A pharmaceutical company is testing the effectiveness of a new drug for lowering cholesterol. As part of this trial, they wish to determine whether there is a difference in effectiveness between women and men.
$\begin{array}{|l|c|c|}
\hline & \text { Women } & \text { Men } \
\hline \text { Sample size } & 60 & 60 \
\hline \text { Mean effect } & 7.5 & 6.65 \
\hline \text { Sample variance } & 2.5 & 4 \
\hline
\end{array}$

A) 7.85
B) 0.33
C) 0.39
D) 2.58

Question

A pharmaceutical company is testing the effectiveness of a new drug for lowering cholesterol. As part of this trial, they wish to determine whether there is a difference in effectiveness between women and men.
$\begin{array}{|l|c|c|}
\hline & \text { Women } & \text { Men } \
\hline \text { Sample size } & 60 & 60 \
\hline \text { Mean effect } & 7.5 & 6.65 \
\hline \text { Sample variance } & 2.5 & 4 \
\hline
\end{array}$

A) 7.85
B) 0.33
C) 0.39
D) 2.58

Quizplus · Accepted Answer

We need to conduct a two-sample t-test to compare the means of the two groups. The formula for the t-statistic is:
$$t=\frac{\bar{X}_1-\bar{X}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$$
where $\bar{X}_1$ and $\bar{X}_2$ are the sample means, $s_1$ and $s_2$ are the sample standard deviations, and $n_1$ and $n_2$ are the sample sizes.

Plugging in the given values, we get:
$$t=\frac{7.5-6.65}{\sqrt{\frac{2.5}{60}+\frac{4}{60}}} \approx 2.58$$
Using a two-tailed t-test with 118 degrees of freedom (total sample size minus 2), at a significance level of 0.05, the critical t-value is approximately 1.98. Since our calculated t-value (2.58) is greater than the critical t-value (1.98), we reject the null hypothesis that there is no difference in effectiveness between women and men.

Therefore, the best choice is D, 2.58.

A Pharmaceutical Company Is Testing the Effectiveness of a New