AP Statistics 3.9: Conditions for Difference in Proportions

Justify a normal sampling distribution for a difference in sample proportions by checking randomization, 10%, and expected counts.

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.B—Justify the appropriateness of conditions for the sampling distribution for the difference between two sample… question 1 — 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.

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3.9.B—Justify the appropriateness of conditions for the sampling distribution for the difference between two sample… question 1