Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 35 of the 40 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 96 of the 113 subjects developed rhinovirus infections. Use a 0.01 significance level to test the claim that echinacea has an effect on rhinovirus infections. Complete parts (a) through (c) below.
a. Test the claim using a hypothesis test.
Consider the first sample to be the sample of subjects treated with echinacea and the second sample to be the sample of subjects treated with a placebo. What are the null and alternative hypotheses for the hypothesis test?
A.
\[
\begin{array}{l}
H_{0}: p_{1} \geq p_{2} \\
H_{1}: p_{1} \neq p_{2}
\end{array}
\]
D.
\[
\begin{array}{l}
H_{0}: p_{1}=p_{2} \\
H_{1}: p_{1}> p_{2}
\end{array}
\]
B.
\[
\begin{array}{l}
H_{0}: p_{1}=p_{2} \\
H_{1}: p_{1}< p_{2}
\end{array}
\]
E.
\[
\begin{array}{l}
H_{0}: p_{1} \leq p_{2} \\
H_{1}: p_{1} \neq p_{2}
\end{array}
\]
C.
\[
\text { C. } \begin{aligned}
H_{0}: p_{1} & \neq p_{2} \\
H_{1} & : p_{1}=p_{2} \\
\text { F. } H_{0}: p_{1}=p_{2} & H_{1}: p_{1} \neq p_{2}
\end{aligned}
\]
Identify the test statistic.
\[
z=0.39
\]
(Round to two decimal places as needed.)
Identify the P-value.
\[
\text { P-value }=\square
\]
(Round to three decimal places as needed.)
Answer
Calculate the P-value. The P-value is the probability of observing a test statistic as extreme as the one we calculated, assuming the null hypothesis is true. If the P-value is less than our significance level, we reject the null hypothesis. The P-value is \(\boxed{0.693}\).
Step 1 ：Identify the null and alternative hypotheses. The null hypothesis is that the proportion of infections in the echinacea group is equal to the proportion of infections in the placebo group, and the alternative hypothesis is that the proportions are not equal. The null and alternative hypotheses for the hypothesis test are: \[\begin{array}{l} H_{0}: p_{1}=p_{2} \\ H_{1}: p_{1} \neq p_{2} \end{array}\]
Step 2 ：Calculate the test statistic. The test statistic is a measure of how far our sample statistic is from the null hypothesis. In this case, the test statistic is given as z=0.39. The test statistic is \(\boxed{0.39}\).
Step 3 ：Calculate the P-value. The P-value is the probability of observing a test statistic as extreme as the one we calculated, assuming the null hypothesis is true. If the P-value is less than our significance level, we reject the null hypothesis. The P-value is \(\boxed{0.693}\).
