n=605 because you already wrote it down from the input screen.

Interpretation: I’m 95% confident that 11.6% to 17.2% of plain M&Ms are red.

You can vary that in several ways. For instance, some people like to put the confidence level last: 11.6% to 17.2% of plain M&Ms are red (95% confidence). Or they may choose more formal language: We’re 95% confident that the true proportion of reds in plain M&Ms is 11.6% to 17.2%.

I’ve already pooh-poohed the margin-of-error form, but sometimes you have to write it that way, for instance if your boss or your thesis advisor demands it. You can get it easily from the TI-83/84 output screen.

The center of the interval, the point estimate, is given: 14.38%. To find the margin of error, subtract that from the upper bound of the interval, or subtract the lower bound from it: .17176−.1438 = .02796, or .1438−.11584 = .02796. Either way it’s 2.8%. You can then express the CI as 14.4%±2.8% with 95% confidence.

Example 3: What about the male doctors who started this section? 104 out of 11037 of the doctors taking nightly aspirin had heart attacks. Assuming that male doctors are representative of adults in general, in terms of heart-attack risk, what can you say about the chance of heart attack for anyone who takes aspirin nightly? Use confidence level 1−α = 95%.

Solution: Requirements check (RC):

• Random sample stated, OK.
• 104 successes in sample, 11037−104 = 10933 failures, OK.
• 10n = 10×11037 = 110370. Without knowing the number of adults in the US or the world, we know it’s a lot more than that; OK.

1-PropZInt 104, 11037, .95

(.00762, .011223), p̂ = .0094228504

Conclusion: People who take nightly aspirin have a 0.76% to 1.12% chance of heart attack (95% confidence).

9A2.  How Big a Sample for Binomial Data?

The equation for margin of error is packed with information:  E = z_α/2

You can see that a larger sample size n means a narrower confidence interval, but the sample size is inside the square-root sign so you don’t get as much benefit as you might hope for. If you take a sample four times as big, the square root of 4 is 2 and so your interval is half as wide, not ¼ as wide.

You can see also that you get a narrower interval if you’re willing to live with a lower confidence level. The lower your confidence interval, the smaller z_α/2 will be, and therefore the narrower your confidence interval.

The bottom line is that there’s a three-way tension among sample size, confidence level, and margin of error. You can choose any two of those, but then you have to live with the third. (p̂ doesn’t come into it. Although p̂ does contribute to the standard error and therefore to the margin of error, you can’t choose what p̂ you’re going to get in a sample.)

If you want to get a confidence interval at your preferred confidence level with (no more than) a specified margin of error, how big a sample do you need? MATH200A Program part 5 will compute this for you, but let’s look at the formula first.

(See Getting the Program for instructions on getting the MATH200A program into your calculator.)

The equation at the start of this section shows the margin of error you get for a given sample size and confidence level. You can solve for the sample size n, like this:

E = z_α/2     ⇒   

In the formula, p̂ is your prior estimate if you have one. This can be the result of a past study, or a reasonable estimate if it has some logical basis. If you don’t have a prior estimate, use 0.5.

Example 4: In a sample of 605 plain M&Ms, 87 were red. The 95% confidence interval had a 2.8% margin of error. How big a sample would you need to reduce the margin of error to 2%?

There’s no requirements check in sample-size problems. These are planning how to take your sample; requirements apply to your sample once you have it.

Example 5: You’re taking the first political poll of the season, and you’d like to know what fraction of adults favor your candidate. You decide you can live with a 90% confidence level and a 3% margin of error. How many adults do you need in your random sample?

Solution: Since you have no prior estimate for p, make p̂ = .5.

9B.  Estimating Population Mean μ When You Know σ

Numeric data are pretty much the same deal as binomial data, though there are a couple of wrinkles:

• The requirements are different. For numeric data you need a sample bigger than about 30. If your sample is smaller, then it must be ND with no outliers. (You need a random sample for every procedure.)
• The standard error is σ_x̅ = σ/√n, so the margin of error is E = z_α/2·σ/√n.

The second one is a problem, because you almost never know the standard deviation of the population. Therefore, we won’t be working any problems for this case. Instead, I’ll give you a little more theory to lay the groundwork for the next section, which explains how we get around this knowledge gap.

9B1.  Confidence Interval

If you know the standard deviation of the population — and you hardly ever do — then your confidence interval is

x̅ − z_α/2 · σ/√n  ≤  μ  ≤  x̅ + z_α/2 · σ/√n

If you’re ever in this situation, you can compute a confidence interval on your TI-83/84 by choosing ZInterval in the STAT TESTS menu.

