Question

Question 20 1 pts The drug Lipitor is mean to reduce cholesterol and LDL cholesterol. In clinical trials 19 out of 863 patients taking 10 mg of Lipitor daily complained of flulike symptoms. Suppose that it is known that 1.9% of patients taking competing drugs complain of flulike symptoms. Is there evidence to conclude that more than 1.9% of Liptor users experience flulike-smptoms as a side effect at a = 0.05, level of significance ? Which of the following is true In the hypothesis testing of the above problem? The conclusion is do not reject The p-value <a = 0.01 The test is two tail test The test statistic is z = 1.56

Answer 1

The correct statement is -

Let p denotes the population proportion. Hypotheses : To test null hypothesis Ho : p=0.019 against alternative hypothesis H :p>0.019 Let o denotes the sample proportion and n denotes the sample size. Here, .. 19 p = 0.0220, n = 863, standard error = 0.019*(1 -0.019) 863 = 0.004647 863 The test statistic can be written as: ; - po) Po*(1-P) n which under H, follows a standard normal distribution. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if ZSTAT > 20.05 = 1.644854 0.4- 0.3 density 0.2 0.1- 0.0 - 3 2 Z Now, Value of test statistic: The value of the test statistic 2 STAT = 0.649019 20.649 (0.0220 - 0.019) 0.019*(1-0.019) 863 The critical value Zcritical = 20.05 = 1.644854 ~ 1.645 P-value = P(Z > ZSTAT) = P(Z >0.649019) = 1 - P(Z <0.649019) = 1-0.741837 = 0.258163 -0.2582 Conclusion : Since p-value > 0.05 and ZSTAT > Zcritical = 1.644854, so We fail to reject H, at 0.05 level of significance Hence, we can conclude that the population proportion is not significantly greater than 0.019

• The conclusion is do not reject