Question

Example 9.13 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 P, 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, P = (rounded to three decimal places) and 2 = (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) + (0.598X0.402) 164 = -0.104: (rounded to three decimal places) = (-0.247, 0.039) At the 99% confidence level, it is plausible that -0.247 <P.-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 is one of the plausible values of P1 - P2 suggesting that neither treatment can be judged superior to the other. Using P1 = 77/156 = 0.494, 41 = 79/156 = 0.506, P2 = 0.596,42 = 0.404 based on sample sizes of 156 and 166, respectively, the "improved" interval here is identical to the earlier interval. You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

Solution Girent x = 76 h = 154 Xz z 98 n2= 164 p = XL (SSDM- LCSESIS 76 154 0.494 X2 n2 - 98 10 169 -0.598 A confidence interal for the difference between proportions based on the traditional Formula with la confidence level of a approximately 997. is = C P - P2 + Zy ✓ PCI-P P2C 1-2 mal In the =(0.494-0.598 +2.58 V0.49411-0.994) 0-598(1-0.598) 154 + 164

= (-0.104 + 2.58 (0.0556)) = (-0.104 0.1434) = (-0.604 – [0.143]) = (-0.247, 0.039) At the gol. confidence lerd, it is plausible that -0.247 <p 1 - P2 < 0.039. This internal is reasonable wide , a reflection of the fact that the sample sizes are not terribly large for this type of interal. Notice that o is one of the plausible values Pi-P2 suggesting neither treatment can be judge superior to the other using P = 77/15 - 0.494 9, = 7/136=0.506 P2=0.596, 92=0.404 based on the sumple sizes 156 and 166 respectively, the "improred" internal here is identical to the earliest interal.