Question

Quel He Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 39 of the 46 subjects treated with echinacea developed rhinovirus infections. In a placebo group. 73 of the 90 subjects developed rhinovirus infections. Use a 0.01 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? O A. Ho: P₂ = P2 H: P1 *P2 OD. Ho: P1 - P2 H:P:P2 OB. Ho: P. *P2 HP - P2 O E. Ho: Pi-P2 HP1 P2 OC. Ho: P2P2 HP *P2 OF. Ho: P: SP2 HP, P2 Identify the test statistic. (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? sufficient evidence to support the claim that The P-value is the null hypothesis. There the significance level of a = 0.01, so Click to select your answer(s).

Answer 1

a) The null and alternative hypothesis

H0:p1=p2

H1: p1 $\neq$ p2

Test statistic

2= pl - p2 VPQ(1/n1+1/n2)

where

n1= 46 , n2= 90

39/46= 0.8478

pl =

73/90= 0.8111

p2

P = (39+73)/(46+90) = 0.8235

Q=1-P= 0.1765

thus ,

$z= \frac{0.8478-0.8111}{\sqrt{0.8235*0.1765(1/46+1/90)}}=0.53$

Therefore , test statistic is

z=0.53

P value = 0.596

The P value is greater than the significance level of 0.01 , so do not reject the null hypothesis > There is not sufficient evidence to support the claim that echinacia treatment has an effect .

Note : P value = P( z < -0.53 ) +P( z > 0.53 ) = 0.2981+0.2981= 0.596 ( from z table)

b) The 99% confidence interval is

$(\hat{p1}-\hat{p2})\pm zc*\sqrt{PQ(1/n1+1/n2)}$

$=(0.8478-0.8111)\pm2.58*\sqrt{0.8235*0.1765(1/46+1/90)}$

= ( -0.1415 , 0.2150)

99% confidence interval is

-0.142 < (p1-p2) < 0.215

Because the confidence interval limits contain 0 , there does not appear to a significant difference between the two proportions . There is not evidence to support the claim that echinacia treatment has an effect .

Note : For 99% confidence , zc = 2.58 ( from z table)

c) Answer is

C) Echinacea does not appear to have a significant effect on infection rate .