Question

Suppose 217 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea. Use a 0.01 significance level to test the claim that more than 20​% of users develop nausea.

Answer 1

H₀: P < 0.2

H₁: P > 0.2

= 54/217 = 0.2488

The test statistic z = ( - P)/sqrt(P(1 - P)/n)

                              = (0.2488 - 0.2)/sqrt(0.2 * 0.8/217)

                              = 1.78

P-value = P(Z > 1.78)

            = 1 - P(Z < 1.78)

            = 1 - 0.9625

            = 0.0375

Since the P-value is greater than the significance level(0.0375 > 0.01), so we should not reject the null hypothesis .

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