Question

Consider a drug that is used to help prevent blood clots in certain patients. In clinical trials, among 6173 patients treated with this drug. 163 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. OA. Ho: p=0.03 Hyp#0.03 O B. Ho: p=0.03 H, p<0.03 O C. Ho: p = 0.03 Hy: p=0.03 OD. Ho: p=0.03 Hyp>0.03 Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is I (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. OA. Fail to reject Ho. There is sufficient evidence to warrant rejection of the claim that 3% of users develop nausea. OB. Reject Ho. There is sufficient evidence to warrant rejection of the claim that 3% of users develop nausea. OC. Reject Ho. There is not sufficient evidence to warrant rejection of the claim that 3% of users develop nausea. OD. Fail to reject Ho. There is not 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 vit a problematic adverse reaction Click to select your answer(s). ?

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H₀ : p = 0.03

H_a : p $\neq$ 0.03

n = 6173

$\hat p$ = x / n = 163 / 6173 = 0.026

P₀ = 0.03

1 - P₀ = 1 - 0.03 = 0.97

Test statistic = z

= $\hat p$ - P₀ / [$\sqrt$P_{0 *} (1 - P₀ ) / n]

= 0.026 - 0.03/ [$\sqrt$0.03 * 0.97/ 6173 ]

= −1.842

Test statistic = z = −1.84

P-value = 0.0657682

P value = 0.0658

$\alpha$ = 0.05

P-value > $\alpha$

0.0658 > 0.05

Fail to Reject the null hypothesis .

There is not sufficient evidence to suggest that

Option D is correct

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