Chapter 16 - 311 Chapter 16 16.1 0 H = 1 p.1 = 2 p.2 = 3 p.3 = 4 p.2 = 5 p.2 1 H At least one i p is not equal to its specified value Cell i i f i

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311 Chapter 16 16.1 : 0 H = 1 p .1, = 2 p .2, = 3 p .3, = 4 p .2, = 5 p .2 : 1 H At least one i p is not equal to its specified value. Cell i i f i e ) ( i i e f - i i i e e f / ) ( 2 - 1 24 300(.1) = 30 -6 1.20 2 64 300(.2) = 60 4 .27 3 84 300(.3) = 90 -6 .40 4 72 300(.2) = 60 12 2.40 5 56 300(.2) = 60 -4 .27 Total 300 300 2 c = 4.54 Rejection region: 2 c > = c = c - a 2 4 , 01 . 2 1 ,k 13.2767 2 c = 4.54, p-value = .3386 (Excel). There is not enough evidence to infer that at least one i p is not equal to its specified value. 16.2 : 0 H = 1 p .1, = 2 p .2, = 3 p .3, = 4 p .2, = 5 p .2 : 1 H At least one i p is not equal to its specified value. Cell i i f i e ) ( i i e f - i i i e e f / ) ( 2 - 1 12 150(.1) = 15 -3 .60 2 32 150(.2) = 30 2 .13 3 42 150(.3) = 45 -3 .20 4 36 150(.2) = 30 6 1.20 5 28 150(.2) = 30 -2 .13 Total 150 150 2 c = 2.26 Rejection region: 2 c > = c = c - a 2 4 , 01 . 2 1 ,k 13.2767 2 c = 2.26, p-value = .6868 (Excel). There is not enough evidence to infer that at least one i p is not equal to its specified value. 312 16.3 : 0 H = 1 p .1, = 2 p .2, = 3 p .3, = 4 p .2, = 5 p .2 : 1 H At least one i p is not equal to its specified value. Cell i i f i e ) ( i i e f - i i i e e f / ) ( 2 - 1 6 75(.1) = 7.5 -1.5 .30 2 16 75(.2) = 15 1 .07 3 21 75(.3) = 22.5 -1.5 .10 4 18 75(.2) = 15 3 .60 5 14 70(.2) = 15 -1 .07 Total 75 75 2 c = 1.14 Rejection region: 2 c > = c = c - a 2 4 , 01 . 2 1 ,k 13.2767 2 c = 1.14, p-value = .8889 (Excel). There is not enough evidence to infer that at least one i p is not equal to its specified value. 16.4 The 2 c statistic decreases. 16.5 : 0 H = 1 p .3, = 2 p .3, = 3 p .2, = 4 p .2 : 1 H At least one i p is not equal to its specified value. Cell i i f i e ) ( i i e f - i i i e e f / ) ( 2 -
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1 38 150(.3) = 45 -7 1.09 2 50 150(.3) = 45 5 0.56 3 38 150(.2) = 30 8 2.13 4 24 150(.2) = 30 -6 1.20 Total 150 150 2 c = 4.98 Rejection region: 2 c > = c = c - a 2 3 , 05 . 2 1 ,k 7.81473 2 c = 4.98, p-value = .1734 (Excel). There is not enough evidence to infer that at least one i p is not equal to its specified value. 313 16.6 : 0 H = 1 p .3, = 2 p .3, = 3 p .2, = 4 p .2 : 1 H At least one i p is not equal to its specified value. Cell i i f i e ) ( i i e f - i i i e e f / ) ( 2 - 1 76 300(.3) = 90 -14 2.18 2 100 300(.3) = 90 10 1.11 3 76 300(.2) = 60 16 4.27 4 48 300(.2) = 60 -12 2.40 Total 300 300 2 c = 9.96 Rejection region: 2 c > = c = c - a 2 3 , 05 . 2 1 ,k 7.81473 2 c = 9.96, p-value = .0189 (Excel). There is enough evidence to infer that at least one i p is not equal to its specified value. 16.7 : 0 H = 1 p .2, = 2 p .2, = 3 p .2, = 4 p .2, = 5 p .2 : 1 H At least one i p is not equal to its specified value. Cell i i f i e ) ( i i e f - i i i e e f / ) ( 2 - 1 28 100(.2) = 20 8 3.20 2 17 100(.2) = 20 -3 0.45 3 19 100(.2) = 20 -1 0.05 4 17 100(.2) = 20 -3 0.45 5 19 100(.2) = 20 -1 0.05 Total 100 100 2 c = 4.20 Rejection region: 2 c > = c = c - a 2 4 , 10 2 1 ,k 7.77944 2 c = 4.20, p-value = .3796 (Excel). There is not enough evidence to infer that at
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This note was uploaded on 06/06/2011 for the course ADMS 2320 taught by Professor Rochon during the Spring '08 term at York University.

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Chapter 16 - 311 Chapter 16 16.1 0 H = 1 p.1 = 2 p.2 = 3 p.3 = 4 p.2 = 5 p.2 1 H At least one i p is not equal to its specified value Cell i i f i

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