Question

Consider a drug that is used to help prevent blood clots in certain patients. In clinical trials, among 6132 patients treated with this drug, 149 developed the adverse reaction of nausea. Use a 0.05 significance level to test the claim that 3% of users develop nausea. Does nausea appear to be a problematic adverse reaction? Identify the null and alternative hypotheses for this test. Choose the correct answer below. O A. Ho: p=0.03 Hp0.03 OB. Ho: p*0.03 Hp0.03 0 C, Ho: p = 0,03 H:p>0.03 OD. Ho: p=0.03 H, p<0.03 Identify the test statistic for this hypothesis test. , The test statistic for this hypothesis testis (Round to two decimal places as needed.) Identify the P-value for this hypothesis test. The P-value for this hypothesis test is (Round to three decimal places as needed.) Identify the conclusion for this hypothesis test, O A Fail to reject Ho. There is not sufficient evidence to warrant rejection of the claim that 3% of users develop nausea, B. Fail to reject H. There is sufficient evidence to warrant rejection of the claim that 3% of users develop nausea O OC Reject Ho. There is not sufficient evidence to warrant rejection of the claim that 3% of users develop nausea, OD. Reject Ho. There is sufficient evidence to warrant rejection of the claim that 3% of users develop nausea. Does nausea appear to be a problematic adverse reaction? Since the rate of nausea appears to be relatively it a problematic adverse reaction

Answer 1

The following null and alternative hypotheses need to be tested:

Ho: p = 0.03

Ha: p ≠​ 0.03

Option A

This corresponds to a two-tailed test for which a z-test for one population proportion needs to be used.

The following information is provided: The sample size is N = 6132, the number of favourable cases is X = 149, and the sample proportion is $\bar p = \frac{X}{N} = \frac{ 149}{ 6132} = 0.0243$ , and the significance level is α=0.05

Based on the information provided, the significance level is α = 0.05, and the critical value for a two-tailed test is zc​ = 1.96.

The z-statistic is computed as follows:

z = 7 - Po - V po(1 – po)/n 0.0243 – 0.03 0.03(1 – 0.03)/6132 = -2.617

The test statistic is

z = -2.62

The p-value is p = 0.0089

Since p = 0.0089 <  0.05(alpha), it is concluded that the null hypothesis is rejected.

Conclusion: (Option D)

Reject H0, there is sufficient evidence to warrant rejection of the claim that 3% of users develop nausea.

Since the rate appears to to be relatively low it does not appear to be a problematic adverse reaction