Chapter 21: What Is a Confidence Interval?

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We used the 95 part of the 68–95–99.7 rule to get a 95% confidence interval for the population proportion. Perhaps you think that a method that works 95% of the time isn’t good enough. You want to be 99% confident. For that, we need to mark off the central 99% of a Normal distribution. For any probability C between 0 and 1, there is a number z* such that any Normal distribution has probability C within z* standard deviations of the mean. Figure 21.5 shows how the probability C and the number z* are related.

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Figure 21.5: Figure 21.5 Critical values z* of the Normal distributions. In any Normal distribution, there is area (probability) C under the curve between –z* and z* standard deviations away from the mean.

Table 21.1 gives the numbers z* for various choices of C. For convenience, the table gives C as a confidence level in percent. The numbers z* are called critical values of the Normal distribution. Table 21.1 shows that any Normal distribution has probability 99% within ±2.58 standard deviations of its mean. The table also shows that any Normal distribution has probability 95% within ±1.96 standard deviations of its mean. The 68–95–99.7 rule uses 2 in place of the critical value z* = 1.96. That is good enough for practical purposes, but the table gives the more exact value.

Table : TABLE 21.1 Critical values of the Normal distributions

Confidence level C	Critical value z*	Confidence level C	Critical value z*
50%	0.67	90%	1.64
60%	0.84	95%	1.96
70%	1.04	99%	2.58
80%	1.28	99.9%	3.29

From Figure 21.5 we see that, with probability C, the sample proportion $\hat{p}$ takes a value within z* standard deviations of p. That is just to say that, with probability C, the interval extending z* standard deviations on either side of the observed $\hat{p}$ captures the unknown p. Using the estimated standard deviation of $\hat{p}$ produces the following recipe.

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Confidence interval for a population proportion

Choose an SRS of size n from a population of individuals of which proportion p are successes. The proportion of successes in the sample is $\hat{p}$ . When n is large, an approximate level C confidence interval for p is

$\hat{p} \pm z* \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}}$

where z* is the critical value for probability C from Table 21.1.

EXAMPLE 5 A 99% confidence interval

The Gallup random sample of 1009 adult Americans in 2015 found that 616 reported actively trying to avoid drinking regular soda or pop. We want a 99% confidence interval for the proportion p of all adult Americans who actively avoided drinking regular soda or pop. Table 21.1 says that for 99% confidence, we must go out z* = 2.58 standard deviations. Here are our calculations:

$\hat{p} = \frac{616}{1009} = 0.611$

$\hat{p} \pm z* \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}} = 0.611 \pm 2.58 \sqrt{\frac{(0.611) (0.389)}{1009}}$

= 0.611 ± (2.58)(0.0153)

= 0.611 ± 0.0395

= 0.5715 to 0.6505

We are 99% confident that the true population proportion is between 57.15% and 65.05%. That is, we got this range of percentages by using a method that gives a correct answer 99% of the time.

Compare Example 5 with the calculation of the 95% confidence interval in Example 3. The only difference is the use of the critical value 2.58 for 99% confidence in place of 2 for 95% confidence. That makes the margin of error for 99% confidence larger and the confidence interval wider. Higher confidence isn’t free—we pay for it with a wider interval. Figure 21.5 reminds us why this is true. To cover a higher percentage of the area under a Normal curve, we must go farther out from the center. Figure 21.6 compares the lengths of the 90%, 95%, and 99% confidence intervals.

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Figure 21.6: Figure 21.6 The lengths of three confidence intervals for the soda consumption example. All three are centered at the estimate

$\hat{p}$ = 0.611. When the data and the sample size remain the same, higher confidence results in a larger margin of error.

NOW IT’S YOUR TURN

Question 21.2

21.2 Gambling. A May 2011 Gallup Poll consisting of a random sample of 1018 adult Americans found that 31% believe that gambling is morally wrong. Find a 99% confidence interval for the proportion of all adult Americans who believe that gambling is morally wrong. How would you interpret this interval?

21.2 The 99% confidence interval for the proportion of all adult Americans who believe gambling is morally wrong is

$\begin{array}{l} \hat{p} \pm 2 \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}} & = & 0.31 \pm 2.58 \sqrt{\frac{0.31 (0.69)}{1018}} \\ = & 0.31 \pm 2.58 (0.014) \\ = & 0.31 \pm 0.036 \\ = & 0.274 to 0.346 \end{array}$

Interpret this result as follows: we got this interval by using a recipe that catches the true unknown population proportion 99% of the time. The shorthand is, we are 99% confident that the true proportion of adult Americans who believe gambling is morally wrong lies between 27.4% and 34.6%.

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Question 21.2

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