ssrs 2012 barcode font Analysis of Variance in Software

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CHAPTER 9 Analysis of Variance
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4. GRAECO-LATIN SQUARES. If it is necessary to control three sources of error or variability, a Graeco-Latin square is used, as indicated in Fig. 9-4. Such a square is essentially two Latin squares superimposed on each other, with Latin letters A, B, C, D used for one square while Greek letters, a, b, g, d are used for the other squares. The additional requirement that must be met is that each Latin letter must be used once and only once with each Greek letter. When this property is met the square is said to be orthogonal.
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One-way classification or one-factor experiments 9.1. Prove that
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We have xjk
(xjk
xj.) # x)2 #
(xj. # a (xjk
x). Then squaring and summing over j and k, we find # xj.)2 # # a (xj.
a (xjk
x)2 #
2 a (xjk
xj.)(xj. # #
x) #
To prove the required result, we must show that the last summation is zero. In order to do this, we proceed as follows.
a (xjk
xj.)(xj. # #
x) #
j 1 a
# a (xj. # a (xj.
x) B a (xjk #
b k 1 b k 1
since xj. #
g b 1xjk. k
x) B a xjk #
xj.) R #
bxj. R #
9.2. Verify that (a) t
(a) (b) (c) Since tj.
abx, (b) tj. #
bxj., (c) g j tj. #
t tj. a xjk
a xjk
ab
abx, using the notation on page 315. #
1 x ab a jk j,k abx # bxj. #
g k xjk, we have a tj.
1 b a xjk b k t
a a xjk
abx #
by part (a). 9.3. Verify the shortcut formulas (10) through (12), page 315.
We have v a (xjk
x)2 # 2x a xjk #
a A x2 jk
2xxjk ##
x2 B #
a x2 jk
abx2 # abx2 #
a x2 jk
2x(abx) # # abx2 # t2 ab
a x2 jk
a x2 jk
CHAPTER 9 Analysis of Variance
using Problem 9.2(a) in the third and last lines above. Similarly vb # a (xj.
x)2 # 2x a xj. # #
# a A xj.
2xxj. ##
x2 B #
# a xj.
1 t2 b2 ja ka j. 1 1 1 t2 b ja j. 1 1 t2 b ja j. 1
tj. 2 a b j,k
j,k a b
abx2 # abx2 # abx2 #
tj. 2x a # b
2x(abx) # #
abx2 # t2 ab
using Problem 9.2(b) in the third line and Problem 9.2(a) in the last line. Finally, (12) follows from the fact that v vb vw or vw v vb.
9.4. Table 9-7 shows the yields in bushels per acre of a certain variety of wheat grown in a particular type of soil treated with chemicals A, B, or C. Find (a) the mean yields for the different treatments, (b) the grand mean for all treatments, (c) the total variation, (d) the variation between treatments, (e) the variation within treatments. Use the long method. Table 9-7 A B C 48 47 49 49 49 51 50 48 50 49 48 50 3 2 4 Table 9-8 4 4 6 5 3 5 4 3 5
To simplify the arithmetic, we may subtract some suitable number, say, 45, from all the data without affecting the values of the variations. We then obtain the data of Table 9-8. (a) The treatment (row) means for Table 9-8 are given, respectively, by x1. # 1 (3 4 4 5 4) 4, x2. # 1 (2 4 4 3 3) 3, x3. # 1 (4 4 6 5 5) 5
Therefore, the mean yields, obtained by adding 45 to these, are 49, 48, and 50 bushels per acre for A, B, and C, respectively. (b) x # 1 (3 12 4 5 4 2 4 3 3 4 6 5 5) 4
Therefore, the grand mean for the original set of data is 45
(c) Total variation v a (xjk
49 bushels per acre.
x)2 (4 4)2 4)2 (4 (6 4)2 4)2 4)2 (5 (3 (5 4)2 4)2 4)2 (4 (3 (5 4)2 4)2 4)2
(3 (2 (4 14 (d)
Variation between treatments
b a (xj. #
x)2 # 4)2 14 (3 8 4)2 6 (5 4)2] 8
4[(4 (e) Variation within treatments vw v vb
CHAPTER 9 Analysis of Variance
Another method
vw a (xjk
xj.)2 # (4 (4 (6 4)2 3)2 5)2 (5 (3 (5 4)2 3)2 5)2 (4 (3 (5 4)2 3)2 5)2
(3 (2 (4 6
4)2 3)2 5)2
9.5. Referring to Problem 9.4, find an unbiased estimate of the population variance s2 from (a) the variation between treatments under the null hypothesis of equal treatment means, (b) the variation within treatments.
(a) (b)
^2 ^2
vb a 1 vw a(b 1)
8 3 3(4 1 6 1)
4 2 3
9.6. Referring to Problem 9.4, can we reject the null hypothesis of equal means at (a) the 0.05 significance level (b) the 0.01 significance level
We have with a 1 3 1 2 and a(b 1) F 3(4
sb sw 1)
4 2>3
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