Question

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, 76 of the 92 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 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?

Identify the test statistic.

Identify the​ P-value.

b. What is the conclusion based on the hypothesis​ test?

Test the claim by constructing an appropriate confidence interval.

What is the conclusion based on the confidence​ interval?

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

Answer 1

here, x1= 35, x2= 76 , N1 = 40 , N2 =92

p1cap= 35/40 = 0.875 and p2cap = 76/92 = 0.8261 and p = (X1 + X2 ) / (n1+n2) = 37/44= 0.8409

a)
Null Hypothesis, H0: p1 = p2
Alternate Hypothesis, Ha: p1 ≠ p2

b)
Test statistic
z = (p1cap - p2cap)/sqrt(p * (1-p) * (1/N1 + 1/N2))
z = (0.875 -0.8261)/sqrt(0.8409*(1-0.8409)*(1/40 + 1/92))
z = 0.70609 = 0.71

c)
P-value Approach
P-value = 1.5223
As P-value >= 0.05, fail to reject null hypothesis.
There is not sufficient evidence to conclude that echinacea has an effect on rhinovirus infections.