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. \[ \begin{array}{l} H_{0}: p_{1} \geq p_{2} \\ H_{1}: p_{1} \neq p_{2} \end{array} \] B. $H_{0}: p_{1}=p_{2}$ C. \[ \begin{array}{l} H_{0}: p_{1}=p_{2} \\ H_{1}: p_{1}>p_{2} \end{array} \] 10. \[ \begin{array}{l} H_{0}: p_{1}=p_{2} \\ H_{1}: p_{1} \neq p_{2} \end{array} \] E. $H_{0}: p_{1} \neq p_{2}$ F. $H_{0}: p_{1} \leq p_{2}$ $H_{1}: p_{1}=p_{2}$ \[ H_{1}: p_{1} \neq p_{2} \] Identify the test statistic. \[ z=1.14 \] (Round to two decimal places as needed.) Identify the P-value. P-value $=$ (Round to three decimal places as needed.) View an example Get more help - Clear all check answer

Step 1 ：The null hypothesis (H0) is that the proportion of subjects who develop rhinovirus infections is the same for both the echinacea and placebo groups (p1 = p2). The alternative hypothesis (H1) is that the proportions are not the same (p1 ≠ p2).

Step 2 ：We are given the number of subjects who developed infections in each group and the total number of subjects in each group. We can use this information to calculate the sample proportions.

Step 3 ：We can then use these sample proportions to calculate the test statistic (z) and the P-value. The test statistic is a measure of how far our sample statistic is from the hypothesized population parameter, in terms of standard errors. The P-value is the probability of obtaining a test statistic as extreme as the one we calculated, assuming the null hypothesis is true.

Step 4 ：If the P-value is less than the significance level (0.05), we reject the null hypothesis and conclude that echinacea has an effect on rhinovirus infections. If the P-value is greater than the significance level, we do not reject the null hypothesis and conclude that echinacea does not have an effect on rhinovirus infections.

Step 5 ：The test statistic (z) is approximately 1.14 and the P-value is approximately 0.252. Since the P-value is greater than the significance level (0.05), we do not reject the null hypothesis. This means that we do not have enough evidence to conclude that echinacea has an effect on rhinovirus infections.

Step 6 ：Final Answer: The null hypothesis is \(H_{0}: p_{1}=p_{2}\) and the alternative hypothesis is \(H_{1}: p_{1} \neq p_{2}\). The test statistic is approximately \(\boxed{1.14}\) and the P-value is approximately \(\boxed{0.252}\). Since the P-value is greater than the significance level (0.05), we do not reject the null hypothesis. This means that we do not have enough evidence to conclude that echinacea has an effect on rhinovirus infections.