AP Statistics 7.9 Carrying Out a Test for the Difference of Two Population Means- MCQs - Exam Style Questions
▶️ Answer/Explanation
\[ t=\frac{\bar x_M-\bar x_W}{\sqrt{\frac{s_M^2}{n_M}+\frac{s_W^2}{n_W}}} =\frac{4.96-4.61}{\sqrt{\frac{1.01^2}{60}+\frac{0.89^2}{45}}}. \]
\(\frac{1.01^2}{60}=0.01700,\;\frac{0.89^2}{45}=0.01760\Rightarrow\) SE \(=\sqrt{0.03460}\approx0.1860\).
\(t=\dfrac{0.35}{0.1860}\approx1.88.\)
Welch degrees of freedom (approx.):
\[ \nu\approx \frac{\left(\frac{s_M^2}{n_M}+\frac{s_W^2}{n_W}\right)^2} {\frac{\left(\frac{s_M^2}{n_M}\right)^2}{n_M-1}+\frac{\left(\frac{s_W^2}{n_W}\right)^2}{n_W-1}} \approx 100. \]
Two-sided \(p\)-value for \(t\approx1.88\) with \(\nu\approx100\) is \(p\approx0.06\text{–}0.07\).
Since \(p<0.10\), we reject \(H_0\) at the 10% level.
✅ Answer: (A) The \(p\)-value is less than \(0.10\), and \(H_0\) should be rejected.
▶️ Answer/Explanation
1. Define Hypotheses:
– \(H_0: \mu_V = \mu_{NV}\) (Mean cholesterol levels are the same).
– \(H_a: \mu_V < \mu_{NV}\) (Vegetarians have a lower mean cholesterol level).
2. Calculate the Test Statistic:
Use a two-sample t-test for the difference of means.
\(t = \frac{(\bar{x}_V – \bar{x}_{NV}) – 0}{\sqrt{\frac{s_V^2}{n_V} + \frac{s_{NV}^2}{n_{NV}}}} = \frac{198 – 207}{\sqrt{\frac{32^2}{62} + \frac{41^2}{231}}} \approx \frac{-9}{4.878} \approx -1.845\)
3. Find the P-value:
Using the conservative degrees of freedom, \(df = \min(61, 230) = 61\).
The p-value is the area to the left of the t-statistic for a one-tailed test: \(P(t_{df=61} < -1.845) \approx 0.035\).
4. Make a Conclusion:
The p-value (\(0.035\)) is less than \(0.05\) but greater than \(0.01\). Therefore, there is a statistically significant difference at the \(5\%\) level, but not at the \(1\%\) level.
✅ Answer: (A)