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25, 000 (c) (0.144, 0.256) 99 % (39.827, 40.073) (0.127 , 0.273 ) = CV* SQRT (0.2* 0-8/200) CV : Norm. Inv ( 0. 99
25, 000 (c) (0.144, 0.256) 99 % (39.827, 40.073) (0.127 , 0.273 ) = CV* SQRT (0.2* 0-8/200) CV : Norm. Inv ( 0. 99 / 2, 0, 1 ) = _ Error bound 20. A college administrator would like to estimate average of GPA of all the currently registered students based on a random sample. He plans to estimate the average GPA =CV .(0.2#0.8 /20 to within 0.05 with a 90% confidence interval. Approximately how many students does he need to include in the sample in order to accomplish that? Assume the standard CV - 0 . 2 = LB deviation of all GPAs in the student population is o = 0.4. CV + 0 2 = VB (a) 105 (b ) 174 (c) 246 (d) 300 (e) 425 2 score = 1. 645 Use the following information for questions 21 to 23. In the past, it is generally agreed that a certain standard treatment yields a mean survival period of 4.2 years for liver cancer patients. Recently, a new treatment is administered to 60 patients and their duration of survival is recorded. The sample mean and standard deviation of the duration is 4.5 years and 0.8 years, respectively. 21. Set up the null and alternative hypotheses to test whether the new treatment increases the mean survival period. (a) Ho : u = 4.5 versus Ha : A > 4.5 (b) Ho : M = 4.5 versus Ha : u # 4.5 (c) Ho : u = 4.2 versus Ha : u # 4.2 (d) Ho : M = 4.2 versus Ha : M > 4.2 (e) Ho : M = 4.2 versus Ha : A > 4.5 above test
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