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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 = = = = where 1 = Wendy, 2 = McDonald, 3 = Checkers, 4 = Burger King, 5 = Long John Silver 1 : Not all are equal j H 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...
View Full Document

Page1 / 34

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online