9B2.  How Big a Sample Do You Need?

The margin of error is E = z_α/2 · σ/√n, so the required sample size for a margin of error E with confidence level 1−α is

n = [ z_α/2 · σ / E]²

(You can also use MATH200A part 5.)

9C.  Estimating Population Mean μ When You Don’t Know σ

“Houston, we have a problem!” A confidence interval is founded on the sampling distribution of the mean or proportion. Everything in Chapter 8 on the sampling distribution of the mean was based on knowing the standard deviation of the population. But you almost never know the standard deviation of the population. How to resolve this?

The solution comes from William Gosset, who worked for Guinness in Dublin as a brewer. (I swear I am not making this up.) In 1908 he published a paper called The Probable Error of a Mean [see “Sources Used” at end of book]. For competitive reasons, the Guinness company wouldn’t let him use his own name, and he chose the pen-name “Student”. The t distribution that he described in his paper has been known as Student’s t ever since.

While looking for Gosset’s original paper, I stumbled on Probable Error of a Mean, The (“Student”) (Moulton [see “Sources Used” at end of book]). It’s a fascinating look at what Gosset did and didn’t accomplish, and how this classic paper was virtually ignored for years. Things didn’t start to happen till Gosset sent a copy of his tables to R. A. Fisher with the remark that Fisher was the only one who would ever use them! It was Fisher who really got the whole world using Student’s t distribution.

9C1.  Student’s t Distribution

Gosset knew that the standard error of the mean is σ/√n, but he didn’t know σ. He wondered what would happen if he estimated the standard error as s/√n, and did some experiments to answer that question. Since s varies from one sample to the next, this new t distribution spreads out more than the ND. Its peak is shallower, and its tails are fatter.

Actually, there’s no such thing as “the” t distribution. There’s a different t for each sample size. The larger the sample, the closer that t distribution is to a normal distribution, but it’s never quite normal.

For technical reasons, t distributions aren’t identified by sample size, but rather by degrees of freedom (symbol df or Greek ν, “nu”). df = n−1. Here are two t distributions:

Solid: standard normal distribution Line: Student’s t for df = 4, n = 5

Solid: standard normal distribution Line: Student’s t for df = 29, n = 30

What do you see? Student’s t for 4 degrees of freedom is quite a bit more spread out than the ND: 12.2% of sample means are more than two standard errors from the mean, versus only 5% for the ND.

At this scale, Student’s t for 29 degrees of freedom looks identical to the ND, but it’s not quite the same. You can see that 6% of sample means are more than two standard errors from the mean, versus 5% for the ND.

You don’t really need a list of properties of Student’s t, because your calculator is going to do the work for you. It’s enough to know this:

• There’s a t distribution for each sample size n. They’re identified by degrees of freedom, ν or df = n−1.
• t distributions show more variability than z (the standard normal curve). As df or n increases, the t distributions get closer and closer to normal. Beyond about df = 30, the difference is slight.
• High t numbers occur more often than high z numbers, but only for small samples is the difference very noticeable.

9C2.  Confidence Interval for μ (Numeric Data)

The logic of confidence intervals for numeric data is the same whether you know the standard deviation of the population or not. Even the requirements are the same. The only difference is between using a z and a t.

Example 6: You’re auditing a bank. You take a random sample of 50 cash deposits and find a mean of $189.56 and standard deviation of $42.17.
(a) Estimate the mean of all cash deposits, with 95% confidence.
(b) The bank’s accounting department tells you that the average cash deposit is over $210.00. Is that believable?

Solution: You want to compute a confidence interval about the mean of all deposits. You have numeric data, and you don’t know the standard deviation of the population, σ. This is Case 1 in Inferential Statistics: Basic Cases. In your sample, n = 50, x̅ = 189.56, and s = 42.17.

First, check the requirements (RC):

• Random sample given, OK.
• Sample size 50 is > 30, OK.
• 10n = 10×50 = 500, and surely the bank has more deposits than that.

Since the sample is large enough, there’s no need to verify normality or check for outliers.)

Now calculate the interval.

On your TI-83/84, in the STAT TESTS menu, select 8:TInterval. The difference between Data and Stats is whether you have all the data points, or just summary statistics. In this case you have only the stats, so cursor onto Stats and press [ENTER]. (The lower part of the screen may change.)

