Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 44 of the 49 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 81 of the 98 subjects developed rhinovirus infections. Use a 0.05 significance level to test the claim that echinacea has an effect on rhinovirus infections. Complete parts (a) through (c) below.
a. Test the claim asing 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. $H_{0}: p_{1} \geq p_{2}$
$\mathrm{H}_{1}: \mathrm{p}_{1} \neq \mathrm{p}_{2}$
D.
\[
\begin{array}{l}
H_{0}: p_{1}=p_{2} \\
H_{1}: p_{1} \neq p_{2}
\end{array}
\]
B.
\[
\begin{array}{l}
H_{0}: p_{1}=p_{2} \\
H_{1}: p_{1}< p_{2}
\end{array}
\]
E.
\[
\begin{array}{l}
H_{0}: p_{1} \neq p_{2} \\
H_{1}: p_{1}=p_{2}
\end{array}
\]
C.
\[
\begin{array}{l}
H_{0}: p_{1}=p_{2} \\
H_{1}: p_{1}> p_{2}
\end{array}
\]
F.
\[
\begin{array}{l}
H_{0}: p_{1} \leq p_{2} \\
H_{1}: p_{1} \neq p_{2}
\end{array}
\]
Identify the test statistic.
\[
\mathrm{z}=
\]
(Round to two decimal places as needed.)
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Answer
The final answer is: The null and alternative hypotheses for the hypothesis test are: \[\begin{array}{l} H_{0}: p_{1}=p_{2} \\ H_{1}: p_{1} \neq p_{2} \end{array}\] And the test statistic is approximately \(\boxed{1.24}\).
Step 1 ：Set up the null and alternative hypotheses. The null hypothesis is that the two proportions are equal, and the alternative hypothesis is that the two proportions are not equal. This corresponds to option D: \[\begin{array}{l} H_{0}: p_{1}=p_{2} \\ H_{1}: p_{1} \neq p_{2} \end{array}\]
Step 2 ：Calculate the test statistic. The test statistic for a hypothesis test for the difference between two proportions is a z-score, which is calculated using the formula: \[z = \frac{(p1 - p2)}{\sqrt{(p1(1 - p1) / n1) + (p2(1 - p2) / n2)}}\] where p1 and p2 are the sample proportions, and n1 and n2 are the sample sizes.
Step 3 ：Substitute the given values into the formula: n1 = 49, n2 = 98, x1 = 44, x2 = 81, p1 = 0.8979591836734694, p2 = 0.826530612244898.
Step 4 ：Calculate the z-score: \[z = 1.2372396188298749\]
Step 5 ：The test statistic z is approximately 1.24. This is the number of standard deviations that the sample proportion is away from the null hypothesis proportion.
Step 6 ：The final answer is: The null and alternative hypotheses for the hypothesis test are: \[\begin{array}{l} H_{0}: p_{1}=p_{2} \\ H_{1}: p_{1} \neq p_{2} \end{array}\] And the test statistic is approximately \(\boxed{1.24}\).
