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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.
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This homework help was uploaded on 04/09/2008 for the course STA 3032 taught by Professor Sapkota during the Spring '08 term at University of Central Florida.
 Spring '08
 Sapkota
 Statistics

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