Question

answer 10 please.

10. Using the results of #8 and #9, test the claim that females have a higher proportion of cat ownership than males. a. State the null and alternative b. State the test statistic c. Use the p-value to make a decision (reject or fail to reject) d. Graph the test statistic, p-value, and a=0.05 e. Write your conclusion in terms of the problem

Answer 1

For sample 1, we have that the sample size is N1 = 30, the number of favorable cases is X1 23, so then the sample proportion is X1 P1 0.7667 N1 = = = 23 30 = - For sample 2, we have that the sample size is N2 = 30, the number of favorable cases is X2 = 5, so then the sample proportion is X2 5 P2 = 0.1667 N2 = - = 30 The value of the pooled proportion is 23+5 computed as p X1 + X2 = 0.4667 Ni+N2 30+30 = Also, the given significance level is a = 0.05. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: p1 = 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 = 0.05, and the critical value for a right-tailed test is ze = 1.64. The rejection region for this right-tailed test is R = {z: 2 > 1.64} (3) Test Statistics The z-statistic is computed as follows: P2 z = P1 Võ(1 - 0) (1/nı + 1/n2) 0.7667 – 0.1667 V0.4667 · (1 0.4667)(1/30 + 1/30) 4.658

(4) Decision about the null hypothesis 4.658 > Zc 1.64, it is then concluded that the null hypothesis is Since it is observed that z = rejected. Using the P-value approach: The p-value is p= 0, and since p = 0 < 0.05, it is concluded that th null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population proportion p1 is greater than p2, at the 0.05 significance level. Confidence Interval The 95% confidence interval for P1 P2 is: 0.398 < P1 – P2 < 0.802.

Z-test for Two Proportions, z = 4.658, p = 0 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0

Thus it may conclude that females have higher proportion of cat ownership than males.

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