Introduction to Probability and Statistics for Engineers and Scientists

10.3One-Way Analysis of Variance 449

The value of the test statistic is thus

TS=

863.3335/2

1991.5785/12

=2.60

Now, from Table A4 in the Appendix, we see thatF2,12,.05=3.89. Hence, because
the value of the test statistic does not exceed 3.89, we cannot, at the 5 percent level of
significance, reject the null hypothesis that the gasolines give equal mileage. ■

Let us now show that

E[SSb/(m−1)]≥σ^2

with equality only whenH 0 is true. So, we must show that

E

[m

∑

i= 1

(Xi.−X..)^2 /(m−1)

]

≥σ^2 /n

with equality only whenH 0 is true. To verify this, letμ.=

∑m
i= 1 μi/mbe the average
of the means. Also, fori=1,...,m, let

Yi=Xi.−μi+μ.

BecauseXi.is normal with meanμi and varianceσ^2 /n, it follows thatYi is normal
with meanμ.and varianceσ^2 /n. Consequently,Y 1 ,...,Ymconstitutes a sample from
a normal population having varianceσ^2 /n. Let

Y ̄ =Y.=

∑m

i= 1

Yi/m=X..−μ.+μ.=X..

be the average of these variables. Now,

Xi.−X..=Yi+μi−μ.−Y.

Consequently,

E

[m

∑

i= 1

(Xi.−X..)^2

]

=E

[m

∑

i= 1

(Yi−Y.+μi−μ.)^2

]

=E

[m

∑

i= 1

[(Yi−Y.)^2 +(μi−μ.)^2 +2(μi−μ.)(Yi−Y.)

]

=E

[m

∑

i= 1

(Yi−Y.)^2

]

+

∑m

i= 1

(μi−μ.)^2 + 2

∑m

i= 1

(μi−μ.)E[Yi−Y.]