One Factor ANOVA F Test Example You want to see if three different golf clubs

# One factor anova f test example you want to see if

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One-Factor ANOVA F Test ExampleYou want to see if three different golf clubs yield different distances. You randomly select five measurements from trials on an automated driving machine for each club. At the .05 significance level, is there a difference in mean distance?Club 1Club 2Club 3 254 234 200 263 218 222 241 235 197 237 227 206 251 216 204
. Chap 12-24 One-Factor ANOVA Example: Scatter Diagram 270 260 250 240 230 220 210 200 190 Distance 227.0 x 205.8 x 226.0 x 249.2 x 3 2 1 = = = = Club 1 Club 2 Club 3 254 234 200 263 218 222 241 235 197 237 227 206 251 216 204 Club 1 2 3 3 x 1 x 2 x x
. Chap 12-25 One-Factor ANOVA Example Computations Club 1 Club 2 Club 3 254 234 200 263 218 222 241 235 197 237 227 206 251 216 204 x 1 = 249.2 x 2 = 226.0 x 3 = 205.8 x = 227.0 n 1 = 5 n 2 = 5 n 3 = 5 n T = 15 k = 3 SSB = 5 [ (249.2 – 227) 2 + (226 – 227) 2 + (205.8 – 227) 2 ] = 4716.4 SSW = (254 – 249.2) 2 + (263 – 249.2) 2 +…+ (204 – 205.8) 2 = 1119.6 MSB = 4716.4 / (3-1) = 2358.2 MSW = 1119.6 / (15-3) = 93.3 25.275 93.3 2358.2 F = =
. Chap 12-26 F = 25.275 One-Factor ANOVA Example Solution H 0 : μ 1 = μ 2 = μ 3 H A : μ i not all equal α = .05 df 1 = 2 df 2 = 12 Test Statistic: Decision: Conclusion: Reject H 0 at α = 0.05 There is evidence that at least one μ i differs from the rest 0 α = .05 F .05 = 3.885 Reject H 0 Do not reject H 0 25.275 93.3 2358.2 MSW MSB F = = = Critical Value: F α = 3.885
. Chap 12-27 SUMMARY Groups Count Sum Average Variance Club 1 5 1246 249.2 108.2 Club 2 5 1130 226 77.5 Club 3 5 1029 205.8 94.2 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 4716.4 2 2358.2 25.275 4.99E-05 3.885 Within Groups 1119.6 12 93.3 Total 5836.0 14 ANOVA -- Single Factor: Excel Output EXCEL: tools | data analysis | ANOVA: single factor
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. Chap 12-29 Tukey-Kramer Critical Range where: q α = Value from standardized range table with k and n T - k degrees of freedom for the desired level of α MSW = Mean Square Within n i and n j = Sample sizes from populations (levels) i and j + = α j i n 1 n 1 2 MSW q Range Critical
. Chap 12-30 The Tukey-Kramer Procedure: Example1. Compute absolute mean differences: x 2. Find the q value from the table in appendix J with k and nT- k degrees of freedom for the desired level of 3.77qα= α
. Chap 12-31 The Tukey-Kramer Procedure: Example 5. All of the absolute mean differences are greater than critical range. Therefore there is a significant difference between each pair of means at 5% level of significance. 16.285 5 1 5 1 2 93.3 3.77 n 1 n 1 2 MSW q Range Critical j i α = + = + = 3. Compute Critical Range: 20.2 x x 43.4 x x 23.2 x x 3 2 3 1 2 1 = - = - = - 4. Compare:
. Chap 12-32 Tukey-Kramer in PHStat
. Chap 12-33 Randomized Complete Block ANOVA Like One-Way ANOVA, we test for equal population means (for different factor levels, for example)... ...but we want to control for possible variation from a second factor (with two or more levels) Used when more than one factor may influence the value of the dependent variable, but only one is of key interest Levels of the secondary factor are called blocks
. Chap 12-34 Partitioning the Variation Total variation can now be split into three parts: SST = Total sum of squares

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• Spring '12
• Srinivas