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# ch9 - 9-1 CHAPTER 9 Section 9-19-1a 25,25:1≠=μμHHYes...

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Unformatted text preview: 9-1 CHAPTER 9 Section 9-19-1a) 25:,25:1≠=μμHHYes, 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=>σσHHNo, because the inequality is in the null hypothesis. c) 50:,50:1≠=xHxHNo, because the hypothesis is stated in terms of the statistic rather than the parameter. d) 3.:,1.:1==pHpHNo, the values in the null and alternative hypotheses do not match and both of the hypotheses are equality statements. e) 30:,30:1>=sHsHNo, because the hypothesis is stated in terms of the statistic rather than the parameter. 9-2 a) α= P(reject H0 when His true) = P(X≤11.5 when μ= 12) =&&¡¢££¤¥-≤-4/5.125.11/nXPσμ= P(Z ≤-2) = 0.02275. The probability of rejecting the null hypothesis when it is true is 0.02275. b) β= P(accept H0 when μ= 11.25) = ()P Xwhen>=1151125..μ= &&¡¢££¤¥->-4/5.25.115.11/nXPσμP(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) =&&¡¢££¤¥-≤-16/5.125.11/nXPσμ= 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) =&&¡¢££¤¥->-16/5.25.115.11/nXPσμ= 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 = &¡¢£¤¥-≤4/5.12cZPWhat Z value will give a probability of 0.01? Using Table 2 in the appendix, Z value is -2.33. Thus, 4/5.12-c= -2.33, c = 11.4175 9-2 9-5. P Zc≤-&¡¢£¤¥12054. /= 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-12054. /= -1.65, c = 11.5875 9-6 a) α= P(X≤98.5) + P(X> 101.5) = ¥¥¤£¢¢¡&-≤-9/21005.989/2100XP+¥¥¤£¢¢¡&->-9/21005.1019/2100XP= 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) =¥¥¤£¢¢¡&-≤-≤-9/21035.1019/21039/21035.98XP= 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) =¥¥¤£¢¢¡&-≤-≤-9/21055.1019/21059/21055.98XP= 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 since 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) =¥¥¤£¢¢¡&-≤-5/21005.985/2100XP+¥¥¤£¢¢¡&->-5/21005.1015/2100XP= 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) =¥¥¤£¢¢¡&-≤-≤-5/2103...
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ch9 - 9-1 CHAPTER 9 Section 9-19-1a 25,25:1≠=μμHHYes...

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