Introduction to Probability and Statistics for Engineers and Scientists

450 Chapter 10:Analysis of Variance

=(m−1)σ^2 /n+

∑m

i= 1

(μi−μ.)^2 + 2

∑m

i= 1

(μi−μ.)E[Yi−Y.]

=(m−1)σ^2 /n+

∑m

i= 1

(μi−μ.)^2

where the next to last equality follows because the sample variance

∑m
i= 1 (Yi−Y.)^2 /(m−1)
is an unbiased estimator of its population varianceσ^2 /nand the final equality because
E[Yi]−E[Y.]=μ.−μ.=0. Dividing bym−1 gives that

E

[m

∑

i= 1

(Xi.−X..)^2 /(m−1)

]

=σ^2 /n+

∑m

i= 1

(μi−μ.)^2 /(m−1)

and the result follows because

∑m

i= 1 (μi−μ.)

(^2) ≥0, with equality only when all theμi
are equal.
Table 10.2 sums up the results of this section.
TABLE 10.2 One-Way ANOVA Table
Source of Degrees of Value of Test
Variation Sum of Squares Freedom Statistic
Between samples SSb=n
∑m
i= 1 (Xi.−X..)^2 m−^1
Within samples SSW=
∑m
i= 1
∑n
j= 1 (Xij−Xi.)^2 nm−m
TS=SSSSWb/(/(nmm−−1)m)
Significance levelαtest:
rejectH 0 ifTS≥Fm−1,nm−m,α
do not reject otherwise
IfTS=v, thenp-value =P{Fm−1,nm−m≥v}

10.3.1 Multiple Comparisons of Sample Means

When the null hypothesis of equal means is rejected, we are often interested in a comparison
of the different sample meansμ 1 ,...,μm. One procedure that is often used for this
purpose is known as theT-method. For a specified value ofα, this procedure gives joint
confidence intervals for all the

(m

2

)
differencesμi−μj,i=j,i,j=1,...,m, such that
with probability 1−αall of the confidence intervals will contain their respective quantities
μi−μj. TheT-method is based on the following result:

With probability 1−α, for everyi=j

Xi.−Xj.−W<μi−μj<Xi.−Xj.+W