Question

Rhino viruses typically cause common colds. In a test of the effectiveness of​ echinacea, 43 of the 49 subjects treated with echinacea developed rhinovirus infections. In a placebo​ group, 79 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.

Test the claim using a hypothesis test.

find test stat

find p value

find conclusion based on hypothesis test

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

Answer 1

Solution:

Here, we have to use two sample z test for the population proportions. The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: Echinacea has no effect on rhinovirus infections.

Alternative hypothesis: H_a: Echinacea has an effect on rhinovirus infections.

H₀: p₁ = p₂ versus H_a: p₁ ≠ p₂

This is a two tailed test.

We are given level of significance = α = 0.05

The test statistic formula is given as below:

Z = (P1 – P2) / sqrt(P*(1 – P)*((1/N1) + (1/N2)))

Where,

X1 = 43

X2 = 79

N1 = 49

N2 = 96

P = (X1+X2)/(N1+N2) = (43 + 79) / (49 + 96) = 0.8414

P1 = X1/N1 = 43/49 = 0.87755102

P2 = X2/N2 = 79/96 = 0.822916667

Z = (P1 – P2) / sqrt(P*(1 – P)*((1/N1) + (1/N2)))

Z = (0.87755102 – 0.822916667) / sqrt(0.8414*(1 – 0.8414)*((1/49) + (1/96)))

Z = 0.054634354 / sqrt(0.8414*(1 – 0.8414)*((1/49) + (1/96)))

Z = 0.8518

P-value = 0.3943

(by using z-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that Echinacea has an effect on rhinovirus infections.