Enter your sample statistics and your desired confidence level. Write down your inputs before you select Calculate:

TInterval 189.56, 42.17, 50, .95

Proceed to the output screen, and write down everything new. There isn’t much:

(177.58, 201.54)

Finally, write your interpretation. I’m 95% confident that the average of all cash deposits is between $177.58 and $201.54.

Caution! Don’t say anything like “95% of deposits are between $177.58 and $201.54.” Your confidence interval is an estimate of the true average of all deposits, and it’s not about the individual deposits. With a standard deviation of $42 and change, you would predict that 95% of deposits are within 2×42.17 = $84.34 either side of the mean, which is a much wider interval.

Now turn to part (b). Management claims that the average of all cash deposits is > $210.00. Is that believable? Well, it’s not impossible, but it’s unlikely. You’re 95% confident that the average of all deposits is between $177.58 and $201.54, which means you’re 95% confident that it’s not < $177.58 or > $201.54. But they’re claiming $210, which is outside your confidence interval. Again, they’re unlikely to be correct — there’s less than a 5% likelihood (100%−95% = 5%).

Example 7: In a random sample from the 237 vehicles on a used-car lot, the following weights in pounds were found:

2500  3250  4000  3500  2900  4500  3800  3000  5000  2200

Estimate the average weight of vehicles on the lot, with 90% confidence.

Solution: Check the requirements first. You have a small sample (n < 30), so you have to verify that the data are ND and there are no outliers. Here are the results of normality check and box-whisker plot in MATH200A:

There’s not much you can write for the box-whisker plot, but you can show the normality test numerically:

• Random sample, OK.
• Box-whisker: no outliers, OK.
• Normality: r(.9936) > crit(.9179). OK.
• 10% of population size is 23.7, and this sample of 10 is smaller than that.

Now proceed to your TInterval. This time you have the actual data, so you choose Data on the screen. Specify your data list. Freq (frequency) should already be set to 1; if not, first press the [ALPHA] key once, and then [1] [ENTER]. Enter your confidence level, and write down your inputs:

TInterval L1, 1, .90

When you have raw data, everything on the output screen is new:

(2956, 3974)

x̅ = 3465, s = 878.1, n = 10

You’re 90% confident that the average weight of all vehicles on the lot is between 2956 and 3974 pounds.

Again, this is an estimate of the average weight of the population (the 237 cars on the lot). In your interpretation, you can’t say anything about the weights of individual vehicles, because you don’t know anything about the weights of individual vehicles, apart from your sample.

9C3.  The Trouble with Outliers

Why do you have to check for outliers? If your sample passes the normality check, isn’t that enough? No! If a sample passes the normality check, it still might have outliers.

Why are outliers a problem? Well, your confidence interval depends on the mean and standard deviation of your sample. But x̅ and s are sensitive to outliers. (That sensitivity goes down as sample size goes up, so you don’t have to worry with samples bigger than about 30.)

To make this clearer, let’s look at an example. I drew these 15 points from a moderately skewed population:

 157   171   182   189   201   208   217   219 
 229   242   247   252   265   279   375 

The normality test shows r > crit. So far so good. But the box plot shows a big honkin’ outlier:

How big a difference does it make? Quite a lot, unfortunately. Here are the 95% confidence intervals for the original sample, and the sample with the outlier removed. The means are different, the standard deviations are really different, and the high ends of the confidence intervals are pretty different too. (The screens don’t show the margins of error, but they too are quite different: (258.45-199.28)/2 = 29.6 and (239.36-197.5)/2 = 20.9.)

95% CI from full sample

95% CI excluding outlier

Do you say that the outlier increased the mean by almost 5% and the SD by almost 50%, moved the confidence interval and made it wider? That’s not really fair — the sample is what it is (assuming you’ve ruled out a mistake in data entry). If you start throwing out points, you no longer have a random sample. On the other hand, that one point does seem to carry an awful lot of weight, and it doesn’t seem right to have results depend so heavily on one point.

So what do you do? If you can, you take another sample, preferably a larger one. Larger samples are less likely to have outliers in the first place, and outliers that do occur have less influence on the results.

But taking a new sample may not be practical. An alternative — not really great, but better than nothing — is to do the analysis both ways, once with the full sample and once with the outlier(s) excluded. That will at least give a sense of how much the outliers affect the results.

9C4.  How Big a Sample for Numeric Data?

What Have You Learned?

Chapter 10 WHYL → ← Chapter 8 WHYL

Exercises for Chapter 9

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 →