Answer+Key+to+Practice+Final

Answer+Key+to+Practice+Final - Chapter 13 13.38 a. Randomly...

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1 Chapter 13 13.38 a. Randomly select finance and marketing MBA graduates and determine their starting salaries. b. Randomly assign some MBA students to major in finance and others to major in marketing. Compare starting salaries after they graduate. c. Better students may be attracted to finance and better students draw higher starting salaries. 13.58 a. 1 / : H 2 2 2 1 0 = σ σ 1 / : H 2 2 2 1 1 σ σ Rejection region: 88 . 1 F F F 29 , 29 , 05 . , , 2 / 2 1 = > ν ν α or 53 . 88 . 1 / 1 F / 1 F / 1 F F 29 , 29 , 05 . , , 2 / , , 2 / 1 1 2 2 1 = = = < ν ν α ν ν α F = 2 2 2 1 s / s = 350/700 =.50, p-value = .0669. There is enough evidence to conclude that the population variances differ. b. Rejection region: 98 . 2 F F F 14 , 14 , 025 . , , 2 / 2 1 = = > ν ν α or 34 . 98 . 2 / 1 F / 1 F / 1 F F 14 , 14 , 025 . , , 2 / , , 2 / 1 1 2 2 1 = = = = < ν ν α ν ν α F = 2 2 2 1 s / s = 350/700 =.50, p-value = .2071. There is not enough evidence to conclude that the population variances differ. c. The value of the test statistic is unchanged and in this exercise the conclusion changed as well. . 13.67 ) p p ( : H 2 1 0 = 0 ) p p ( : H 2 1 1 0 a. + = 2 1 2 1 n 1 n 1 ) p ˆ 1 ( p ˆ ) p ˆ p ˆ ( z = + 100 1 100 1 ) 425 . 1 ( 425 . ) 40 . 45 (. = .72, p-value = 2P(Z > .72) = 2(1 – .7642) = .4716.
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2 b. + = 2 1 2 1 n 1 n 1 ) p ˆ 1 ( p ˆ ) p ˆ p ˆ ( z = + 400 1 400 1 ) 425 . 1 ( 425 . ) 40 . 45 (. = 1.43, p-value = 2P(Z > 1.43) = 2(1 – .9236) = .1528. c. The p-value decreases. 13.69 a. 2 2 2 1 1 1 2 / 2 1 n ) p ˆ 1 ( p ˆ n ) p ˆ 1 ( p ˆ z ) p ˆ p ˆ ( + ± α = (.18–.22 ) 100 ) 22 . 1 ( 22 . 100 ) 18 . 1 ( 18 . 645 . 1 + ± = –.040 ± .0929 b. 2 2 2 1 1 1 2 / 2 1 n ) p ˆ 1 ( p ˆ n ) p ˆ 1 ( p ˆ z ) p ˆ p ˆ ( + ± α = (.48–.52 ) 100 ) 52 . 1 ( 52 . 100 ) 48 . 1 ( 48 . 645 . 1 + ± = –.040 ± .1162 c. The interval widens. 13.70 ) p p ( : H 2 1 0 = 0 > ) p p ( : H 2 1 1 0 + = 2 1 2 1 n 1 n 1 ) p ˆ 1 ( p ˆ ) p ˆ p ˆ ( z = , 70 . 1 178 1 229 1 ) 177 . 1 ( 177 . ) 140 . 205 (. = + p-value = P(Z > 1.70) = 1 – .9554 = .0446. There is enough evidence to conclude that those who paid the regular price are more likely to buy an extended warranty.
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3 Chapter 15 15.2 : H 0 = 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 Total 150 150 5 28 150(.2) = 30 -2 .13 2 χ = 2.26 Rejection region: 2 χ > = χ = χ α 2 4 , 01 . 2
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This note was uploaded on 11/11/2011 for the course ECON 100 taught by Professor Chandle during the Spring '11 term at UC Irvine.

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Answer+Key+to+Practice+Final - Chapter 13 13.38 a. Randomly...

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