# hw8 - STAT 350 – Spring 2009 Homework#8 – SOLUTION...

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Unformatted text preview: STAT 350 – Spring 2009 Homework #8 – SOLUTION covers through Lecture L Note: to obtain a p-value for an F-statistic for this assignment, you should use Excel. For example, if your F-statistic is 12.39 and df1 = 3 and df2 = 5. In Excel, type: =Fdist(12.39,3,5) Excel will return: 0.009434 Your p-value is 0.009434 Text Exercises Chapter 9, Problems #6(a-f), 10, 24(b) 9.6 a. 3.69 d. 10.29 b. 4.82 e. 3.10 c. 6.63 f. 0.95 9.10 Source Brand Error Total df 4 45 49 SS 39.2192 28.8 68.0192 MS 9.8048 0.64 F 15.32 9.24 b. The variances for the three groups are: 5.708, 2.799, 7.990. The ratio of the largest to smallest is 7.990/2.799 = 2.855. This ratio is less than 4, so the assumption of equal variances is reasonable. Homework #8 - Solution Page 1 of 1 Then answer the following problems: 1. For this problem, use the data summarized below. Group ni xi 5 59.2 A 5 64.9 B 5 65.8 C 5 69.5 D 5 70.8 E si 5.2 5.1 6.1 5.9 5.5 a. Give the ANOVA table for this data (including the p-value). Be sure to show your work. 5 ⎛n ⎞ = ∑ ⎜ i ⎟ xi i =1 ⎝ n ⎠ ⎛5⎞ = ⎜ ⎟ 59.2 ⎝ 25 ⎠ = 66.04 x ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ + ⎜ ⎟ 64.9 + ⎜ ⎟ 65.8 + ⎜ ⎟ 69.5 + ⎜ ⎟ 70.8 ⎝ 25 ⎠ ⎝ 25 ⎠ ⎝ 25 ⎠ ⎝ 25 ⎠ 5 SSTr = ∑ ni ( xi − x ) 2 i =1 = 5 ( 59.2 − 66.04 ) + 5 ( 64.9 − 66.04 ) + 5 ( 65.8 − 66.04 ) 2 2 +5 ( 69.5 − 66.04 ) + 5 ( 70.8 − 66.04 ) 2 2 2 = 413.86 5 SSE = ∑ ( ni − 1) si2 i =1 = ( 5 − 1) 5.22 + ( 5 − 1) 5.12 + ( 5 − 1) 6.12 + ( 5 − 1) 5.92 + ( 5 − 1) 5.52 = 621.28 ANOVA Table Source Trt error total df 4 20 24 SS MS F p 413.86 103.465 3.330704 0.030317147 621.28 31.064 1035.14 b. What is the critical value of the test statistic? The numerator degrees of freedom are 4 The denominator degrees of freedom are 20 For α = 0.05, the critical value is 2.87 (see Appendix Table VIII, page 575) Homework #8 - Solution Page 2 of 2 c. Conduct Tukey’s Multiple Comparison Procedure and present your results using the line method as given in class. qα = 4.23 (see Table IXa on p. 577; there are 5 groups and the error df = 20) T = qα MSE 31.064 = 4.23 = 10.5435 ni 5 The following table gives the absolute value of the differences in the sample means for each pair of groups. I have highlighted those cells that exceed 10.5435 (T) GROUP A B C D E Mean 59.2 64.9 65.8 69.5 70.8 A 59.2 0 5.7 6.6 10.3 11.6 B 64.9 5.7 0 0.9 4.6 5.9 C 65.8 6.6 0.9 0 3.7 5 D 69.5 10.3 4.6 3.7 0 1.3 E 70.8 11.6 5.9 5 1.3 0 Answer: A 59.2 Homework #8 - Solution B 64.9 C 65.8 D 69.5 E 70.8 Page 3 of 3 2. For this problem, use the data summarized below. Group ni xi 4 59.2 A 6 64.9 B 5 65.8 C 7 69.5 D 7 70.8 E si 5.2 5.1 6.1 5.9 5.5 a. Give the ANOVA table for this data (including the p-value). Be sure to show your work. 5 ⎛n ⎞ = ∑ ⎜ i ⎟ xi x i =1 ⎝ n ⎠ ⎛4⎞ ⎛6⎞ ⎛5⎞ ⎛7⎞ ⎛7⎞ = ⎜ ⎟ 59.2 + ⎜ ⎟ 64.9 + ⎜ ⎟ 65.8 + ⎜ ⎟ 69.5 + ⎜ ⎟ 70.8 ⎝ 29 ⎠ ⎝ 29 ⎠ ⎝ 29 ⎠ ⎝ 29 ⎠ ⎝ 29 ⎠ = 66.80345 5 SSTr = ∑ ni ( xi − x ) 2 i =1 = 4 ( 59.2 − 66.80345 ) + 6 ( 64.9 − 66.80345 ) + 5 ( 65.8 − 66.80345 ) 2 2 +7 ( 69.5 − 66.80345 ) + 7 ( 70.8 − 66.80345 ) 2 2 2 = 420.7297 5 SSE = ∑ ( ni − 1) si2 i =1 = ( 4 − 1) 5.22 + ( 6 − 1) 5.12 + ( 5 − 1) 6.12 + ( 7 − 1) 5.92 + ( 7 − 1) 5.52 = 750.37 ANOVA Table Source Trt error total df 4 24 28 SS MS F p 420.7297 105.1824 3.364178 0.025431136 750.37 31.26542 1171.1 b. What is the critical value of the test statistic? The numerator degrees of freedom are 4 The denominator degrees of freedom are 24 For α = 0.05, the critical value is 2.78 (see Appendix Table VIII, page 575) Homework #8 - Solution Page 4 of 4 c. Assume that Group A is a control. Conduct Dunnett’s Multiple Comparison Procedure to determine whether any of the other 4 treatments have significantly different means than the control (Group A). Critical value for Dunnett’s Method is 2.61 (Appendix Table X, page 579, k=5, n = 29) Group n xi T xi − x A Significantly Different A 4 59.2 n/a n/a 5.7 6.6 10.3 *** 11.6 *** ⎛1 1⎞ 2.61 31.26542 ⎜ + ⎟ ⎝6 4⎠ B 6 64.9 = 9.420349 ⎛1 1⎞ 2.61 31.26542 ⎜ + ⎟ ⎝5 4⎠ C 5 65.8 = 9.789914 ⎛1 1⎞ 2.61 31.26542 ⎜ + ⎟ ⎝7 4⎠ D 7 69.5 = 9.147237 ⎛1 1⎞ 2.61 31.26542 ⎜ + ⎟ ⎝7 4⎠ E 7 70.8 = 9.147237 Based on Dunnett’s Multiple Comparison Procedure, the means for Groups D and E are significantly different from the mean of Group A. Homework #8 - Solution Page 5 of 5 ...
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