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Keith_Richards [23]
3 years ago
6

It has been observed that some persons who suffer acute heartburn, again suffer acute heartburn within one year of the first epi

sode. 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 163 people in the first group and this group will be administered the new drug. there are 160 people in the second group and this group will be administered a placebo. after one year, 13% of the first group has a second episode and 14% of the second group has a second episode. select a 90% confidence interval for the difference in true proportion of the two groups.
Business
1 answer:
topjm [15]3 years ago
8 0

Confidence interval at 90% level of confidence :


CI = -0.052751325 ≤ p ≤ 0.072751325

CI = 0.01 ± 0.062751325

We arrive at the Confidence Interval with the following formula:

(p₁ - p₂ ) ± Z × √{[p₁ (1-p₁)] /n₁} + {[p₂ (1-p₂)] /n₂}

Substituting, we get,

(0.13-0.14) ± 1.65 × √[(0.13×0.87)/163] + [(0.14×0.86)/160]

0.01 ± ( 1.65 ×√0.001446365)

0.01 ± (1.65 × 0.038031106)

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