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G-2WAY_f03

# G-2WAY_f03 - 2-Way Factorial Design ANOVA Example 10.9...

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2-Way Factorial Design ANOVA Example 10.9 Brand A is 1, B is 2, C is 3 and D is 4. Club 1 is the driver is the 5-iron. MTB > read 'mcclave/c10t11.dat' c1-c3 Entering data from file: mcclave/c10t11.dat 32 rows read. MTB > name c1 'brand' c2 'club' c3 'distance' MTB > anova 'distance' = 'brand'|'club'; SUBC> means 'brand'|'club'. Factor Type Levels Values brand fixed 4 1 2 3 4 club fixed 2 1 2 Analysis of Variance for distance Source DF SS MS F P brand 3 800.7 266.9 7.79 0.001 club 1 32093.1 32093.1 936.75 0.000 brand*club 3 766.0 255.3 7.45 0.001 Error 24 822.2 34.3 Total 31 34482.0 MEANS brand N distance 1 8 199.86 2 8 208.20 3 8 205.14 4 8 195.12 club N distance 1 16 233.75 2 16 170.41 brand club N distance 1 1 4 228.43 1 2 4 171.30 2 1 4 233.73 2 2 4 182.68 3 1 4 243.10 3 2 4 167.18 4 1 4 229.75 4 2 4 160.50 MINITAB does not compute SST for us. We have to do it ourselves. SST = SSA + SSB + SSA(AB) = 800.7 + 32093.1 + 766.0 = 33659.8. MST = 33659.8/7 = 4808.5 and F treatment = 4898.5/34.3 = 140.2

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Since F treatment is 140.2 which is greater than F .10, 7, 24 which is 1.98, there is enough evidence to indicate that the mean distances differ for at least 2 of the 8 brand and club combinations. A test to determine whether the brand of ball and the type of club interact to affect the distance should be done. Since F brand*club = 7.45 with p-value = 0.001, there is enough evidence to indicate that the brand of golf ball and the type of club used interact to affect the distance traveled. Therefore, we need to compare all possible pairs of means. Since p = 8, there are 28 possible pairs to test. Since this would be a very boring and time consuming task to do by hand, I have analyzed the data using a one-way ANOVA with 8 treatments and used Bonferroni with α /c = .10/8 to compare the 28 possible pairs of means. MTB > let c4 = 4*c2 +c1 -4 MTB > name c4 'treat' MTB > print c1 c2 c4 ROW brand club treat 1 1 1 1 Treatment 1 is brand A with a driver. (Ad) 5 1 2 5 Treatment 5 is brand A with a 5-iron. (A5) 9 2 1 2 Treatment 2 is brand B with a driver. (Bd)
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G-2WAY_f03 - 2-Way Factorial Design ANOVA Example 10.9...

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