In randomized, double-blind clinical trials of Prevnar, infants were randomly divided into two groups. Subjects in group 1 received Prevnar, while subjects in group 2 received a control vaccine. Aft er the second dose, 137 of 452 subjects in the experimental group (group 1) experienced drowsiness as a side effect. After the second dose, 31 of 99 subjects in the control group (group 2) experienced drowsiness as a side effect. Does the evidence suggest that a lower proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the αα=0.05 level of significance?

Step-by-step explanation:

From the summary of the given data;

After the second dose, 137 of 452 subjects in the experimental group (group 1) experienced drowsiness as a side effect.

Let consider to be the probability of those that experience the drowsiness in group 1

=

= 0.3031

After the second dose, 31 of 99 subjects in the control group (group 2) experienced drowsiness as a side effect.

Let consider to be the probability of those that experience the drowsiness in group 1

=

= 0.3131

The objective is to be able to determine if the evidence suggest that a lower proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the α=0.05 level of significance.

In order to do that; we have to state the null and alternative hypothesis; carry out our test statistics and make conclusion based on it.

So; the null and the  alternative hypothesis can be computed as:

The test statistics is computed as follows:

Z = - 0.1946

At the level of significance ∝ = 0.05

From the standard normal table;

the critical value for Z(0.05) = -1.645

Decision Rule: Reject the null hypothesis if Z-value is lesser than the critical value.

Conclusion: We do not reject the null hypothesis because the Z value is greater than the critical value. Therefore, we cannot conclude that a lower proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2