AP Statistics 6.6 Concluding a Test for a Population Proportion- MCQs - Exam Style Questions
▶️ Answer/Explanation
The reported result means a left-tailed test with
\[ H_0: p = 0.40 \quad \text{vs} \quad H_a: p < 0.40 \] at level \(\alpha=0.05\) rejected \(H_0\). Therefore the p-value satisfies \(p\text{-value} \le 0.05\).
If we use the same data with a left-tailed test at \(\alpha=0.10\), the rejection rule is easier (\(0.10 > 0.05\)). Since \(p\text{-value} \le 0.05 < 0.10\), that test will also reject \(H_0\). Thus, it will again conclude \(p<0.40\).
• Option (B) must be true. ✅
• (A) need not be true because a two-sided test uses \(p\text{-value}_{2\text{-sided}} \approx 2\,p\text{-value}_{\text{one-sided}}\), which may exceed \(0.05\).
• (C) need not be true: \(\alpha=0.01\) is more stringent; we only know \(p\text{-value} \le 0.05\), not that it is \(\le 0.01\).
• (D) and (E) are false: rejecting the left-tailed test at \(\alpha=0.05\) implies a two-sided \(95\%\) CI for \(p\) will lie entirely below \(0.40\); a \(90\%\) CI will also not include \(0.40\).
✅ Answer: (B)
\[ H_0: p = 0.40 \quad \text{vs} \quad H_a: p < 0.40 \] at level \(\alpha=0.05\) rejected \(H_0\). Therefore the p-value satisfies \(p\text{-value} \le 0.05\).
If we use the same data with a left-tailed test at \(\alpha=0.10\), the rejection rule is easier (\(0.10 > 0.05\)). Since \(p\text{-value} \le 0.05 < 0.10\), that test will also reject \(H_0\). Thus, it will again conclude \(p<0.40\).
• Option (B) must be true. ✅
• (A) need not be true because a two-sided test uses \(p\text{-value}_{2\text{-sided}} \approx 2\,p\text{-value}_{\text{one-sided}}\), which may exceed \(0.05\).
• (C) need not be true: \(\alpha=0.01\) is more stringent; we only know \(p\text{-value} \le 0.05\), not that it is \(\le 0.01\).
• (D) and (E) are false: rejecting the left-tailed test at \(\alpha=0.05\) implies a two-sided \(95\%\) CI for \(p\) will lie entirely below \(0.40\); a \(90\%\) CI will also not include \(0.40\).
✅ Answer: (B)
▶️ Answer/Explanation
Detailed solution
Hypotheses: \(H_0: p=0.41\) vs. \(H_a: p\neq 0.41\).
Sample proportion: \(\hat{p}=\dfrac{245}{600}\approx 0.4083\).
Test statistic: \[ z=\frac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}} =\frac{0.4083-0.41}{\sqrt{\dfrac{0.41(0.59)}{600}}} \approx -0.083. \] Two-sided \(p\)-value: \(p=2\big(1-\Phi(|z|)\big)\approx 2(1-\Phi(0.083))\approx 0.934\).
Since \(p\approx 0.934\) is greater than any reasonable significance level (e.g., \(0.10, 0.05, 0.01\)), we fail to reject \(H_0\).
There is not convincing evidence that the proportion differs from \(0.41\).
✅ Answer: (B)