Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 37 of the 44 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 91 of the 104 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?
(b) Identify the test statistic.
(c) Identify the P-value.

This is a test of 2 population proportions. Let 1 and 2 be the subscript for the subjects treated with echinacea and subjects treated with a placebo. The population proportion of subjects treated with echinacea and subjects treated with a placebo would be p1 and p2 respectively.

P1 - P2 = difference in the proportion of sample of subjects treated with echinacea the sample of subjects treated with a placebo.

The null hypothesis is

H0 : p1 = p2

p1 - p2 = 0

The alternative hypothesis is

Ha : p1 ≠ p2

p1 - p2 ≠ 0

it is a two tailed test

Sample proportion = x/n

Where

x represents number of success(number of complaints)

n represents number of samples

For women

x1 = 37

n1 = 44

P1 = 37/44 = 0.84

For men,

x2 = 91

n2 = 104

P2 = 91/104 = 0.88

The pooled proportion, pc is

pc = (x1 + x2)/(n1 + n2)

pc = (37 + 91)/(44 + 104) = 0.86

1 - pc = 1 - 0.86 = 0.14

z = (P1 - P2)/√pc(1 - pc)(1/n1 + 1/n2)

z = (0.84 - 0.88)/√(0.86)(0.14)(1/44 + 1/104) = - 0.04/0.045

z = - 0.64

Since it is a two tailed test, the curve is symmetrical. We will look at the area in both tails. Since it is showing in one tail only, we would double the area

From the normal distribution table, the area below the test z score in the left tail 0.26

We would double this area to include the area in the right tail of z = 0.64 Thus

p = 0.26 × 2 = 0.52

By using the p value,

Since 0.01 < 0.52, we would accept the null hypothesis.