# ch09 - CHAPTER 9 Section 9-1 9-1 a H 0 = 25 H1 25 H 0 > 10...

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CHAPTER 9 Section 9-1 9-1 a) 25 : , 25 : 1 0 = µ µ H H Yes, because the hypothesis is stated in terms of the parameter of interest, inequality is in the alternative hypothesis, and the value in the null and alternative hypotheses matches. b) 10 : , 10 : 1 0 = > σ σ H H No, because the inequality is in the null hypothesis. c) 50 : , 50 : 1 0 = x H x H No, because the hypothesis is stated in terms of the statistic rather than the parameter. d) No, the values in the null and alternative hypotheses do not match and 3 . 0 : , 1 . 0 : 1 0 = = p H p H both of the hypotheses are equality statements. e) No, because the hypothesis is stated in terms of the statistic rather than the 30 : , 30 : 1 0 > = s H s H parameter. 9-2 a) α = P(reject H 0 when H 0 is true) = P( X 11.5 when µ = 12) = 4 / 5 . 0 12 5 . 11 / n X P σ µ = P(Z 2) = 0.02275. The probability of rejecting the null hypothesis when it is true is 0.02275. b) β = P(accept H 0 when µ = 11.25) = ( ) P X when > = 115 1125 . . µ = > 4 / 5 . 0 25 . 11 5 . 11 / n X P σ µ P(Z > 1.0) = 1 P(Z 1.0) = 1 0.84134 = 0.15866 The probability of accepting the null hypothesis when it is false is 0.15866. 9-3 a) α = P( X 11.5 | µ = 12) = 16 / 5 . 0 12 5 . 11 / n X P σ µ = P(Z 4) = 0. The probability of rejecting the null, when the null is true, is approximately 0 with a sample size of 16. b) β = P( X > 11.5 | µ =11.25) = > 16 / 5 . 0 25 . 11 5 . 11 / n X P σ µ = P(Z > 2) = 1 P(Z 2) = 1 0.97725 = 0.02275. The probability of accepting the null hypothesis when it is false is 0.02275. 9-4 Find the boundary of the critical region if α = 0.01: 0.01 = 4 / 5 . 0 12 c Z P What Z value will give a probability of 0.01? Using Table 2 in the appendix, Z value is 2.33. Thus, 4 / 5 . 0 12 c = 2.33, = 11.4175 c 9-1

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9-5. P Z c 12 05 4 . / = 0.05 What Z value will give a probability of 0.05? Using Table 2 in the appendix, Z value is 1.65. Thus, c 12 05 4 . / = 1.65, = 11.5875 c 9-6 a) α = P( X 98.5) + P( X > 101.5) = 9 / 2 100 5 . 98 9 / 2 100 X P + > 9 / 2 100 5 . 101 9 / 2 100 X P = P(Z 2.25) + P(Z > 2.25) = (P(Z - 2.25)) + (1 P(Z 2.25)) = 0.01222 + 1 0.98778 = 0.01222 + 0.01222 = 0.02444 b) β = P(98.5 X 101.5 when µ = 103) = 9 / 2 103 5 . 101 9 / 2 103 9 / 2 103 5 . 98 X P = P( 6.75 Z 2.25) = P(Z 2.25) P(Z 6.75) = 0.01222 0 = 0.01222 c) β = P(98.5 X 101.5 when µ = 105) = 9 / 2 105 5 . 101 9 / 2 105 9 / 2 105 5 . 98 X P = P( 9.75 Z 5.25) = P(Z 5.25) P(Z 9.75) = 0 0 = 0 The probability of accepting the null hypothesis when it is actually false is smaller in part (c) because the true mean, µ = 105, is further from the acceptance region. A larger difference exists. 9-7 Use n = 5, everything else held constant (from the values in exercise 9-6): a) P( X 98.5) + P( X >101.5) = 5 / 2 100 5 . 98 5 / 2 100 X P + > 5 / 2 100 5 . 101 5 / 2 100 X P = P(Z 1.68) + P(Z > 1.68) = P(Z 1.68) + (1 P(Z 1.68)) = 0.04648 + (1 0.95352) = 0.09296 b) β = P(98.5 X 101.5 when µ = 103) = 5 / 2 103 5 . 101 5 / 2 103 5 / 2 103 5 . 98 X P = P( 5.03
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