Question

The authors of the article "Adjuvant Radiotherapy and Chemotherapy in Node-Positive Premenopausal Women with Breast Cancer"† reported on the results of an experiment designed to compare treating cancer patients with chemotherapy only to treatment with a combination of chemotherapy and radiation. Of the 154 individuals who received the chemotherapy-only treatment, 76 survived at least 15 years, whereas 98 of the 164 patients who received the hybrid treatment survived at least that long. With p1 denoting the proportion of all such women who, when treated with just chemotherapy, survive at least 15 years and p2 denoting the analogous proportion for the hybrid treatment, p̂1 = ________ (rounded to three decimal places) and p̂2 = ________ (rounded to three decimal places). A confidence interval for the difference between proportions based on the traditional formula with a confidence level of approximately 99% is 0.494 − 0.598 ± (2.58) (0.494)(0.506) 154 + (0.598)(0.402) 164 = −0.104 ± ______ (rounded to three decimal places) = (−0.247, 0.039) At the 99% confidence level, it is plausible that −0.247 < p1 − p2 < 0.039. This interval is reasonably wide, a reflection of the fact that the sample sizes are not terribly large for this type of interval. Notice that 0 is one of the plausible values of p1 − p2 suggesting that neither treatment can be judged superior to the other. Using 1 = 77/156 = 0.494, 1 = 79/156 = 0.506, 2 = 0.596, 2 = 0.404 based on sample sizes of 156 and 166, respectively, the "improved" interval here is identical to the earlier interval.

Answer 1

Ans Given n = 154 X = 76 n2 = 164 X2 98 Sample proportion 1 = 76 0.494 n 154 a Sample proportion 2, P2 0.598 98 164 n2 99 % confidence interval for difference between two pooportion's is, i - P2 I Zc PCI-E P2 (1-22) n + n2 0.494 -0.598 +2.576 0.494(1-0.494), 0.598(1-0.598) 154 164 (si 0.104 £ 2.576 0.001623142 +0.001465829 + 0.1432 ) (-0.10.4 -0.247 0.039 99 confidence interval for difference between two population proportions is (-0.247 0.039 >