Question

Suppose 224 subjects are treated with a drug that is used to treat pain and 53 of them develop nausea. Use a 0.01 significance level to test the claim that more than 20% of users develop nausea

Answer 1

Solution :

This is the right tailed test .

The null and alternative hypothesis is

H₀ : p = 0.20

H_a : p > 0.20

n = 224

x = 53

= x / n = 53 / 220 = 0.2366

P₀ = 0.20

1 - P₀ = 1 - 0.20 = 0.80

z = - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.2366 - 0.20 / [(0.20 * 0.80) / 224]

= 1.37

Test statistic = 1.37.

P(z > 1.37) = 1 - P(z < 1.37) = 1 - 0.9147 = 0.0853

P-value = 0.0853

= 0.01

P-value >

Fail to reject the null hypothesis .

There is not sufficient evidence to test the claim that more than 20% of users develop nausea .