AP Statistics 3.9: Difference in Proportions

Check whether a difference between two sample proportions has an approximately normal sampling distribution using design and count conditions.

Covered syllabus points

3.9.A—Calculate the mean and standard deviation of the sampling distribution for the difference between two sample…3.9.B—Justify the appropriateness of conditions for the sampling distribution for the difference between two sample…3.9.C—Interpret the mean, standard deviation, and probabilities for the sampling distribution for the difference…

A large company produces an equal number of brand-name lightbulbs and generic lightbulbs. The director of

quality control sets guidelines that production will be stopped if there is evidence that the proportion of all

lightbulbs that are defective is greater than 0.10. The director also believes that the proportion of brand-name

lightbulbs that are defective is not equal to the proportion of generic lightbulbs that are defective. Therefore, the

director wants to estimate the average of the two proportions.

To estimate the proportion of brand-name lightbulbs that are defective, a simple random sample of

400 brand-name lightbulbs is taken and 44 are found to be defective. Let X represent the number of

brand-name lightbulbs that are defective in a sample of 400, and let $p_{X}$ represent the proportion of all

brand-name lightbulbs that are defective. It is reasonable to assume that X is a binomial random variable.

Consider $\hat{D}$ , the point estimate of D.

Calculate $s_{\hat{D}}$ , the standard error of $\hat{D}$ .

The standard error of $\hat{p}_{X}=0.0156$ is obtained from part (b). The standard error of $\hat{p}_{Y}$ is

$\sqrt{\frac{\hat{p}_{Y}\left(1-\hat{p}_{Y}\right)}{n}}=\sqrt{\frac{(0.26)(0.74)}{400}}=0.0219$ . So the standard error of $\hat{D}$ is

$S_{\hat{D}}=\sqrt{\frac{1}{4}\left(0.0156^{2}+0.0219^{2}\right)}=0.0134$ .

Part (e)

W=\frac{0.185-0.10}{0.0134}=6.34 .

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Question 6 (continued)

Part (f)

Suppose the true mean D is 0.10. Then the observed value of $\hat{D}=0.185$ is 6.34 standard errors from

the mean D. Using Chebyshev's inequality, the probability of observing a value of $\hat{D}$ within 6.34 standard

errors of the mean of 0.10 is at least $1-\frac{1}{6.34^{2}}=0.975$ . So the probability of observing a value as far from

0.10 as the one observed, or farther, is at most 0.025 if the true mean really is 0.10. Therefore, the p-value

for this test is at most 0.025, which is less than 0.05, so the null hypothesis can be rejected. There is

sufficient statistical evidence at the 0.05 level to conclude that the average proportion for all products that

are defective is greater than 0.10.

Psychologists interested in the relationship between meditation and health conducted a study with a random sample of 28 men who live in a large retirement community. Of the men in the sample, 11 reported that they participate in daily meditation and 17 reported that they do not participate in daily meditation.
The researchers wanted to perform a hypothesis test of

\begin{aligned} & \mathrm{H}_{0}: p_{m}-p_{c}=0 \\ & \mathrm{H}_{a}: p_{m}-p_{c}<0, \end{aligned}

where $\quad p_{m}$ is the proportion of men with high blood pressure among all the men in the retirement community who participate in daily meditation and $p_{c}$ is the proportion of men with high blood pressure among all the men in the retirement community who do not participate in daily meditation.

Let $\hat{p}_{m}$ represent the proportion of men with high blood pressure among those in a random sample of 11 who meditate daily, and let $\hat{p}_{c}$ represent the proportion of men with high blood pressure among those in a random sample of 17 who do not meditate daily. Why is it not reasonable to use a normal approximation for the sampling distribution of $\hat{p}_{m}-\hat{p}_{c}$ ?

Although a normal approximation cannot be used, it is possible to simulate the distribution of $\hat{p}_{m}-\hat{p}_{c}$ . Under the assumption that the null hypothesis is true, 10,000 values of $\hat{p}_{m}-\hat{p}_{c}$ were simulated. The histogram below shows the results of the simulation.

Figure for Question 3.9 Sampling Distributions for the Difference Between Two Population Proportions question 2 — AP Statistics

Part (b):

The sample sizes were too small, relative to the overall sample proportion of successes, to justify using a normal approximation. One way to check this is to note that the combined sample proportion of successes is $\hat{p}=\frac{0+8}{11+17}=\frac{8}{28} \approx 0.286$ , so neither $n_{m} \hat{p}=11 \times \frac{8}{28} \approx 3.143$ nor $n_{c} \hat{p}=17 \times \frac{8}{28} \approx 4.857$ is at least 10.

[Maximum number: 1]

An international polling agency estimates that 36 percent of adults from Country X were first married between the ages of 18 and 32, and 26 percent of adults from Country Y were first married between the ages of 18 and 32. Based on the estimates, which of the following is closest to the probability that the difference in proportions between a random sample of 60 adults from Country X and a random sample of 50 adults from Country Y (Country X minus Country Y) who were first married between the ages of 18 and 32 is greater than 0.15 ?

0.1398

0.2843

0.4315

0.5685

0.7157

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Covered syllabus points

3.9 Sampling Distributions for the Difference Between Two Population Proportions question 1

3.9 Sampling Distributions for the Difference Between Two Population Proportions question 2

3.9 Sampling Distributions for the Difference Between Two Population Proportions question 3