ISM Chapter 16

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

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Unformatted text preview: Chapter 16 16.1 : H = 1 p .1, = 2 p .2, = 3 p .3, = 4 p .2, = 5 p .2 : H 1 At least one i p is not equal to its specified value. Cell i i f i e ) e f ( i i- i 2 i i e / ) e f (- 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 χ = 4.54 Rejection region: 2 χ > = χ = χ- α 2 4 , 01 . 2 1 k , 13.2767 2 χ = 4.54, p-value = .3386. There is not enough evidence to infer that at least one i p is not equal to its specified value. 16.2 : H = 1 p .1, = 2 p .2, = 3 p .3, = 4 p .2, = 5 p .2 : H 1 At least one i p is not equal to its specified value. Cell i i f i e ) e f ( i i- i 2 i i e / ) e f (- 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 χ = 2.26 Rejection region: 2 χ > = χ = χ- α 2 4 , 01 . 2 1 k , 13.2767 2 χ = 2.26, p-value = .6868. There is not enough evidence to infer that at least one i p is not equal to its specified value. 16.3 : H = 1 p .1, = 2 p .2, = 3 p .3, = 4 p .2, = 5 p .2 : H 1 At least one i p is not equal to its specified value. 127 Cell i i f i e ) e f ( i i- i 2 i i e / ) e f (- 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 χ = 1.14 Rejection region: 2 χ > = χ = χ- α 2 4 , 01 . 2 1 k , 13.2767 2 χ = 1.14, p-value = .8889. There is not enough evidence to infer that at least one i p is not equal to its specified value. 16.4 The 2 χ statistic decreases. 16.5 : H = 1 p .3, = 2 p .3, = 3 p .2, = 4 p .2 : H 1 At least one i p is not equal to its specified value. Cell i i f i e ) e f ( i i- i 2 i i e / ) e f (- 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 χ = 4.98 Rejection region: 2 χ > = χ = χ- α 2 3 , 05 . 2 1 k , 7.81473 2 χ = 4.98, p-value = .1734. There is not enough evidence to infer that at least one i p is not equal to its specified value. 16.6 16.5 : H = 1 p .3, = 2 p .3, = 3 p .2, = 4 p .2 : H 1 At least one i p is not equal to its specified value. Cell i i f i e ) e f ( i i- i 2 i i e / ) e f (- 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 χ = 9.96 128 Rejection region: 2 χ > = χ = χ- α 2 3 , 05 . 2 1 , k 7.81473 2 χ = 9.96, p-value = .0189. There is enough evidence to infer that at least one i p is not equal to its specified value. 16.7 : H = 1 p .2, = 2 p .2, = 3 p .2, = 4 p .2, = 5 p .2 : H 1 At least one i p is not equal to its specified value. Cell i i f i e ) e f ( i i- i 2 i i e / ) e f (- 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 χ = 4.20 Rejection region: 2 χ > = χ = χ- α 2 4 , 10 ....
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This homework help was uploaded on 04/18/2008 for the course STAT 212 taught by Professor Holt during the Spring '08 term at UVA.

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

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