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9 degrees of freedom. 4.26. Since F 6 F0.95, we can reject F0.99, we cannot
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(a) Referring to Appendix F, with n1 2 and n2 9, we see that F0.95 the null hypothesis of equal means at the 0.05 level.
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(b) Referring to Appendix F, with n1 2 and n2 9, we see that F0.99 8.02. Since F 6 reject the null hypothesis of equal means at the 0.01 level. The analysis of variance table for Problems 9.4 through 9.6 is shown in Table 9-9.
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Table 9-9
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Variation Between Treatments, vb 8 Within Treatments, vw v vb 14 8 6 Total, v 14 Degrees of Freedom a 1 2 Mean Square
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(3)(3) (3)(4) 11
6 with 2, 9 degrees of freedom
9.7. Use the shortcut formulas (10) through (12) to obtain the results of Problem 9.4.
(a) We have a xjk
Also t 3 4 5 4 2 4 3 3 4 6 5 5 48
CHAPTER 9 Analysis of Variance
t2 ab (48)2 (3)(4) 206 192 14
Therefore,
a x2 jk
206 (b) The totals of the rows are t1. t2. t3. Also Then vb t 1 t2 b a j. j 1 (162 4 (c) vw 3 2 4 16 4 4 6 12 t2 ab 122 v 5 3 5
4 3 5 20
16 12 20 48
202) vb 14
(48)2 (3)(4) 8 6
It is convenient to arrange the data as in Table 9-10. Table 9-10 tj. A B C 3 2 4 4 4 6 a x2 jk
t2 j. 256 144 400 a tj.
5 3 5 206
4 3 5 t
16 12 20 a tj.
v vb
206 1 (800) 4
(48)2 (3)(4) (48)2 (3)(4)
14 8
The results agree with those obtained in Problem 9.4 and from this point the analysis proceeds as before.
9.8. A company wishes to purchase one of five different machines A, B, C, D, E. In an experiment designed to decide whether there is a difference in performance of the machines, five experienced operators each work on the machines for equal times. Table 9-11 shows the number of units produced. Test the hypothesis that there is no difference among the machines at the (a) 0.05, (b) 0.01 level of significance.
Table 9-11 A B C D E 68 72 60 48 64 72 52 82 61 65 75 63 65 57 70 42 55 77 64 68 53 48 75 50 53
CHAPTER 9 Analysis of Variance
Table 9-12 tj. A B C D E 8 12 0 12 4 12 8 22 1 5 a x2 jk 15 3 6 3 10 2356 18 5 17 4 8 7 2 15 10 7 10 0 60 20 20 70 t2 j. 100 0 3600 400 400 4500
Subtract a suitable number, say, 60, from all the data to obtain Table 9-12. Then v vb We now form Table 9-13. 2356 1 (4500) 5 (70)2 (5)(4) (70)2 (5)(4) 2356 900 245 245 2111 655
Table 9-13
Variation Between Treatments, vc 655 Within Treatments, vw 1456 Total, 2111 Degrees of Freedom
Mean Square sb 655 4 1456 (5)(4) 163.75 F
F sb sw
a a(b
1 1)
4 5(4) 20
For 4, 20 degrees of freedom we have F0.95 2.87. Therefore, we cannot reject the null hypothesis at a 0.05 level and therefore certainly cannot reject it at a 0.01 level.
Modifications for unequal numbers of observations 9.9. Table 9-14 shows the lifetimes in hours of samples from three different types of television tubes manufactured by a company. Using the long method, test at (a) the 0.05, (b) the 0.01 significance level whether there is a difference in the three types.
Table 9-14
Sample 1 Sample 2 Sample 3 407 404 410 411 406 408 409 408 406 405 408 402
CHAPTER 9 Analysis of Variance
Table 9-15 Total Sample 1 Sample 2 Sample 3 7 4 10 11 6 8 x # 9 8 6 5 8 grand mean 84 12 2 27 25 32 7 Mean 9 5 8
It is convenient to subtract a suitable number, say, 400, obtaining Table 9-15. In this table we have indicated the row totals, the sample or group means, and the grand mean. We then have v vb a (xjk
x)2 # x)2 #
(11 x)2 # 5(7
# a (xj.
# a nj(xj.
3(9 vw v vb 72 36
7)2 36
We can also obtain vw directly by observing that it is equal to (7 9)2 (11 (2 9)2 5)2 (9 (10 9)2 8)2 (4 (8 5)2 8)2 (6 (6 5)2 8)2 (8 (8 5)2 8)2 (5 5)2
The data can be summarized in the analysis of variance table, Table 9-16.
Table 9-16
Variation vb vw 36 36 Degrees of Freedom a n 1 a 2 9 Mean Square
F sb sw
36 2 36 9
18 4
18 4 4.5
Now for 2 and 9 degrees of freedom we find from Appendix F that F0.95 4.26, F0.99 8.02. Therefore, we can reject the hypothesis of equal means (i.e., there is no difference in the three types of tubes) at the 0.05 level but not at the 0.01 level.
9.10. Work Problem 9.9 by using the shortcut formulas included in (24), (25), and (26).
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