Part 5 of 6 Polnts: 0 of 1 In randomized, double-blind clinical trials of a new vaccine, rats were randomly divided into two groups. Subjects in group 1 received the new vaccine while subjects in group 2 received a control vaccine. After the second dose, 108 of 713 subjects in the experimental group (group 1) experienced drowsiness as a side effect. After the second dose, 65 of 594 of the subjects in the control group (group 2) experienced drowsiness as a side effect. Does the evidence suggest that a higher proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the $\alpha=0.10$ level of significance? Interpret the P-value. If the poputation proportions are equal, one would expect a sample difference proportion greater than the one observed in about 13 out of 1000 repetitions of this experiment. (Round to the nearest integer as needed.) State the conclusion for this hypothesis test. A. Reject $\mathrm{H}_{0}$. There is not sufficient evidence to conclude that a highe group 2 at the $\alpha=0.10$ level of significance. B. Reject $\mathrm{H}_{0}$. There is sufficient evidence to conclude that a higher proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the $\alpha=0.10$ level of significance. C. Do not reject $\mathrm{H}_{0}$. There is sufficient evidence to conclude that a higher proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the $\alpha=0.10$ level of significance. D. Do not reject $\mathrm{H}_{0}$. There is not sufficient evidence to conclude that a higher proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the $\alpha=0.10$ level of significance. View an example Get more help - Clear all Final check
Solution
Step 1 ：Define the null hypothesis (H0) as the proportions of subjects experiencing drowsiness in group 1 and group 2 are equal, and the alternative hypothesis (H1) as the proportion in group 1 is higher than in group 2.
Step 2 ：Calculate the sample proportions for group 1 and group 2. For group 1, \(p1 = \frac{108}{713} = 0.1514726507713885\). For group 2, \(p2 = \frac{65}{594} = 0.10942760942760943\).
Step 3 ：Calculate the pooled proportion, which is the proportion of subjects experiencing drowsiness in both groups combined. \(pooled\_p = \frac{108+65}{713+594} = 0.13236419280795717\).
Step 4 ：Calculate the standard error (se) using the formula \(\sqrt{pooled\_p*(1-pooled\_p)*[\frac{1}{713}+\frac{1}{594}]}\), which gives \(se = 0.018825815478792758\).
Step 5 ：Calculate the z-score using the formula \(\frac{p1-p2}{se}\), which gives \(z = 2.2333715844151727\).
Step 6 ：Calculate the p-value, which is the probability of observing a z-score as extreme as the one calculated, assuming the null hypothesis is true. The p-value is \(0.012762225219572487\).
Step 7 ：Compare the p-value to the significance level (α=0.10). Since the p-value is less than the significance level (0.0127 < 0.10), we reject the null hypothesis.
Step 8 ：Conclude that there is sufficient evidence to suggest that a higher proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the α=0.10 level of significance. \(\boxed{\text{B. Reject } H_{0}. \text{There is sufficient evidence to conclude that a higher proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the } \alpha=0.10 \text{ level of significance.}}\)
