Question

Consider a drug that is used to help prevent blood clots in certain patients. In clinical trials, among 5809 patients treated with this drug. 162 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? A. Ho: p=0.03 Hyp#0.03 O B. Ho: p=0.03 He: p > 0.03 Oc. Họ p= 0,03 He: p=0.03 O D. Họ p= 0,03 Hyp<0.03 Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is (Round to two decimal places as needed.) Click to select your answer(s). 11:37 PM

Answer 1

Here we want to test the claim that 3% of users developed nausea at 5% level of Significance.

In the given example:

n=sample size=5809

X= number users developed the adverse reaction of nausea=162

p= sample proportion of users developed the adverse reaction of nausea=X/n=162/5809=0.0279

P=Population proportion of users developed nausea

P ₀ =0.03 & Q₀ =1- P₀ =0.97

Hypothesis:

H₀:The proportion of users developed nausea is equal to 3%.

i.e. H₀ : P =0.03

Against

H₁ : The proportion of users developed nausea is not equal to 3%

Hypothesis: Ho: P = 0,03 against Hi: P+0.03 Test statistic: To test the above hypothesis the test statistic is large sample z-test is given by: z= p - Po N N Cool Pole z = 0.02789 -0.03 (0.03 X 0.97 5809 z = -0.9427 callz)=0.94 The p-value for this hypothesis test at s% Level of significance of two o tailed test is: P-value = 0.345 table value or critical value Zalzalog6 at 5% L.o. call2lc 2012 g p-value > 0.05

Conclusion: Failed to reject H ₀ .There is not sufficient evidence to warrant rejection of claim that 3% of users develop nausea.