- Chapter 17 Analysis of Variance 17.1 Given the Analysis of Variance table compute mean squares for between and for within groups Compute the F

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Unformatted text preview: Chapter 17: Analysis of Variance 17.1 Given the Analysis of Variance table, compute mean squares for between and for within groups. Compute the F ratio and test the hypothesis that the group means are equal. 1 2 3 4 5 1 : , : H H otherwise μ μ μ μ μ = = = = , , 1 SSG SSW MSG MSG MSW F k n k MSW = = =-- 1000 750 250 , , 4 15 50 MSG MSW F = = = = 5.00 4,15,.05 4,15,.01 3.06, 4.89 F F = = , therefore, reject H at the 1% level. 17.2 Given the Analysis of Variance table, compute mean squares for between and for within groups. Compute the F ratio and test the hypothesis that the group means are equal. 1 2 3 4 1 : , : H H otherwise μ μ μ μ = = = , , 1 SSG SSW MSG MSG MSW F k n k MSW = = =-- 879 798 293 , , 3 16 49.875 MSG MSW F = = = = 5.875 3,16,.05 3,16,.01 3.24, 5.29 F F = = , Therefore, reject H at the 1% level. 17.3 Given the Analysis of Variance table, compute mean squares for between and for within groups. Compute the F ratio and test the hypothesis that the group means are equal. 1 2 3 1 : , : H H otherwise μ μ μ = = , , 1 SSG SSW MSG MSG MSW F k n k MSW = = =-- 1000 743 500 , , 2 15 49.5333 MSG MSW F = = = = 10.09 2,15,.05 2,15,.01 3.68, 6.36 F F = = , Therefore, reject H at the 1% level. 142 Statistics for Business & Economics, 6 th edition 17.4 a. 1 2 3 62, 53, 52 x x x = = = n=16, SSW = 1028 + 1044 + 1536 = 3608 SSG = 7(62 – 56.0625) 2 + 6(53 – 56.0625) 2 + 6(52 – 56.0625) 2 = 340.9375 SST = 3948.9375 b. Complete the anova table One-way ANOVA: SodaSales versus CanColor Analysis of Variance for SodaSale Source DF SS MS F P CanColor 2 341 170 0.61 0.556 Error 13 3608 278 Total 15 3949 Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev -+---------+---------+--------- +----- 1 6 62.00 14.34 (------------*-----------) 2 5 53.00 16.16 (------------*-------------) 3 5 52.00 19.60 (------------*-------------) -+---------+---------+--------- +----- Pooled StDev = 16.66 36 48 60 72 1 2 3 1 : , : H H otherwise μ μ μ = = 2,13,.05 3.81 F = , do not reject H at the 5% level. 17.5 a. 1 2 3 4 73, 86, 72, 72 x x x x = = = = n=23, SSW = 228 + 312 + 374 + 428 = 1342 SSG = 5(73 – 75.8696) 2 + 6(86 – 75.8696) 2 + 6(72 – 75.8696) 2 + 6(72 – 75.8696) 2 = 836.6087, SST = 2178.6087 b. Complete the anova table One-way ANOVA: Scores versus TA Analysis of Variance for Scores Source DF SS MS F P TA 3 836.6 278.9 3.95 0.024 Error 19 1342.0 70.6 Total 22 2178.6 Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ---------+---------+--------- +------- 1 5 73.000 7.550 (---------*---------) 2 6 86.000 7.899 (-------*--------) 3 6 72.000 8.649 (--------*--------) 4 6 72.000 9.252 (--------*--------) ---------+---------+--------- +------- Pooled StDev = 8.404 72.0 80.0 88.0 1 2 3 1 : , : H H otherwise μ μ μ = = Chapter 17: Analysis of Variance 143 3,19,.05 3.13 F = , reject H 0 at the 5% level. 144 Statistics for Business & Economics, 6 th edition 17.6 a....
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This note was uploaded on 04/26/2010 for the course ECO 329 taught by Professor K during the Spring '08 term at University of Texas at Austin.

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- Chapter 17 Analysis of Variance 17.1 Given the Analysis of Variance table compute mean squares for between and for within groups Compute the F

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