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9.14. Work Problem 9.13 if the 0.01 level is used.
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At this level there is still no appreciable interaction, so we can proceed further.
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CHAPTER 9 Analysis of Variance
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Since F0.99 4.47 for 3, 32 df, and since the computed F for rows is 6.42, we can conclude that even at the 0.01 level the machines are not equally effective. Since F0.99 7.51 for 1, 32 df, and since the computed F for columns is 3.06, we can conclude that at the 0.01 level there is no significant difference in shifts
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Latin squares 9.15. A farmer wishes to test the effects of four different fertilizers, A, B, C, D, on the yield of wheat. In order to eliminate sources of error due to variability in soil fertility, he uses the fertilizers in a Latin square arrangement as indicated in Table 9-24, where the numbers indicate yields in bushels per unit area. Perform an analysis of variance to determine if there is a significant difference between the fertilizers at the (a) 0.05, (b) 0.01 levels of significance.
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Table 9-25 TOTALS Table 9-24 A 18 D 22 B 15 C 22 C 21 B 12 A 20 D 21 D 25 A 15 C 23 B 10 B 11 C 19 D 24 A 17 TOTALS A 18 D 22 B 15 C 22 77 C 21 B 12 A 20 D 21 74 D 25 A 15 C 23 B 10 73 B 11 C 19 D 24 A 17 71 75 68 82 70 295
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Table 9-26 A TOTAL 70 B 48 C 85 D 92 295
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We first obtain totals for rows and columns as indicated in Table 9-25. We also obtain total yields for each of the fertilizers as shown in Table 9-26. The total variation and the variations for rows, columns, and treatments are then obtained as usual. We find Total variation v (18)2 5769 Variation between rows vr (21)2 5439.06 (25)2 c (10)2 (17)2 (295)2 16
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329.94 (295)2 16
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(75)2 (68)2 (82)2 (70)2 4 4 4 4 5468.25 5439.06 29.19 (77)2 (74)2 (73)2 (71)2 4 4 4 4 5443.75 5439.06 4.69 (70)2 (48)2 (85)2 (92)2 4 4 4 4 5723.25 5439.06 284.19
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Variation between columns
(295)2 16
Variation between treatments
(295)2 16
The analysis of variance is now shown in Table 9-27. (a) Since F0.95,3,6 4.76, we can reject at the 0.05 level the hypothesis that there are equal row means. It follows that at the 0.05 level there is a difference in the fertility of the soil from one row to another. Since the F value for columns is less than 1, we conclude that there is no difference in soil fertility in the columns. Since the F value for treatments is 47.9 4.76, we can conclude that there is a difference between fertilizers.
CHAPTER 9 Analysis of Variance
Table 9-27 Variation Rows, 29.19 Columns, 4.69 Treatments, 284.19 Residuals, 11.87 Total, 329.94 Degrees of Freedom 3 3 3 6 15 Mean Square 9.73 1.563 94.73 1.978 F 4.92 0.79 47.9
(b) Since F0.99,3,6 9.78, we can accept the hypothesis that there is no difference in soil fertility in the rows (or the columns) at a 0.01 level of significance. However, we must still conclude that there is a difference between fertilizers at the 0.01 level.
Graeco-Latin squares 9.16. It is of interest to determine if there is any difference in mileage per gallon between gasolines A, B, C, D. Design an experiment using four different drivers, four different cars, and four different roads.
Since the same number (four) of gasolines, drivers, cars, and roads are involved, we can use a Graeco-Latin square. Suppose that the different cars are represented by the rows and the different drivers by the columns, as indicated in Table 9-28. We now assign the different gasolines A, B, C, D to rows and columns at random, subject only to the requirement that each letter appear just once in each row and just once in each column. Therefore, each driver will have an opportunity to drive each car and use each type of gasoline (and no car will be driven twice with the same gasoline). We now assign at random the four roads to be used, denoted by a, b, g, d, subjecting them to the same requirement imposed on the Latin letters. Therefore, each driver will have the opportunity to drive along each of the roads also. One possible arrangement is that given in Table 9-28. Table 9-28 DRIVERS 1 2 3 1 CARS 2 3 4 Bg Aa Da Cb Ab Ba Cd Dg Dd Cg Bb Aa