Question

Q2) A drug manufacturer claims that fewer than 10% of patients who take its new drug for treating Alzheimer’s disease will experience nausea. In a random sample of 500 patients, 47 experienced nausea. Perform a significance test at the 5% significance level to test this claim.

(a) State the null and alternative hypothesis.

(b) Calculate the test statistic.

(c) Calculate the p-value. Round to 4 decimal places

(d) Make a decision at the 0. 05 significance level to reject Ho or fail to reject Ho.

(e) Write your conclusion in the context of the claim.

Answer 1

Solution :

Given that,

n = 500

x = 47

Point estimate = sample proportion = = x / n = 47/500 = 0.094

= 10% = 0.10

1 - = 1 - 0.10 = 0.90

= 0.05

a)

This is a left (One) tailed test,

Ho: 0.10

Ha: 0.10

b)

The test statistics

z = ( - )/( *( 1- ) )/n

= ( 0.094 - 0.10)/( 0.10*0.90)/500

= 0.447

c)

p-value = P(Z < z )

= P( Z < 0.447)

= 0.6726

d)

The p-value is p = 0.3274 > = 0.05, it is concluded that the null hypothesis is fails to reject.

e)

Conclusion:

It is concluded that the null hypothesis Ho is fails to reject. Therefore, there is not enough evidence to claim that the population proportion  is less than = 0.05 level of significance.