Question

In a clinical​ trial, 22 out of 832 patients taking a prescription drug daily complained of flulike symptoms. Suppose that it is known that 2.3 % of patients taking competing drugs complain of flulike symptoms. Is there sufficient evidence to conclude that more than 2.3​% of this​ drug's users experience flulike symptoms as a side effect at the alpha equals 0.05α= level of​ significance?

What are the null and alternative​ hypotheses?

Find the test​ statistic z 0z0.

Find the​ P-value.

A. Since ​P-valuegreater thanalpha​, do not reject the null hypothesis and conclude that there is not sufficient evidence that more than 2.3​% of the users experience flulike symptoms.

B. Since ​P-valueless thanalpha​, reject the null hypothesis and conclude that there is sufficient evidence that more than 2.3​% of the users experience flulike symptoms.

C. Since ​P-valueless thanalpha​, do not reject the null hypothesis and conclude that there is sufficient evidence that more than 2.3​% of the users experience flulike symptoms.

D. Since ​P-valuegreater thanalpha​, reject the null hypothesis and conclude that there is not sufficient evidence that more than 2.3​% of the users experience flulike symptoms.

Answer 1

Stepl: We need to test the true popuation proportion using z - test and the test is right-tailed test at the 5% level of significance The null and alternative hypotheses are Ho: p 0.023(The populati on proportion is equal to 0.023) versus Hi p >0.023(The population proportion is greater than 0.023 Step2: The given information is The population proporti on isp-0.023 The level of significance is a 0.05 The sample proportion is 832 -0.026442 p0.026442 Step3: The formula for calculating the test statistic is given by 0.026442-0.023 0.023(1-0.023) 832 0.003442 0.023 (0.977 832 0.003442 0.022471 832 0.003442 0,000027 0.003442 0.005196962 0.662369232 0.662369232 Step4: The P- value approach P-value - Pz 0.6624) -(1-P(z < 06624)) (1-0.746142548) 0.253857452 P-value 0.2539a0.05. We do not reject the null hypothesis Ho Step5: Conclusion The z -test statistic is z 0.6624 > zo0s-1.645 We do not reject the null hypothesis Ho Using, the P- value approach, because the P- value 0.6624 > a 0.05 We do not reject the null hypothesis H. Since, p-value is greater than alpha, P-value= 0.6624 > α=0.05. Do not reject the null hypothesis Ho and conclude that there is not sufficient evidence that more than 2.3% of the users experience flulike symptoms.