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Homework Chapter 9

# Homework Chapter 9 - 9.40 To test the null hypothesis that...

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Unformatted text preview: 9.40 To test the null hypothesis that handicap and performance are independent against the alternative that they are not independent at the 5% level, we use the x2 statistic with 4 degrees of freedom and reject the null hypothesis when X2 > x205 = 9.488. In Table 9.6, the expected frequency and the contribution to the X2 statistic of each cell are given in the parentheses and brackets respectively. Table 9.6. Exercise 9.40. Above average ererage Below average Total Blind 21 64 17 102 (20.34) (63.48) (18.18) [.02] [.00] [.08] Deaf 16 49 14 79 (15.75) (49.17) (14.08) [.00] {.00} [.00] Neither 29 93 1 28 150 (29.91) (93.35) (26.74) [.031 [.001 1.061 LEtal 66 206 59 331 The )8 statistic is 2 .552 .522 +1.182 X : 20.34+63.48 18.18 .252 .172 .082 + + 15.75 49.17 14.08 .912 .352 1.262 + + 29.91 93.35 25.74 = .196. + Thus, we cannot reject the null hypothesis. v.42 ‘lhe data are summarized in the table: Table 9.8. Ezemise 9.42. No. needing No. of Binomial prob. Expected no. adjustments days p=.l, n=4 of days I 79 .2916 58.32 2 19 .0486 9.72 3 1 20 .0036 _72 10.46 4 U .0001 .02 l r we 1 th 5153. > X 9 e BC: the 111111 11 H e S nce 2 m , 9 43 The mean of the observed distribumon 15 9.45 The mean of the Poisson distribution is 2.0. Using Table 2, the data and expected frequencies are: Table 9.10. Exercise 9.45. Poisson prob. Expected Number Frequency A22 ,0 frequency 0 52 .135 67.5 1 151 .271 135.5 2 130 .271 135.5 3 102 .180 90.0 4 45 .090 45.0 5 12 .036 18.0 6 5 .012 6.0 7 1 8 .004 2.0 8.5 8 2 .001 .5 We use the X2 statistic with 7—2 = 5 degrees of freedom to test the null hypothesis that the arrival distribution is Poisson. Thus, we reject the null hypothesis at the 5% level when X7 > fog, : 11.070. Now, 2 ! (52 — 57.5)2 (151 — 135.5)2 (130 — 135.5)? + (102 — 90.0)2 X _ 67.5 135.5 + 135.5 90.0 (45 — 45.0)2 (12 ~18.0)2 (8 — 5.5)2 1 = .185. 45.0 + 18.0 8.5 9 Thus, we cannot reject the null hypothesis. 9.48 The distribution function of the exponential distribution with [.L = 40 is F(t) : 1 — exp (—t/40). Thus, we can summarize the data and the expected frequencies in the table. Table 9.13. Eze'rcise 9.48. X = Service Exponential Expected life Frequency probability frequency X < 20 46 .393 39.3 20 g X < 40 19 .239 23.9 40 g X < 60 17 .145 14.5 60 S X < 80 1'2 .088 8.8 X Z 80 6 .135 13.5 We use the x2 statistic with 5 — 1 = 4 degrees of freedom to test the null hypoth esis that the data are from an exponential distribution with u = 40 against the alternative that they are from some other distribution. Thus, we reject the null hypothesis at the 1% level when x2 > x201 = 13.227. Now, 2 _ (46 — 39.3)2 (19 - 23.9)2 (17 — 14.5)2 X _ 39.3 + 23.9 + 14.5 (12 — as)2 (6 — 13.5)2 ~——— ————— : _ 1. 8.8 + 13.5 7 9 Since 7.91 < x2 2 13.227, we cannot reject the null hypothesis. .01 n An I'm... M1MI‘T‘AR nntnut is- ...
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