This Question: 1 pt 2 of 12 (0 complete. ? This Quiz: 12 pts possible Rhino viru

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This Question: 1 pt 2 of 12 (0 complete. ? This Quiz: 12 pts possible Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 44 of the 49 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 83 of the 97 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 aliternative hypotheses for the hypothesis test? ? A. Ho:p1-p2 H1 P1 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? The P-value is the significance level of ?:005, so ?| the null hypothesis. There | ? sufficient evidence to support the claim that echinacea treatment has an effect. b. Test the claim by constructing an appropriate confidence interval The 95% confidence interval is ?

Explanation / Answer

let f1 denotes the number of subjects treated with echinacea developed rhinovirus infection out of n1=49 subjects and f2 denotes the number of subjects of placebo group developed rhinovirus infection out of n2=97 subjects.

let p1 be the proportion of subjects from the first sample developed rhinovirus infection and p2 be the same for second sample

then f1~Bin(n1,p1) and f2~Bin(n2,p2) independently.

claim is echinacea has an effect on rhinovirus infection.

significance level is alpha=0.05

a) then the hyptheses are

null hypothesis H0: p1=p2 and alternative hypothesis H1:p1<p2 [option E]

now E[f1/n1]=p1 and E[f2/n2]=p2 V[f1/n1]=p1*(1-p1)/n1 V[f2/n2]=p2*(1-p2)/n2

since n1 and n2 are greater than 30, hence the distribution of f1/n1 and f2/n2 can be considered as normal distribution.

f1/n1~N(p1,p1*(1-p1)/n1) and f2/n2~N(p2,p2*(1-p2)/n2) independently.

then f1/n1-f2/n2~N(p1-p2,p1*(1-p1)/n1+p2*(1-p2)/n2)

then under H0, f1/n1-f2/n2~N(0,p*(1-p)*[1/n1+1/n2]) [assuming p1=p2=p]

but p is unknwon, so it is estimated by phat=(f1+f2)/(n1+n2)

so the test statistic is z=(f1/n1-f2/n2)/sqrt[phat*(1-phat)*(1/n1+1/n2)]

now n1=49 n2=97 f1=44 f2=83 phat=(44+83)/(49+97)=0.86986

so z=(44/49-83/97)/sqrt[0.86986*(1-0.86986)*(1/49+1/97)]=0.72 [answer]

since the alternative hypothesis is less than type, it is a left tailed test

hence the p value is p=P[Z<0.72]=0.764 where Z~N(0,1)

so p value >0.05=level of significane

so conclusion is

the p value is greater than the significance level alpha=0.05, so failed to reject null hypothesis. there is no sufficient evidence to support the claim that echinacea teatment has an effect.

b) we have f1/n1-f2/n2~N(p1-p2,p1*(1-p1)/n1+p2*(1-p2)/n2)

so P[|(f1/n1-f2/n2)-(p1-p2)|/sqrt[p1*(1-p1)/n1+p2*(1-p2)/n2]<taoalpha/2]=1-alpha

or, P[f1/n1-f2/n2-taoalpha/2*sqrt[p1*(1-p1)/n1+p2*(1-p2)/n2]<p1-p2<f1/n1-f2/n2+taoalpha/2*sqrt[p1*(1-p1)/n1+p2*(1-p2)/n2]]=1-alpha

but p1 and p2 are unknown. they are replaced by p1hat=f1/n1 and p2hat=f2/n2

so 100*(1-alpha)% confidence interval of p1-p2 is

[f1/n1-f2/n2-taoalpha/2*sqrt[p1hat*(1-p1hat)/n1+p2hat*(1-p2hat)/n2,f1/n1-f2/n2+taoalpha/2*sqrt[p1hat*(1-p1hat)/n1+p2hat*(1-p2hat)/n2]

so for 95% confidence interval alpha=0.05 so tao0.05/2=1.97

so the required confidence interval is

[44/49-83/97-1.97*sqrt[(44/49)*(1-44/49)/49+(83/97)*(1-83/97)/97,44/49-83/97+1.97*sqrt[(44/49)*(1-44/49)/49+(83/97)*(1-83/97)/97]

=[-0.068,0.098] [answer]

conclusion is

because the confidence interval limits contain 0, there is appear to be a significant difference between the two proportions. there is evidence to support the claim that echinacea treatment has an effect.

c. the results are inconclusive since the two results show two different conclusions

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