Question

A doctor released the results of clinical trials for a vaccine to prevent a disease. In these clinical trials, 20,000 children were randomly divided in two groups. The subjects in group 1 (the experimental group) were given the vaccine, while the subjects in group 2 (the control group) were given a placebo. Of the 10,000 children in the experimental group, 230 developed the disease. Of the 10,000 children in the control group, 480 developed the disease. Estimate the difference in the proportion of children who got the disease when having the vaccine versus those who did not.

a. Construct a 99% confidence interval for the difference in the proportions.

b. Interpret the confidence interval, if appropriate.

c. What is the margin of error?

Answer 1

We have

P1 = 0.023
X1 = 230
n1 = 10000
P2 = 0.048
X2 = 480
n2 = 10000
Sample Diff = P1-P2. = -0.025
P0=(x1+x2)/(n1+n2) = 0.0355
Std. Err. = sqrt(P0*(1-P0)*(1/n1 +1/n2)) = 0.0026

=0.01

Z/2 = Z0.005 =2.5758

margin of error=Z/2 *Std. Err =2.5758*0.0026=0.0067

99% confidence interval for the difference in the proportions.=(P1-P2 +/- margin of error)

=(-0.025-0.0067,-0.025+0.0067)

=(-0.0317,-0.0183)

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a) 99% confidence interval for the difference in the proportions=(-0.0317,-0.0183)

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b) The confidence interval suggests that the difference in the proportions is  between -0.0317 and -0.0183 with 99% confidence. If we were to take repeated SRS's of size 10,000 children for each group   we would expect the resulting confidence interval to contain the true difference in the proportions about 99 times out of 100.

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c) Margin of error=Z/2 *Std. Err =2.5758*0.0026=0.0067

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