ch9 - 9-1 CHAPTER 9 Section 9-19-1a) 25:,25:1=HHYes,...

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Unformatted text preview: 9-1 CHAPTER 9 Section 9-19-1a) 25:,25:1=HHYes, 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=>HHNo, 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) = P(reject H0 when His true) = P(X11.5 when = 12) =&&--4/5.125.11/nXP= P(Z -2) = 0.02275. The probability of rejecting the null hypothesis when it is true is 0.02275. b) = P(accept H0 when = 11.25) = ()P Xwhen>=1151125..= &&->-4/5.25.115.11/nXPP(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) = P(X11.5 | = 12) =&&--16/5.125.11/nXP= 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) = P(X > 11.5 | =11.25) =&&->-16/5.25.115.11/nXP= 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 = 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) = P(X98.5) + 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) = P(98.5 X101.5 when = 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) = P(98.5 X101.5 when = 105) =&---9/21055.1019/21059/21055.98XP= P(-9.75Z -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, = 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(X98.5) + 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) = P(98.5 X101.5 when = 103) =&---5/2103...
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ch9 - 9-1 CHAPTER 9 Section 9-19-1a) 25:,25:1=HHYes,...

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