In randomized, double-blind clinical trials of a new vaccine, infants were rando

Question

In randomized, double-blind clinical trials of a new vaccine, infants were randomly divided into two groups. Subjects in group 1 received the new vaccine while subjects in group 2 received a control vaccine. After the first dose, 111 of 704 subjects in the experimental group (group 1) experienced vomiting as a side effect. After the first dose, 54 of 639 of the subjects in the control group (group 2) experienced vomiting as a side effect. Construct a 99 % confidence interval for the difference between the two population proportions, p 1 minus p 2 . Use x 1 equals 111, n 1 equals 704, x 2 equals 54, and n 2 equals 639.

Explanation / Answer

let p1hat be the sample proportion of group1 and p2hat be the sample proportion of group2.

the actual proportions be p1 and p2

x1 be the number of subjects in group1 experienced vomiting as a side effect.

x2 be the number of subjects in group2 experienced vomiting as a side effect.

n1=size of group1 n2=size of group2

so x1~Bin(p1,n1)    and x2~Bin(p2,n2)

where p1 and p2 are the two population proportions.

now n1=704 and n2=639 are very large.

hence the distribution of p1hat and p2hat can be approximated by normal distributions

now p1hat=x1/n1   and p2hat=x2/n2

so E[p1hat]=E[x1]/n1=n1*p1/n1=p1   similarly E[p2hat]=p2

V[p1hat]=V[x1]/n12=n1*p1*(1-p1)/n12=p1(1-p1)/n1 similarly   V[p2hat]=p2(1-p2)/n2

so p1hat~N(p1,p1(1-p1)/n1) and p2hat~N(p2,p2(1-p2)/n2)

since the groups are randomly divided p1hat and p2hat are independent.

so p1hat-p2hat~N(p1-p2,p1(1-p1)/n1+p2(1-p2)/n2)

so T=p1hat-p2hat-(p1-p2)/sqrt[p1(1-p1)/n1+p2(1-p2)/n2]~N(0,1)

so P[|p1hat-p2hat-(p1-p2)/sqrt[p1(1-p1)/n1+p2(1-p2)/n2]|<taoalpha/2]=1-alpha

where alpha=level of significance and taoalpha/2 is the upper alpha/2 point of a N(0,1) distribution.

or,P[ -taoalpha/2<p1hat-p2hat-(p1-p2)/sqrt[p1(1-p1)/n1+p2(1-p2)/n2]<taoalpha/2]=1-alpha

or,P[p1hat-p2hat-sqrt[p1(1-p1)/n1+p2(1-p2)/n2]*taoalpha/2<p1-p2<p1hat-p2hat+sqrt[p1(1-p1)/n1+p2(1-p2)/n2]*taoalpha/2]=1-alpha

hence [p1hat-p2hat-sqrt[p1(1-p1)/n1+p2(1-p2)/n2]*taoalpha/2,p1hat-p2hat+sqrt[p1(1-p1)/n1+p2(1-p2)/n2]*taoalpha/2] is the 100(1-alpha)% confidence interval of p1-p2

now since p1 and p2 are unknown they are replaced by p1hat and p2hat.

hence the 100(1-alpha)% confidence interval of p1-p2 is

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

we need to construct a 99% confidence interval.

we have alpha=0.01 x1=111 n1=704 x2=54 n2=639 so p1hat=111/704=0.1577 p2hat=54/639=0.0892

hence tao0.005=2.575829

hence the 99% confidence interval of p1-p2 is

[0.1577-0.0892-sqrt[0.1577(1-0.1577)/704+0.0892(1-0.0892)/639]*2.575829,0.1577-0.0892+sqrt[0.1577(1-0.1577)/704+0.0892(1-0.0892)/639]*2.575829]

=[0.02272,0.11427] [answer]

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