AP Statistics 5.6 Sampling Distributions for Differences in Sample Proportions - MCQs - Exam Style Questions
▶️ Answer/Explanation
Population proportion \(p=0.36\), sample size \(n=40\).
Normality check for the sampling distribution of \(\hat{p}\): \(np=40(0.36)=14.4\) and \(n(1-p)=40(0.64)=25.6\), both \(\ge 10\). ⇒ The distribution of \(\hat{p}\) is approximately normal.
Mean: \(\mu_{\hat{p}}=p=0.36\).
Standard deviation: \(\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{(0.36)(0.64)}{40}}=\sqrt{0.00576}\approx 0.076\).
✅ Answer: (C)
(A) Skewed is incorrect; conditions support normality.
(B) & (E) use mean \(0.64\) (should be \(p=0.36\)).
(D) Uses an incorrect standard deviation (\(0.006\) instead of \(\approx 0.076\)).
▶️ Answer/Explanation
1. Define Parameters and Goal:
– Country X: $p_X = 0.36$, $n_X = 60$
– Country Y: $p_Y = 0.26$, $n_Y = 50$
We want to find the probability $P(\hat{p}_X – \hat{p}_Y > 0.15)$.
2. Find the Mean of the Sampling Distribution:
The mean of the difference in sample proportions is the difference in population proportions.
$\mu_{\hat{p}_X – \hat{p}_Y} = p_X – p_Y = 0.36 – 0.26 = 0.10$
3. Find the Standard Deviation of the Sampling Distribution:
$\sigma_{\hat{p}_X – \hat{p}_Y} = \sqrt{\frac{p_X(1-p_X)}{n_X} + \frac{p_Y(1-p_Y)}{n_Y}}$
$\sigma = \sqrt{\frac{0.36(0.64)}{60} + \frac{0.26(0.74)}{50}}$
$\sigma = \sqrt{0.00384 + 0.003848} = \sqrt{0.007688} \approx 0.0877$
4. Calculate the Z-score:
We standardize the value 0.15.
$z = \frac{\text{value} – \text{mean}}{\text{std. dev.}} = \frac{0.15 – 0.10}{0.0877} = \frac{0.05}{0.0877} \approx 0.57$
5. Find the Probability:
We need to find $P(Z > 0.57)$.
Using a z-table, $P(Z < 0.57) \approx 0.7157$.
$P(Z > 0.57) = 1 – P(Z < 0.57) = 1 – 0.7157 = 0.2843$.
✅ Answer: (B)