Question

Suppose 237 subjects are treated with a drug that is used to treat pain and 51 of them developed nausea. Use a 0.01

significance level to test the claim that more than20​% of users develop nausea

The test statistic for this hypothesis test is??? round three decimal

The​ P-value for this hypothesis test is???

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H₀ : p = 0.20

H_a : p > 0.20

= x / n = 51/237 = 0.2152

P₀ = 0.20

1 - P₀ = 1 - 20 = 0.80

Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

=0.2152 - 0.20 / [0.20(1- 0.20) /237 ]

= 0.585

P(z >0.585 ) = 1 - P(z <0.585 ) = 0.2794

P-value = 0.2794

= 0.01

p = 0.2794 ≥ 0.01, it is concluded that the null hypothesis is not rejected