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The total variation for all data of Table 9-21 is v 62 42 52 c 72 102 (268)2 40 1946 1795.6 150.4
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In order to consider the two main factors, Machine and Shift, we limit our attention to the total of replication values corresponding to each combination of factors. These are arranged in Table 9-22, which thus is a twofactor table with single entries. Table 9-22 First Shift Second Shift A B C D TOTALS 24 41 32 28 125 30 44 31 38 143
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TOTALS 54 85 63 66 268
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The total variation for Table 9-22, which we shall call the subtotal variation vs, is given by vs (24)2 5 1861.2 (41)2 5 1795.6 (32)2 5 65.6 (28)2 5 (30)2 5 (44)2 5 (31)2 5 (38)2 5 (268)2 40
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The variation between rows is given by vr (54)2 10 (85)2 10 (63)2 10 (66)2 10 (268)2 40 1846.6 1795.6 51.0
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The variation between columns is given by vc (125)2 20 (143)2 20 (268)2 40 1803.7 1795.6 8.1
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CHAPTER 9 Analysis of Variance
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If we now subtract from the subtotal variation vs the sum of the variations between rows and columns (vr we obtain the variation due to interaction between rows and columns. This is given by vi vs vr vc 65.6 51.0 8.1 6.5
vc),
Finally, the residual variation, which we can think of as the random or error variation ve (provided that we believe that the various days of the week do not provide any important differences), is found by subtracting the sum of the row, column, and interaction variations (i.e., the subtotal variation) from the total variation v. This yields ve v (vr vc vi) v vs 150.4 65.6 84.8
These variations are indicated in the analysis of variance, Table 9-23. The table also gives the number of degrees of freedom corresponding to each type of variation. Therefore, since there are 4 rows in Table 9-22, the variation due to rows has 4 1 3 degrees of freedom, while the variation due to the 2 columns has 2 1 1 degrees of freedom. To find the degrees of freedom due to interaction, we note that there are 8 entries in Table 9-22. Therefore, the total degrees of freedom is 8 1 7. Since 3 of these are due to rows and 1 to columns, the remainder, 7 (3 1) 3, is due to interaction. Since there are 40 entries in the original Table 9-21, the total degrees of freedom is 40 1 39. Therefore, the degrees of freedom due to random or residual variation is 39 7 32. Table 9-23 Variation Rows (Machines), vr 51.0 Column (Shifts), vc 8.1 Interaction, vi 6.5 Subtotal, vs 65.6 Random or Residual, ve 84.8 v Total, 150.4 Degrees of Freedom 3 1 3 7 Mean Square
F 17.0 2.65 8.1 2.65 2.167 2.65 6.42 3.06 0.817
17.0 8.1 2.167
32 39
To proceed further, we must first determine if there is any significant interaction between the basic factors (i.e., rows and columns of Table 9-22). From Table 9-23 we see that for interaction F 0.817, which shows that interaction is not significant, i.e., we cannot reject hypothesis H(3) of page 323. Following the rules on 0 page 323, we see that the computed F for rows is 6.42. Since F0.95 2.90 for 3, 32 degrees of freedom we can reject the hypothesis H(1) that the rows have equal means. This is equivalent to saying that at the 0.05 level, we 0 can conclude that the machines are not equally effective. For 1, 32 degrees of freedom F0.95 4.15. Then since the computed F for columns is 3.06, we cannot reject the hypothesis H(2) that the columns have equal means. This is equivalent to saying that at the 0.05 level there is 0 no significant difference between shifts. If we choose to analyze the results by pooling the interaction and residual variations as recommended by some statisticians, we find for the pooled variation and pooled degrees of freedom (df) vi ve 6.5 84.8 91.3 and 3 32 35, respectively, which lead to a pooled variance of 91.3>35 2.61. Use of this value instead of 2.65 for the denominator of F in Table 9-23 does not affect the conclusions reached above.
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