Home / AP® Exam / AP® Statistics / 6.11 Carrying Out a Test for the Difference of Two Population Proportions- MCQs

AP Statistics 6.11 Carrying Out a Test for the Difference of Two Population Proportions- MCQs - Exam Style Questions

AP Statistics -MCQs and Answers - All Topics

Question

A group of \(80\) people who had been diagnosed as prediabetic because of high blood glucose levels volunteered to participate in a study designed to investigate the use of cinnamon to reduce blood glucose to a normal level. Of the \(80\) people, \(40\) were randomly assigned to take a cinnamon tablet each day and the other \(40\) were assigned to take a placebo each day. The people did not know which tablet they were taking. Their blood glucose levels were measured at the end of one month. The results showed that \(14\) people in the cinnamon group and \(10\) people in the placebo group had normal blood glucose levels. For people similar to those in the study, do the data provide convincing statistical evidence that the proportion who would be classified as normal after one month of taking cinnamon is greater than the proportion who would be classified as normal after one month of not taking cinnamon?
(A) No conclusion can be made about the use of cinnamon because the people in the study were volunteers.
(B) There is convincing statistical evidence at the level of \(0.01\).
(C) There is convincing statistical evidence at the level of \(0.05\) but not at the level of \(0.01\).
(D) There is convincing statistical evidence at the level of \(0.10\) but not at the level of \(0.05\).
(E) There is not convincing statistical evidence at any reasonable significance level.
▶️ Answer/Explanation
Detailed solution

Let group \(1\) be cinnamon and group \(2\) be placebo. We test \(H_0\!:\, p_1-p_2=0\) vs. \(H_a\!:\, p_1-p_2>0\).
Sample counts: \(x_1=14,\; n_1=40 \Rightarrow \hat p_1=\dfrac{14}{40}=0.35;\quad x_2=10,\; n_2=40 \Rightarrow \hat p_2=\dfrac{10}{40}=0.25.\)
Pooled proportion for \(H_0\): \(\hat p_c=\dfrac{x_1+x_2}{n_1+n_2}=\dfrac{24}{80}=0.30.\)
Standard error: \(\mathrm{SE}=\sqrt{\hat p_c(1-\hat p_c)\!\left(\dfrac{1}{n_1}+\dfrac{1}{n_2}\right)}=\sqrt{0.30\cdot 0.70\!\left(\dfrac{1}{40}+\dfrac{1}{40}\right)}\approx 0.1025.\)
Test statistic: \(z=\dfrac{(\hat p_1-\hat p_2)-0}{\mathrm{SE}}=\dfrac{0.35-0.25}{0.1025}\approx 0.976.\)
One-sided \(p\)-value: \(p=\Pr(Z\ge 0.976)\approx 0.1645.\)
Since \(p\approx 0.1645\) is greater than \(0.10\), \(0.05\), and \(0.01\), we fail to reject \(H_0\). There is not convincing statistical evidence that the cinnamon group has a larger proportion with normal glucose after one month.
✅ Answer: (E)

Question

A pollster is interested in comparing the proportions of women and men in a particular town who are in favor of a ban on fireworks within town borders. The pollster plans to test the hypothesis that the proportion of women in favor of the ban is different from the proportion of men in favor of the ban. There are 4,673 women and 4,502 men who live in the town. From a simple random sample of 40 women in the town, the pollster finds that 38 favor the ban. From an independent simple random sample of 50 men in the town, the pollster finds that 27 favor the ban. Which of the following statements is true about this situation?
(A) Because the samples are from normal populations, a two-proportion z-test would be valid.
(B) Because the size of each sample is greater than 30, a two-proportion z-test would be valid.
(C) Because the number who favored the ban is greater than 10 in both groups, a two-proportion z-test would be valid.
(D) Because of the relative sizes of the populations and samples, a two-proportion z-test would be valid.
(E) A two-proportion z-test would not be valid for these data.
▶️ Answer/Explanation
Detailed solution

To determine if a two-proportion z-test is valid, we must check the necessary conditions. The key condition here is the Large Counts Condition, which ensures the sampling distribution is approximately Normal.

The Large Counts Condition requires that the number of successes and failures in each sample are all at least 10. The formulas are $n\hat{p} \ge 10$ and $n(1-\hat{p}) \ge 10$.

Let’s check the counts for each group:
–   Women’s Sample:
– Sample size $n_w = 40$. – Successes (favor ban): 38. (This is $\ge 10$) – Failures (do not favor ban): $40 – 38 = 2$. (This is not $\ge 10$)
–   Men’s Sample:
– Sample size $n_m = 50$. – Successes (favor ban): 27. (This is $\ge 10$) – Failures (do not favor ban): $50 – 27 = 23$. (This is $\ge 10$)

Because the number of “failures” in the women’s sample is only 2, the Large Counts Condition is not met. Therefore, a two-proportion z-test is not a valid procedure for these data.
✅ Answer: (E)

More AP Statistics MCQ Questions..
Scroll to Top