STEPS FOR CONDUCTING AN ANOVA FOR A TWO-FACTOR FACTORIAL EXPERIMENT (PART II)
1. Create an ANOVA table for the treatments (only for the main effects)
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2. Use the F-ratio to test the hypothesis that the treatment means are equal
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3. Create an ANOVA table for the main and interaction effects
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4. Use the F-ratio to test the hypothesis that the factors do not interact |
5.Use the F-ratio to test the two hypotheses that the mean response is the same at each level of factor A and factor B. |
6. Compare the means (multiple comparison) |
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ACTIVITY
Suppose that USGA tests four different brands (A, B, C, D) of galf balls and two different clubs (driver, five-iron) in a completely randomized design. Each of the eight Brand-Club combinations (treatments) is randomly and independently assigned to four experimental units, each experimental unit consisting of a specific position in the sequence of hits by Iron Byron. The distance response is recorded for each of the 32 hits and the results are shown below.
Brand
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Club
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A
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B
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C
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D
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Driver
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226.4
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238.3
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240.5
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219.8
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232.6
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231.7
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246.9
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228.7
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234.0
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227.7
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240.3
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232.9
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220.7
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237.2
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244.7
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237.6
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Five-iron
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163.8
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184.4
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179.0
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157.8
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179.4
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180.6
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168.0
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161.8
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168.6
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179.5
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165.2
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162.1
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173.4
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186.2
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156.5
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160.3
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Use the following steps and Statlets window given above, to carry out ANOVA procedure to appropriate for this two-factor factorial experiment.
Next we will carry out two-way ANOVA.
STEP 3, 4, & 5 of the Conduction ANOVA for the Two-factor Factorial Experiment:
Distance Brand Club 226.4 A DRIVER 232.6 A DRIVER 234.0 A DRIVER 220.7 A DRIVER 163.8 A FIVE-IRON 179.4 A FIVE-IRON 168.6 A FIVE-IRON 173.4 A FIVE-IRON 238.3 B DRIVER 231.7 B DRIVER 227.7 B DRIVER 237.2 B DRIVER 184.4 B FIVE-IRON 180.6 B FIVE-IRON 179.5 B FIVE-IRON 186.2 B FIVE-IRON 240.5 C DRIVER 246.9 C DRIVER 240.3 C DRIVER 244.7 C DRIVER 179.0 C FIVE-IRON 168.0 C FIVE-IRON 165.2 C FIVE-IRON 156.5 C FIVE-IRON 219.8 D DRIVER 228.7 D DRIVER 232.9 D DRIVER 237.6 D DRIVER 157.8 D FIVE-IRON 161.8 D FIVE-IRON 162.1 D FIVE-IRON 160.3 D FIVE-IRON