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is always a best (unbiased) estimate of s2 regardless of whether H0 is true or not. On the other hand, from (16) and (18) we see that only if H0 is true will we have E Vb a 1 s2 E S2
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Vb a 1
V ab 1
(22)
CHAPTER 9 Analysis of Variance
provide unbiased estimates of s2. If H0 is not true, however, then we have from (16) E(S 2) b
1a j
a2 j
(23)
Distributions of the Variations
Using Theorem 4-4, page 115, we can prove the following fundamental theorems concerning the distributions of the variations Vw, Vb, and V. Theorem 9-1 Vw >s2 is chi-square distributed with a(b 1) degrees of freedom. 1 and ab 1 Theorem 9-2 Under the null hypothesis H0, Vb >s2 and V>s2 are chi-square distributed with a degrees of freedom, respectively.
It is important to emphasize that Theorem 9-1 is valid whether or not we assume H0, while Theorem 9-2 is valid only if H0 is assumed.
The F Test for the Null Hypothesis of Equal Means
If the null hypothesis H0 is not true, i.e., if the treatment means are not equal, we see from (23) that we can ex^ pect S 2 to be greater than s2, with the effect becoming more pronounced as the discrepancy between means inb ^ creases. On the other hand, from (19) and (20) we can expect S 2 to be equal to s2 regardless of whether the w ^ ^ means are equal or not. It follows that a good statistic for testing the hypothesis H0 is provided by S 2>S 2 . If this b w is significantly large, we can conclude that there is a significant difference between treatment means and thus reject H0. Otherwise we can either accept H0 or reserve judgment pending further analysis. In order to use this statistic, we must know its distribution. This is provided in the following theorem, which is a consequence of Theorem 5-8, page 159. Theorem 9-3 The statistic F S 2>S 2 has the F distribution with a b w
1 and a(b
1) degrees of freedom.
Theorem 9-3 enables us to test the null hypothesis at some specified significance level using a one-tailed test of the F distribution.
Analysis of Variance Tables
The calculations required for the above test are summarized in Table 9-2, which is called an analysis of variance table. In practice we would compute v and vb using either the long method, (3) and (8), or the short method, (10) and (11), and then compute vw v vb. It should be noted that the degrees of freedom for the total variation, i.e., ab 1, is equal to the sum of the degrees of freedom for the between-treatments and within-treatments variations. Table 9-2 Variation Between Treatments, vb b a (xj. #
Degrees of Freedom
Mean Square vb a vw a(b 1) 1
F sb sw with a 1, a(b 1) degrees of freedom
^2 ^2
x)2 #
Within Treatments, vw v vb Total, v vb
vw x)2 #
a (xjk
CHAPTER 9 Analysis of Variance
Modifications for Unequal Numbers of Observations
In case the treatments 1, . . . , a have different numbers of observations equal to n1, . . . , na, respectively, the above results are easily modified. We therefore obtain v vb vw a (xjk
x)2 # x)2 #
a x2 jk
t2 n x)2 # t2 j. a nj
(24) t2 n
# a (xj.
# a nj (xj.
(25) (26)
where g j,k denotes the summation over k from 1 to nj and then over j from 1 to a, n g j nj is the total number of observations in all treatments, t is the sum of all observations, tj. is the sum of all values in the jth treatment, and g j is the sum from j 1 to a. The analysis of variance table for this case is given in Table 9-3.
Table 9-3 Variation Between Treatments, vb # a nj (xj.
Degrees of Freedom
Mean Square vb a 1
F sb sw with a 1, n a degrees of freedom
^2 ^2
x)2 #
Within Treatments, vw v vb Total, v vb
vw n a
vw x)2 #
a (xjk
Two-Way Classification or Two-Factor Experiments
The ideas of analysis of variance for one-way classification or one-factor experiments can be generalized. We illustrate the procedure for two-way classification or two-factor experiments.
EXAMPLE 9.2 Suppose that an agricultural experiment consists of examining the yields per acre of 4 different varieties of wheat, where each variety is grown on 5 different plots of land. Then a total of (4)(5) 20 plots are needed. It is convenient in such case to combine plots into blocks, say, 4 plots to a block, with a different variety of wheat grown on each plot within a block. Therefore, 5 blocks would be required here. In this case there are two classifications or factors, since there may be differences in yield per acre due to (i) the particular type of wheat grown or (ii) the particular block used (which may involve different soil fertility, etc.).
By analogy with the agricultural experiment of Example 9.2, we often refer to the two classifications or factors in an experiment as treatments and blocks, but of course we could simply refer to them as Factor 1 and Factor 2, etc.
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