Question

Rhinoviruses 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, 83 of the 96 subjects developed rhinovirus infections. Use a 0.05 significance level to test the claim that echinacea has an effect on rhinovirus infections.

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?

H0: p1 __?__ p2

H1::p1___?__p2

B. Identify the test statistic. z = ________ ​(Round to two decimal places as​ needed.)

C. Identify the P-value. P-value = _________ ​(Round to two decimal places as​ needed.)

D. Based on the results, does echinacea appear to have any effect on the infection rate?

1. Echinacea does appear to have a significant effect on the infection rate. There is evidence that it increases the infection rate.

2. Echinacea does not appear to have a significant effect on the infection rate.

3. Echinacea does appear to have a significant effect on the infection rate. There is evidence that it lowers the infection rate.

4. The results are inconclusive.

PLEASE INCLUDE ALL WORK, I WOULD LIKE TO SEE EACH STEP AND UNDERSTAND AND LEARN HOW TO WORK THIS QUESTION. THANK YOU!

Answer 1

n1=40

n2=96

x1=35

x2=83

level of significance = 0.05

A)

null and alternate hypotheses.

H0: p1 = p2

H1: p1 > p2

B)

Test statistic

Formula

C)

P-value = 0.4351 ( using z-table )

P-value = 0.44

P-value , Failed to Reject H0

D)

2. Echinacea does not appear to have a significant effect on the infection rate.