Question

Some people think that the attitude of cancer patients can influence the progress of their disease. We can't experiment with humans, but here is a rat experiment on this theme. Inject 60 rats with tumor cells and then divide them at random into two groups of 30. All the rats receive electric shocks, but rats in Group 1 can end the shock by pressing a lever. (Rats learn this sort of thing quickly.) The rats in Group 2 cannot control the shocks, which presumably makes them feel helpless and unhappy. We suspect that the rats in Group 1 will develop fewer tumors. The results: 16 of the Group 1 rats and 24 of the Group 2 rats developed tumors. (Use a = 0.05.) Let p and py be the population proportions of rats that develop tumors, among injected rats that are able to control the shocks and among those unable to control them, respectively. The null and alternative hypotheses for this investigation are which of the following? Ho: P1 P2, Ha: P1 = P2 Ho: P1 = P2, Ha: P1 > P2 Ho: P1 = P2, Ha: P1 # P2 Ho: P1 = P2, Ha: P1 P2 The z-test rather than the chi-square test for a 2 x 2 table is the proper test because of which of the following? We need a confidence interval for the difference P1 - P2, and only the z-test is related to confidence intervals. The chi-square does not test one sided hypothesis while the z-test does. The chi-square test demands more computations than the 2-test, so we prefer the 2-test. One of the expected counts in the 2 x 2 table is less than 5. Find the P-value of the z-test. (Round your answer to four decimal places.) P-value = Helplessness raises the chances of rats that were injected with tumor cells, to develop tumors. True or False? True False You may need to use the appropriate Appendix Table to answer this question.

Answer 1

Solution:

For sample 1, we have that the sample size is N1 = 30, the number of favorable cases is X1 = 16, so then the sample proportion is pi = * = 16 = 0.5333 For sample 2, we have that the sample size is N2 = 30, the number of favorable cases is X2 = 24, so then the sample proportion is P2 = * = 34 = 0.8 The value of the pooled proportion is computed as p = X1 + X2 = 36+30 = 0.6667 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 left-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 left-tailed test is ze = -1.64. The rejection region for this left-tailed test is R = {z:z< -1.64} (3) Test Statistics The z-statistic is computed as follows: @1 - P2 Võ(1 – 7)(1/nı +1/n2) 0.5333 – 0.8 70.6667 - (1 – 0.6667)(1/30+ 1/30) = -2.191 (4) Decision about the null hypothesis Since it is observed that z = –2.191 < zc = -1.64, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0142, and since p= 0.0142 < 0.05, it is concluded that the 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 pı is less than P2, at the 0.05 significance level.