Question

A drug manufacturer markets a brand of pain reliever that it claims to be 96​% effective in relieving headache pain. A consumer group decides to select a sample of 21 people at random who suffer from headaches and give them the medicine. They will then count the number of people who indicate to have experienced relief from their pain. Assume that the effect of the drug on each person is independent of the effect on all other people receiving the drug.

​a) If p​ = 0.96​, as the manufacturer​ claims, what is the expected number of people in the sample who experience pain​ relief? (Answer is 20.2)

​b) If the number of people in the sample who experience pain relief is 19 or​ fewer, the consumer group will publish a report questioning the claim that the drug is 96​% effective. ​ Otherwise, no results will be published. If p​ = 0.96 as​ claimed, what is the probability that the sample of 21 people will have 19 or fewer who experience pain​ relief? ​

c) Suppose the​ manufacturer's claim is not​ true, and in fact p​ = 0.91. What is the probability​ that, based on the results in the​ sample, the consumer group will publish a report questioning the claim made by the​ manufacturer?

Answer 1

Here Pr(Effective in releiving pain) = 0.96

sample size = n = 21

a) If p​ = 0.96​, as the manufacturer​ claims, what is the expected number of people in the sample who experience pain​ relief? (Answer is 20.2)

Answer : Here expected number of people in the sample who experience pain​ relief = 21 * 0.96 = 20.2

​b) If the number of people in the sample who experience pain relief is 19 or​ fewer, the consumer group will publish a report questioning the claim that the drug is 96​% effective. ​ Otherwise, no results will be published. If p​ = 0.96 as​ claimed, what is the probability that the sample of 21 people will have 19 or fewer who experience pain​ relief? ​

Answer : probability that the sample of 21 people will have 19 or fewer who experience pain​ relief

Pr(x <= 19) = BINOMDIST(x < = 19; 21 ; 0.96) = BINOMDIST(19, 21, 0.96, True) = 0.2044

(c) Suppose the​ manufacturer's claim is not​ true, and in fact p​ = 0.91. What is the probability​ that, based on the results in the​ sample, the consumer group will publish a report questioning the claim made by the​ manufacturer?

Answer : Here true probability p = 0.91

The consumer group will publish a report questioning the claim made by the​ manufacturer if there are less than 19 or fewer.

Pr(x <= 19) = BINOMDIST(x < = 19 ; 21 ; 0.91) = BINOMDIST(19, 21, 0.91, true) = 0.5754