Question

Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 44 of the 51 subjects treated with echinacea developed thinovirus infections. In a placebo group, 87 of the 104 subjects developed thinovirus infections. Use a 0.05 significance level to test the claim that echinacea has an effect on thinovirus 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? OA HOPP2 Hp : P1 P2 OD. Ho: P2P2 HP, Py OB. Hy: 2 HP: O E HO PSP HD, OC. He:D Hy: P, >P2 OF HP, P2 НР Ра Identify the test statistic zu (Round to two decimal places as needed) Identify the P-value P-value- (Round to three decimal places as needed) What is the conclusion based on the hypothesis test? The P value is the significance level of has an effect 0.05, the nut hypothesis. There sufficient evidence to support the claim that echinacea treatment

Answer 1

For Sample 1:

n1 = 51

x1 = 44

p̂1 = x1/n1 = 0.8627

For Sample 2:

n2 = 104

x2 = 87

p̂2 = x2/n2 = 0.8365

α = 0.05

a)

Null and Alternative hypothesis:

Ho : p1 = p2

H1 : p1 ≠ p2

Pooled proportion:

p̄ = (x1+x2)/(n1+n2) = (44+87)/(51+104) = 0.8452

Test statistic:

z = (p̂1 - p̂2)/√ [p̄*(1-p̄)*(1/n1+1/n2)] = (0.8627 - 0.8365)/√[0.8452*0.1548*(1/51+1/104)] = 0.42

p-value = 2*(1-NORM.S.DIST(ABS(0.4238), 1)) = 0.672

Conclusion:

The p-value > α = 0.05,so do not reject the null hypothesis. There is not sufficient evidence to support the claim.

b)

95% Confidence interval for the difference:

At α = 0.05, two tailed critical value, z_c = NORM.S.INV(0.05/2) = 1.960

Lower Bound = (p̂1 - p̂2) - z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]

= (0.8627 - 0.8365) - 1.96*√[(0.8627*0.1373/51) + (0.8365*0.1635/104)] = -0.092

Upper Bound = (p̂1 - p̂2) + z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]

= (0.8627 - 0.8365) + 1.96*√[(0.8627*0.1373/51) + (0.8365*0.1635/104)] = 0.144

-0.092 < p1 -p2 < 0.144

because the confidence interval contain 0. There do not appear to be a significant difference between the two proportions.

There is not sufficient evidence to support the claim.

c) Answer C.