ISM_chapter13 - Chapter 13: The Analysis of Variance 13.1 2...

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264 Chapter 13: The Analysis of Variance 13.1 The summary statistics are: 1 y = 1.875, 2 1 s = .6964286, 2 y = 2.625, 2 2 s = .8392857, and n 1 = n 2 = 8. The desired test is: H 0 : μ 1 = μ 2 vs. H a : μ 1 μ 2 , where μ 1 , μ 2 represent the mean reaction times for Stimulus 1 and 2 respectively. a. SST = 4(1.875 – 2.625) 2 = 2.25, SSE = 7(.696428) + 7(.8392857) = 10.75. Thus, MST = 2.25/1 = 2.25 and MSE = 10.75/14 = .7679. The test statistic F = 2.25/.7679 = 2.93 with 1 numerator and 14 denominator degrees of freedom. Since F .05 = 4.60, we fail to reject H 0 : the stimuli are not significantly different. b. Using the Applet, p –value = P ( F > 2.93) = .109. c. Note that 2 p s = MSE = .7679. So, the two–sample t –test statistic is | t | = 8 2 7679 . 625 . 2 875 . 1 = 1.712 with 14 degrees of freedom. Since t .025 = 2.145, we fail to reject H 0 . The two tests are equivalent, and since F = T 2 , note that 2.93 (1.712) 2 (roundoff error). d. We assumed that the two random samples were selected independently from normal populations with equal variances. 13.2 Refer to Ex. 10.77. The summary statistics are: 1 y = 446, 2 1 s = 42, 2 y = 534, 2 2 s = 45, and n 1 = n 2 = 15. a. SST = 7.5(446 – 534) 2 = 58,080, SSE = 14(42) + 14(45) = 1218. So, MST = 58,080 and MSE = 1218/28 = 1894.5. The test statistic F = 58,080/1894.5 = 30.64 with 1 numerator and 28 denominator degrees of freedom. Clearly, p –value < .005. b. Using the Applet, p –value = P ( F > 30.64) = .00001. c. In Ex. 10.77, t = –5.54. Observe that (–5.54) 2 30.64 (roundoff error). d. We assumed that the two random samples were selected independently from normal populations with equal variances. 13.3 See Section 13.3 of the text. 13.4 For the four groups of students, the sample variances are: 2 1 s = 66.6667, 2 2 s = 50.6192, 2 3 s = 91.7667, 2 4 s = 33.5833 with n 1 = 6, n 2 = 7, n 3 = 6, n 4 = 4. Then, SSE = 5(66.6667) + 6(50.6192) + 5(91.7667) + 3(33.5833) = 1196.6321, which is identical to the prior result. 13.5 Since W has a chi–square distribution with r degrees of freedom, the mgf is given by 2 / ) 2 1 ( ) ( ) ( r tW W t e E t m = = . Now, W = U + V , where U and V are independent random variables and V is chi–square with s degrees of freedom. So, 2 / 2 / ) ( ) 2 1 ( ) 2 1 )( ( ) ( ) ( ) ( ) ( ) ( r s tU tV tU V U t tW W t t e E e E e E e E e E t m + = = = = = . Therefore, 2 / ) ( 2 / 2 / ) 2 1 ( ) 2 1 ( ) 2 1 ( ) ( ) ( s r s r tU U t t t e E t m = = = . Since this is the mgf for a chi– square random variable with r s degrees of freedom, where r > s , by the Uniqueness Property for mgfs U has this distribution.
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Chapter 13: The Analysis of Variance 265 Instructor’s Solutions Manual 13.6 a. Recall that by Theorem 7.3, 2 2 / ) 1 ( σ i i S n is chi–square with n i – 1 degrees of freedom. Since the samples are independent, by Ex. 6.59, = σ = σ k i i i S n 1 2 2 2 / ) 1 ( / SSE is chi–square with n k degrees of freedom. b. If H 0 is true, all of the observations are identically distributed since it was already assumed that the samples were drawn independently from normal populations with common variance. Thus, under H 0 , we can combine all of the samples to form an estimator for the common mean, Y , and an estimator for the common variance, given by TSS/( n – 1). By Theorem 7.3, TSS/ σ 2 is chi–square with n – 1 degrees of freedom.
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This note was uploaded on 03/18/2009 for the course STA 4321 taught by Professor Staff during the Spring '08 term at University of Florida.

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ISM_chapter13 - Chapter 13: The Analysis of Variance 13.1 2...

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