Question

Rhino viruses typically cause common colds. In a test of the effectiveness of​ echinacea, 37 of the 42 subjects treated with echinacea developed rhinovirus infections. In a placebo​ group, 84 of the 101 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. ​(Round to two decimal places as​ needed.)

b. 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 greater than the significance level of ​, so fail to reject the null hypothesis. There is not sufficient evidence to support the claim that echinacea treatment has an effect.

c. Test the claim by constructing an appropriate confidence interval. The ​% confidence interval is negative ?_? ​(Round to three decimal places as​ needed.) What is the conclusion based on the confidence​ interval? Because the confidence interval limits include or does not include ​0, there does/does not appear to be a significant difference between the two proportions. There is evidence/not enough evidence to support the claim that echinacea treatment has an effect. Based on the​ results, does echinacea appear to have any effect on the infection​ rate?

Answer 1

Ginn that echinacea n = 42 x=37 Placebo 12=101 12=84 x=005 To test null & Alternative hypothesis dre Hoi P = P2 PLP HY pooled proportion is P= X1 + X2 nitn, 37+84 42tlol P=0.846153 A Test statistics is P P₂ z = PCI-P) (2 + 1 h 37/42 84/101 0.84615 0.15385 x tol 4 4 2 + + ñ 0.74 ✓ z = 0 + 7437 v © p-value= PLZ 20.7437] = 0.7715 v n 0.772 v

6 The p-value is greater than the greater than the significance level of 0.05 © 95% confidence interval (l) is Ett Zairt P - Pr + Zalz PCI-P) - [ + +1] IL 37 84. I 1.96 0.84615 X0:15385 Lt + I Jo] lol Mo 95% (I is = [ 0.0493 I 0.12984 (-0.0805 0.17914 ) not negative v Confidence interval include zero, there does not appear to be a significant difference He two proportions. There is not enough evidence to support the claim that echnacea treatment has an effect between there does not have effect on the infection rate,