Answer to Question #308732 in Statistics and Probability for OLIVIA

Question

Side effects A drug manufacturer claims that less

than 10% of patients who take its new drug for treat-

ing Alzheimer’s disease will experience nausea. To

test this claim, researchers conduct an experiment.

They give the new drug to a random sample of 300

out of 5000 Alzheimer’s patients whose families have

given informed consent for the patients to participate

in the study. In all, 25 of the subjects experience

nausea. Use these data to perform a test of the drug

manufacturer’s claim at the a = 0.05 significance

level.

Answer 1

We have,

n = 300 x = 25 p = 0.1 q = 1 - p = 0.9  , level of significance=0.05, hence

p^ =x/n = 25/300 = 0.0833

The hypotheses are:

H₀: P=0.1

H₁:p<0.1 which is left tailed

   

The critical value is defined as below

   Z_0.05 =1.64 its from the Z- table such that P(Z>1.64) = 0.05

But test is left-tailed, therefore the critical value is -1.64

From the above we may define Reject Ho if Z < -1.64

Next, we define the test statistic as Z = (P^ - P) / ( (pq/n)^1/2) which becomes

z = (0.0833 - 0.1) / ( (0.1*0.9)/300 )^1/2

solving the above we get that z= -0.96 which is our test statistic.

from the above we see that z > -1.64, so we do not reject H₀.

Hence, we conclude that the data does not provide convincing evidence to support the manufacturers claim.