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9.17. Suppose that, in carrying out the experiment of Problem 9.16, the numbers of miles per gallon are as given in Table 9-29. Use analysis of variance to determine if there are any significant differences at the 0.05 level.
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We first obtain row and column totals as shown in Table 9-30. Table 9-29 DRIVERS 2 3 Ab 16 Ba 18 Cd 11 Dg 16 Dd 16 Cg 11 Bb 21 Aa 15 Table 9-30 TOTALS 4 Ca 14 Db 15 Ag 16 Bd 23 TOTALS Bg 19 Ad 15 Da 14 Cb 16 64 Ab 16 Ba 18 Cd 11 Dg 16 61 Dd 16 Cg 11 Bb 21 Aa 15 63 Ca 14 Db 15 Ag 16 Bd 23 68 65 59 62 70 256
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1 1 CARS 2 3 4 Bg 19 Ad 15 Da 14 Cb 16
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CHAPTER 9 Analysis of Variance
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Then we obtain totals for each Latin letter and for each Greek letter, as follows: A total: 15 B total: 19 C total: 16 D total: 14 a total: 14 b total: 16 g total: 19 d total: 15 16 18 11 16 18 16 16 11 15 21 11 16 15 21 11 16 16 23 14 15 14 15 16 23 62 81 52 61 61 68 62 65
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We now compute the variations corresponding to all of these, using the shortcut method. Rows: Columns: Gasolines: (A, B, C, D) Roads: (a, b, g, d) The total variation is (19)2 (16)2 (16)2 c (15)2 (23)2 (256) 16 7.50 4244 4096 148.00 (65)2 4 (64)2 4 (62)2 4 (61)2 4 (59)2 4 (61)2 4 (81)2 4 (68)2 4 (62)2 4 (63)2 4 (52)2 4 (62)2 4 (70)2 4 (68)2 4 (61)2 4 (65)2 4 (256)2 16 (256)2 16 (256)2 16 (256)2 16 4112.50 4102.50 4207.50 4096 4096 4096 16.50 6.50 111.50
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so that the variation due to error is 148.00 16.50 6.50 111.50 6.00
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The results are shown in the analysis of variance, Table 9-31. The total number of degrees of freedom is n2 1 for an n n square. Rows, columns, Latin letters, and Greek letters each have n 1 degrees of freedom. Therefore, the degrees of freedom for error is n2 1 4(n 1) (n 1)(n 3). In our case n Table 9-31 Variation Rows (Cars), 16.50 Columns (Drivers), 6.50 Gasolines (A, B, C, D), 111.50 Roads (a, b, g, d), 7.50 Error, 6.00 Total, 148.00 Degrees of Freedom 3 3 3 3 3 15 Mean Square 5.500 2.167 37.167 2.500 2.000 5.500 2.000 2.167 2.000 37.167 2.000 2.500 2.000 F 2.75 1.08 18.6 12.5
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CHAPTER 9 Analysis of Variance
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We have F0.95,3,3 9.28 and F0.99,3,3 29.5. Therefore, we can reject the hypothesis that the gasolines are the same at the 0.05 level but not at the 0.1 level.
Miscellaneous problems 9.18. Prove that a aj 0 [(15), page 316].
The treatment population means are given by mj
a a a
m am
aj. Hence,
a a a
a mj
a aj
a aj
a mj
a aj
where we have used the definition m
(gmj)>a. It follows that gaj
9.19. Derive (a) equation (17), (b) equation (16), on page 316.
(a) By definition we have Vw a (Xjk
j,k a b
# Xj.)2 # Xj.)2 R
1 b a B a (Xjk bk 1 j 1
b a S2 j
where S2 is the sample variance for the jth treatment, as defined by (15), 5. Then, since the sample j size is b,
E(Vw)
b a E(S2) j ba
j 1 a j 1
a(b using (16) of 5. (b) By definition, Vb
1)s2
s2
# b a (Xj.
# X)2 # # 2bX a Xj.
j 1 a
# j. b a X2
# abX2
# j. b a X2
# abX2
since # j. a X2 # X Then, omitting the summation index, we have (1) E(Vb) # j. b a E(X2) # abE(X)2
CHAPTER 9 Analysis of Variance
Now for any random variable U, E(U2) (2) (3)
Var(U)
[E(U)]2. Therefore, # [E(Xj.)]2 # [E(X)]2
# j. E(X2) # E(X2)
# Var (Xj.) # Var (X)
But since the treatment populations are normal, with means mj and common variance s2, we have from Theorem 5-4, page 156: (4) (5) (6) (7) baB (a (a # Var (Xj.) # Var (X) # E(Xj.) mj # E(X) s2 b s2 ab m m aj)2 R ab B m2 R
Using the results (2) through (7), plus the result of Problem 9.18, in (1) we have E(Vb) s2 b (m s2 ab
b a (m
aj)2
abm2
1)s2 1)s2
abm2
2bm a aj
b a a2 j
abm2
b a a2 j
9.20. Prove Theorem 9-1, page 317.
As shown in Problem 9.19(a),
b a S2 j
Vw s2
a j 1
a s2
bS2 j
where S2 is the sample variance for samples of size b drawn from the population of treatment j. By Theorem 5-6, j page 158, bS2 >s2 has a chi-square distribution with b 1 degrees of freedom. Then, since the variances S2 are j j independent, we conclude from Theorem 4-4, page 121, that Vw >s2 is chi square distributed with a(b 1) degrees of freedom.
9.21. In Problem 9.13 we assumed that there were no significant differences in replications, i.e., the different days of the week. Can we support this conclusion at a (a) 0.05, (b) 0.01 significance level
If there is any variation due to the replications, it is included in what was called the residual or random, error, ve 84.8, in Table 9-23. To find the variation due to replication, we use the column totals in Table 9-21, obtaining vrep (57)2 8 1807 (51)2 8 1795.6 (54)2 8 11.4 (47)2 8 (59)2 8 (268)2 40
Since there are 5 replications, the number of degrees of freedom associated with this variation is 5 1 4. The residual variation after subtracting variation due to replication is vre 84.8 11.4 73.4. The other variations are the same as in Table 9-23. The final analysis of variance table, taking into account replications, is Table 9-32. From the table we see that the computed F for replication is 1.09. But since F0.95 2.71 for 4, 28 degrees of freedom, we can conclude that there is no significant variation at the 0.05 level (and therefore at the 0.01 level) due to replications, i.e., the days of the week are not significant. The conclusions concerning Machines and Shifts are the same as those obtained in Problem 9.13.
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