In: Statistics and Probability
Rhinoviruses typically cause common colds. In a test of the effectiveness of echinacea, 40 of the 45 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 88 of the 103 subjects developed rhinovirus infections. We want to 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.
Group of answer choices
A.P value=0.572; fail to reject H0
B.P value=0.508; fail to reject H0
C.P value=0.605; fail to reject H0
D.P value=0.501; fail to reject H0
x1 = | 40 | x2 = | 88 |
p̂1=x1/n1 = | 0.8889 | p̂2=x2/n2 = | 0.8544 |
n1 = | 45 | n2 = | 103 |
estimated prop. diff =p̂1-p̂2 = | 0.0345 | ||
pooled prop p̂ =(x1+x2)/(n1+n2)= | 0.8649 | ||
std error Se=√(p̂1*(1-p̂1)*(1/n1+1/n2) = | 0.0611 | ||
test stat z=(p̂1-p̂2)/Se = | 0.565 | ||
P value = | 0.5721 |
option A is correct
A.P value=0.572; fail to reject H0