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Unformatted text preview: 910 (a) The critical values are determined by 0 /2 /2 = 2.25, = 2(2.25) = 0.0244 = (2.25)  (6.75) = 0.0122
98.5105 2/9 98.5103 101.5103 (b) = (98.5 101.5) = 2/ 9 2/ 9 = (6.75 2.25) . It follows that /2 IE 300/GE 331 Discussion Problem Set 12
2 9 = 1.5, (c) = 912 (a) We reject the null hypothesis when is outside of the interval [0  /2 942 (a) Since the true mean (105) is further away from the acceptance region, the probability of failing to reject the false null hypothesis is smaller (in discussion sections, draw a picture to show this). With = 9, = 2, 0 = 100, /2 = 0.005 = 2.57, the critical values are 98.29 and 101.71. 101.5105 2/9 = (5.25)  (9.75) = 7.6  8 , 0 + /2 ]. The parameter of interest is the true mean hole diameter . 0 = 1.50, 1 1.50 = 0.01, 0 = / = 1.4975, = 25, = 0.01, 0 = / (c) = 0  /2 = 1.49481.495 /  0 Since 0  2.58, we fail to reject the null hypothesis and conclude that there is no strong evidence that the true mean hole diameter is different from 1.5 at the significant level = 0.01. 0 + /2
1.50511.495 / , reject 0 if 0  > /2 ,where 0.005 = 2.58  0 = 1.25 (e) The 99% CI on the true mean hole diameter is given by:  /2 0.01 25 = (5.076)  (0.076) = 0.5303. So 1  = 0.4697 + /2 = (1.494848 1.505152) That is 1.4923 1.5027. Since 0 = 1.5 falls within this interval, we fail to reject the null hypothesis. 1.4975  0.005 1.4975 + 0.005 0.01 25 ...
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This note was uploaded on 05/04/2010 for the course GE 331 taught by Professor Negarkayavash during the Spring '09 term at University of Illinois at Urbana–Champaign.
 Spring '09
 NegarKayavash

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