A drug manufa cturer claims that fewer than 20 % of patients who take its new dr

Question

A drug manufa cturer claims that fewer than 20 % of patients who take its new drug for treating Alzheimer’s disease will experience nausea. To test this claim, a significance tes t is carried out on a random sample of 100 patients where H 0 : p = 0 . 2 0 a n d H A : p < 0 . 2 0 . a) Find ! z * that corres ponds to the alpha level of 0.05 . Wh at is the proportion of patients ( ˆ p ) from our sample that would give us this z - score? b) Find if p = 0 . 1 5 ( s o t h a t H 0 i s f a l s e ) . c) What is the corresponding power of the test

Explanation / Answer

a) z* at alpha=0.05 is -1.645 (left tailed test)

b) z=-1.645, Pu=0.20,N=100. Substitute the values in following equation to obtain Ps.

z=(Ps-Pu)/sqrt(Pu(1-Pu)/N)

-1.645=(Ps-0.20)/sqrt (0.20*0.80)/100

-1.645=(Ps-0.20)/0.04

Ps=-0.0658+0.20=0.2658 [ans]

b) Substitute Ps with 0.15 to find the required z value.

z=(0.15-0.20)/sqrt(0.20*0.80)/100=-1.25

Beta is 0.1056 (area corresponding to -1.25)

c) power of test is 1-beta=0.8944

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