Question

in randomized, double - blind clinical trials of a new vaccine, infants 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, 111 of 714 subjects in the experimental group (group 1) experienced fever as a side effect. after the second dose, 68 of 595 of the subjects in the control group (group 2) experienced fever as a side effect. does the evidence suggest that a higher proportion of subjects in group 1 experienced fever as a side effect than subjects in group 2 at the \\( \alpha=0.05 \\) level of significance?
c. the sample size is less than \\( 5 \\% \\) of the population size for each sample.
d. the sample size is more than \\( 5 \\% \\) of the population size for each sample.
e. the samples are dependent.
f. the samples are independent.
determine the null and alternative hypotheses.
\\( h_{0}: p_{1}=p_{2} \\)
\\( h_{1}: p_{1}>p_{2} \\)
find the test statistic for this hypothesis test.
2.16 (round to two decimal places as needed.)
determine the p - value for this hypothesis test.
(round to three decimal places as needed.)

Explanation:

Step1: Calculate the pooled proportion

The pooled proportion $\hat{p}=\frac{x_1 + x_2}{n_1 + n_2}$, where $x_1 = 111$, $n_1=714$, $x_2 = 68$, $n_2 = 595$.
$\hat{p}=\frac{111+68}{714 + 595}=\frac{179}{1309}\approx0.137$.

Step2: Calculate the standard error

The standard error $SE=\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1}+\frac{1}{n_2})}$.
Substitute $\hat{p}=0.137$, $n_1 = 714$, $n_2=595$ into the formula:
$SE=\sqrt{0.137\times(1 - 0.137)\times(\frac{1}{714}+\frac{1}{595})}\approx\sqrt{0.137\times0.863\times(\frac{595 + 714}{714\times595})}\approx\sqrt{0.118\times\frac{1309}{424830}}\approx\sqrt{\frac{154.462}{424830}}\approx0.019$.

Step3: Calculate the P - value

Since the test statistic $z = 2.16$ (given) and the alternative hypothesis is $H_1:p_1>p_2$ (right - tailed test).
The P - value is $P(Z>2.16)$.
Using the standard normal distribution table or a calculator, $P(Z>2.16)=1 - P(Z\leq2.16)$.
From the standard normal table, $P(Z\leq2.16)=0.9846$.
So $P(Z>2.16)=1 - 0.9846 = 0.0154\approx0.015$.

Answer:

$0.015$