Question

mapep reventing heart attacks: Medical researchers performed a comparison of two drugs, clopidogrel and ticagrelor, which are esigned to reduce the risk of heart attack or stroke in coronary patients. A total of 6740 patients were given clopidogrel, and 662 were given ticagrelor. Of the clopidogrel patients, 571 suffered a heart attack or stroke within one year, and of the ticagrelor atients, 656 suffered a heart attack or stroke. Can you conclude that the proportion of patients suffering a heart attack or stroke greater for ticagrelor? Let P, denote the proportion of patients suffering a heart attack or stroke with clopidogrel and p2 denote ne proportion of patients suffering a heart attack or stroke with ticagrelor. Use the a = 0.01 level of significance and the P-value nethod with the TI-84 Plus calculator. Part: 0 / 4 Part 1 of 4 State the null and alternate hypotheses. HO H: This hypothesis test is a (Choose one) test.

a. State the null and alternate hypotheses.
b. Compute the P-value.
C. Determine whether to reject Ho.
D. State a conclusion.

Answer 1

For sample 1, we have that the sample size is N1 = 6740, the number of favorable cases is X1 = 571, so then the sample proportion is ội = * = 571 6740 = 0.0847 For sample 2, we have that the sample size is N2 662, the number of favorable cases is X2 656, so then the sample proportion is Ộ2 = x 0.9909 656 662 = The value of the pooled proportion is computed as p 0.1658 Xi+X2 Ni+N2 571+656 6740+662 Also, the given significance level is a = 0.01. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho:pi = P2 Ha: p1 > P2 This corresponds to a right-tailed test, for which a z-test for two population proportions needs to be conducted. (2) Rejection Region Based on the information provided, the significance level is a = 0.01, and the critical value for a right-tailed test is ze = 2.33. The rejection region for this right-tailed test is R = {z:z> 2.33} (3) Test Statistics The Z-statistic is computed as follows: 0.0847 – 0.9909 2 = p1 - Ô2 Vp(1-7)(1/n1 + 1/n2) -59.831 0.1658 - (1 – 0.1658)(1/6740 + 1/662) (4) Decision about the null hypothesis Since it is observed that z = -59.831 < zc = 2.33, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p=1, and since p=1> 0.01, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion pı is greater than p2, at the 0.01 significance level.

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