Question

HW#9 - Chapter 18 (20pts for turned in on time) (16pts new idea for treating advanced melanoma, the most series in cancer trough Genetically engineer white blood cells to better recent and destroy cancer cells then w e these cells intonatients. The subjects in a man initialudy were 11 patients who melanoma had not responded to existing treatment. One question was how rapidly the new cells would multiply after infusion, as measure by the doubling time in das Here are the data collected for doubling time 14 10 13 10 13 2.0 06 TON 107 o 19 (1) ( Provide a 95% confidence interval for the average obling time X" 3 1172727 S14 063 n df.10 95 to 2.228 1.10+ 2.228 *0.4606 1.4794 1.17 :1.1383 -2.228 0.flach (2) pts)One researcher claimed that the average doubling time is 1 day. Perform a statistical analysis to justify his claim based on 0.05 significance level. 9- Hon=1 Hs Mt! b PL0.05 C X1170727 S=0.460632 n = 11 X=1.172727 1.179721 = 1.244 I. i-u I sin 0.460630/ df = 11-1=10 P-value=2P (T>11.2441)

Answer 1

1) Assumption: The sample come from a normal population, . We have to find the confidence interval for . . f = average doubling time. Let the sample be x,,X2 ,... xm where n=11 and Xin N(6.82) ; r ? - variance 1. So, y ~ N(M, Th) Heuce Fn(x H) ~ N(0.1) :> (x-7 m2 ;X + = 942), is the 95%. си от н а As the population variance is not known, we replace r by s= .. sample deviation standard Hence required I for r is (* I ting, n-1 5 2 )

= 104 Now + 1 + .. +1 = 1/72+27 n-1 so and 31*-3)2 = 10:212182 - 0.460632 at d=0:05, tindo, nos to:975, 10 is Bee = 2:238 Hence CI for Mis : x 2.228 79997 + 0.460632 2 - ( 0.863289148216) one researches claim that the average doubling time is 1 day. Hence at x = 0.05 we test Ho: M = 1 vs Hit & using a sample t test .. As the population variance is unknown bis hin 7 (2-Min :-(1172727-1) vi Now Test statistie e = 0.460632 Critical region for je . 3 1.24366 rejecting Ho is Tl>tit, nne. ITI 7 2:228 ITI = 1:24 366 < 2.28 = tica n Here -) Hence double we time accept is Ho 1 day and ie. conclude that average M=1 is true.