Question

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?

Answer 1

Answer:

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 $p_1$ to be the probability of those that experience the drowsiness in group 1

$p_1$ = $\dfrac{137}{452}$

$p_1$ = 0.3031

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

Let consider $p_2$ to be the probability of those that experience the drowsiness in group 1

$p_2$ = $\dfrac{31}{99}$

$p_2$ = 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:

$H_o :p_1 =p_2$

$H_a= p_1<p_2$

The test statistics is computed as follows:

$Z = \dfrac{p_1-p_2}{\sqrt{p_1 *\dfrac{1-p_1}{n_1} +p_2 *\dfrac{1-p_2}{n_2}} }$

$Z = \dfrac{0.3031-0.3131}{\sqrt{0.3031 *\dfrac{1-0.3031}{452} +0.3131 *\dfrac{1-0.3131}{99}} }$

$Z = \dfrac{-0.01}{\sqrt{0.3031 *\dfrac{0.6969}{452} +0.3131 *\dfrac{0.6869}{99}} }$

$Z = \dfrac{-0.01}{\sqrt{0.3031 *0.0015418 +0.3131 *0.0069384} }$

$Z = \dfrac{-0.01}{\sqrt{4.6731958*10^{-4}+0.00217241304} }$

$Z = \dfrac{-0.01}{0.051378 }$

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