Question

It has been observed that some persons who suffer colitis, again suffer colitis within one year of the first episode. This is due, in part, to damage from the first episode. The performance of a new drug designed to prevent a second episode is to be tested for its effectiveness in preventing a second episode. In order to do this two groups of people suffering a first episode are selected. There are 114 people in the first group and this group will be administered the new drug. There are 165 people in the second group and this group wil be administered a placebo. After one year, 19% of the first group has a second episode and 21% of the second group has a second episode. Select a 99% confidence interval for the difference in true proportion of the two groups.

a) [−0.170, 0.130]

b) [−0.645, 0.605]

c) [−0.130, 0.170]

d) [−0.105, 0.022]

e) [−0.145, 0.105]

f) None of the above

Answer 1

Formula for confidence interval for the difference in true proportion of the two groups

Number people in the first group : sample size of first group : n₁ = 114

Number of people in the second group : sample size of second group : n₂ = 165

Proportion of the first group who has a second episode : = 19/100 =0.19

Proportion of the second group who has a second episode : = 21/100 =0.21

for 99% confidence level = (100-99)/100 =0.01

/2 =0.01/2 =0.005

Z_/2 = Z_0.005 = 2.5758

99%  confidence interval for the difference in true proportion of the two groups

99%  confidence interval for the difference in true proportion of the two groups = (-0.145,0.105)

Ans :

e) [−0.145, 0.105]