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We therefore have v vb vw a x2 jk
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Using these, the analysis of variance then proceeds as in Problem 9.9.
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CHAPTER 9 Analysis of Variance
Two-way classification or two-factor experiments 9.11. Table 9-17 shows the yields per acre of four different plant crops grown on lots treated with three different types of fertilizer. Using the long method, test at the 0.01 level of significance whether (a) there is a significant difference in yield per acre due to fertilizers, (b) there is a significant difference in yield per acre due to crops.
Table 9-17 Crop I Fertilizer A Fertilizer B Fertilizer C 4.5 8.8 5.9 Crop II 6.4 7.8 6.8 Crop III 7.2 9.6 5.7 Crop IV 6.7 7.0 5.2
Compute the row totals and row means, as well as the column totals and column means and grand mean, as shown in Table 9-18. Table 9-18 Crop I Fertilizer A Fertilizer B Fertilizer C Column Totals 4.5 8.8 5.9 19.2 Crop II 6.4 7.8 6.8 21.0 Crop III 7.2 9.6 5.7 22.5 Crop IV 6.7 7.0 5.2 18.9 Row Totals 24.8 33.2 23.6 Row Means 6.2 8.3 5.9
Grand total 81.6 Grand mean 6.8
Column Means
variation of row means from grand mean 4[(6.2 6.8)2 (8.3 6.8)2 (5.9 6.8)2] 13.68
variation of column means from grand mean 3[(6.4 6.8)2 (7.0 6.8)2 (7.5 6.8)2 (6.3 6.8)2] 2.82
total variation (4.5 (8.8 (5.9 23.08 6.8)2 6.8)2 6.8)2 (6.4 (7.8 (6.8 6.8)2 6.8)2 6.8)2 (7.2 (9.6 (5.7 6.8)2 6.8)2 6.8)2 (6.7 (7.0 (5.2 6.8)2 6.8)2 6.8)2
random variation
This leads to the analysis of variance in Table 9-19. At the 0.05 level of significance with 2, 6 degrees of freedom, F0.95 5.14. Then, since 6.24 5.14, we can reject the hypothesis that the row means are equal and conclude that at the 0.05 level there is a significant difference in yield due to fertilizers. Since the F value corresponding to differences in column means is less than 1, we can conclude that there is no significant difference in yield due to crops.
CHAPTER 9 Analysis of Variance
Table 9-19
Variation vr 13.68 Degrees of Freedom 2 Mean Square
^ s r >s 2 6.24 e df: 2, 6 ^2
vc ve v
2.82 6.58 23.08
3 6 11
0.94 1.097
^ s c >s 2 0.86 e df: 3, 6 ^2
9.12. Use the short computational formulas to obtain the results of Problem 9.11.
We have from Table 9-18: a x2 jk
(4.5)2 24.8 (24.8)2 (19.2)2
(6.4)2 33.2
c 23.6
(5.2)2 8.16
2 a tj.
(33.2)2 (21.0)2
(23.6)2 (22.5)2
2274.24 (18.9)2 1673.10
a t2 .k Then v vr vc ve
a x2 jk
t2 ab t2 ab t2 ab vc
23.08 13.68 2.82 6.58
1 t2 b a j. 1 2 a a t.k v vr
1 (2274.24) 4 1 (1673.10) 3 23.08 13.68
554.88 554.88 2.82
in agreement with Problem 9.11.
Two-factor experiments with replication 9.13. A manufacturer wishes to determine the effectiveness of four types of machines, A, B, C, D, in the production of bolts. To accomplish this, the number of defective bolts produced by each machine on the days of a given week are obtained for each of two shifts. The results are indicated in Table 9-20. Perform an analysis of variance to test at the 0.05 level of significance whether there is (a) a difference in machines, (b) a difference in shifts.
Table 9-20
FIRST SHIFT Mon A B C D 6 10 7 8 Tues 4 8 5 4 Wed 5 7 6 6 Thurs 5 7 5 5 Fri 4 9 9 5 Mon 5 7 9 5 SECOND SHIFT Tues 7 9 7 7 Wed 4 12 5 9 Thurs 6 8 4 7 Fri 8 8 6 10
CHAPTER 9 Analysis of Variance
The data can be equivalently organized as in Table 9-21. In this table the two main factors, namely, Machine and Shift, are indicated. Note that for each machine two shifts have been indicated. The days of the week can be considered as replicates or repetitions of performance of each machine for the two shifts.
Table 9-21
FACTOR I Machine A B C D FACTOR II Shift b 1 2 Mon 6 5 10 7 7 9 8 5 57 Tues 4 7 8 9 5 7 4 7 51 REPLICATES Wed 5 4 7 12 6 5 6 9 54 Thurs 5 6 7 8 5 4 5 7 47 Fri 4 8 9 8 9 6 5 10 59 TOTALS 24 30 41 44 32 31 28 38 268
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