# Chap10 - Solutions to End-of-Section and Chapter Review...

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Unformatted text preview: Solutions to End-of-Section and Chapter Review Problems 201 CHAPTER 10 10.1 (a) df A = c – 1 = 5 – 1 = 4 (b) df W = n – c = 35 – 5 = 30 (c) df T = n – 1 = 35 – 1 = 34 10.2 (a) SSW = SST – SSA = 210 – 60 = 150 (b) MSA = SSA c –1 = 60 5 – 1 = 15 (c) MSW = SSW n – c = 150 35 – 5 = 5 (d) F = MSA MSW = 15 5 = 3 10.3 (a) Source df SS MS F Among groups 4 60 15 3.00 Within groups 30 150 5 Total 34 210 (b) F 4, 30 = 2.69 (c) Decision rule: If F > 2.69, reject H . (d) Decision: Since F calc = 3.00 is above the critical bound F = 2.69, reject H . 10.4 (a) df A = c – 1 = 3 – 1 = 2 (b) df W = n – c = 21 – 3 = 18 (c) df T = n – 1 = 21 – 1 = 20 10.5 Source df SS MS F Among groups 4 – 1 =3 80 x (3) = 240 80 80 ÷ 20 = 4 Within groups 32 – 4 = 28 560 560 ÷ 28 = 20 Total 32 – 1 = 31 240 + 560 = 800 201 202 Chapter 10: Analysis of Variance 10.6 (a)-(b) Decision rule: If F > 2.95, reject H . (c) Decision: Since F calc = 4.00 is above the critical bound of F = 2.95, reject H . There is enough evidence to conclude that the four group means are not all the same. (d) To perform the Tukey-Kramer procedure, we use c = 4 degrees of freedom in the numerator and n – c = 32 – 4 = 28 degrees of freedom in the denominator. (e) Since Table E.7 does not contain a value for 4 and 28 degrees of freedom, we will use the next larger (and more conservative) value for 4 and 24 degrees of freedom as an upper bound. That value is 3.90. (f) critical range = Q U ( c , n – c ) ⋅ MSW 2 ⋅ 1 n j + 1 n j ' = 3.90 ⋅ 20 2 ⋅ 1 8 + 1 8 = 6.166 10.7 (a) 1 2 3 4 5 : H μ μ μ μ μ = = = = where 1 = Wendy, 2 = McDonald, 3 = Checkers, 4 = Burger King, 5 = Long John Silver 1 : Not all are equal j H μ where j = 1, 2, 3, 4, 5 Decision Rule: If p-value < 0.05, reject H . One-Way ANOVA Source DF SS MS F Statistic p Value Between 4 6536 1634 12.51148545 3.24067E-08 Within 95 12407 130.6 Total 99 18943 Since the p- value is virtually 0, reject H 0. There is sufficient evidence to conclude that there is a difference in the average drive-through time of the 5 chains. (b) To determine which of the means are significantly different from one another, we use the Tukey-Kramer procedure to establish the critical range: Q U(c, n – c) = Q U (5, 95) = 3.92 critical range = ( , ) ' 1 1 130.6 1 1 3.92 2 2 20 20 U c n c j j MSW Q n n- + = + = 10.017 202 Solutions to End-of-Section and Chapter Review Problems 203 10.7 (b) Partial PHStat output for Tukey-Kramer multiple comparision: cont. Sample Sample Absolute Group Mean Size Comparison Difference Results 1 150 20 Group 1 to Group 2 17 Means are different 2 167 20 Group 1 to Group 3 19 Means are different 3 169 20 Group 1 to Group 4 21 Means are different 4 171 20 Group 1 to Group 5 22 Means are different 5 172 20 Group 2 to Group 3 2 Means are not different Group 2 to Group 4 4 Means are not different Group 2 to Group 5 5 Means are not different Group 3 to Group 4 2...
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Chap10 - Solutions to End-of-Section and Chapter Review...

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