Ch14_selected_ChiSquare_Tests

Ch14_selected_ChiSqu - 10.24 10.24 4.76 The test statistic value is 3 2 X =(5 – 4.76 2/4.76(12 – 10.24 2/10.24(9 – 10.24 2/10.24(4 – 4.76

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Chapter 14 - Nonparametric Methods and Chi-Square Tests CHAPTER 14 NONPARAMETRIC METHODS AND CHI-SQUARE TESTS – CHI- SQUARE TESTS 46 ) 4 ( 2 X = 20 ) 20 10 ( ) 20 40 ( ) 20 34 ( ) 20 12 ( ) 20 4 ( 2 2 2 2 2 - + - + - + - + - = 50.8 Reject H 0 ; p -value < .005 47 The expected counts are as follows: NB: 0.729(1,000) = 729 PL: 0.23(1,000) = 230 G: 0.041(1,000) = 41 The observed counts are, respectively: 610, 290, and 100. We now compute the value of the test statistic: - = E E O X / ) ( 2 2 = 119.97. Strongly reject the null hypothesis. 49 For a normal distribution with mean 11 and standard deviation 2, we expect 0.3413 of the observations to be between 11 and 13; 0.3413 of the observations to be between 9 and 11; 0.1587 of them below 9; and the same proportion above 13. [Here we use one standard deviation.] The observed counts are as follows: Below 9: 5; 9 to 11: 12; above 11 and up to 13: 9; above 13: 4. Expected counts are (respectively): 30(.1587) = 4.76; 30(.3413) =
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Unformatted text preview: 10.24; 10.24; 4.76. The test statistic value is: ) 3 ( 2 X = (5 – 4.76) 2 /4.76 + (12 – 10.24) 2 /10.24 + (9 – 10.24) 2 /10.24 + (4 – 4.76) 2 /4.76 = 0.586. Do not reject H . [Note that two expected counts are slightly below 5. However, other partitions also lead to non-rejection of the null hypothesis.] 50 The test statistic now has 2 fewer degrees of freedom: df = 4 – 3 = 1. We have: x = 11.21 and s = 2.71. Forming a partition using z = 0.44 and z = –0.44 leads to: . 88 . 2 082 . 24 . 86 . 7 . 1 ) 1 ( 2 = + + + = X At α = 0.05, we cannot reject H . 58 X 2 = 4.23 df = 3 Do not reject H . 61 X 2 = 24.36 df = 6 Reject H . 62 Median = 15 X 2 = 4.159 df = 2 Do not reject H . 14-1...
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This note was uploaded on 11/24/2010 for the course MATH BAOI taught by Professor Phong during the Spring '10 term at Wash. College.

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