Introduction to Probability and Statistics for Engineers and Scientists

442 Chapter 10:Analysis of Variance

estimator, then the resulting expression remains chi-square but with a degree of freedom
that is reduced by 1 for each parameter that is replaced by its estimator.
For an illustration of the preceding, consider the case where all the means are known
to be equal; that is,

E[Xi]=μ, i=1,...,N

Thusk=1, because there is only one parameter that needs to be estimated. Substituting
X, the estimator of the common meanμ, forμiin Equation 10.2.1, results in the quantity

∑N

i= 1

(Xi−X)^2 /σ^2 (10.2.2)

and the conclusion is that this quantity is a chi-square random variable withN− 1
degrees of freedom. But in this case where all the means are equal, it follows that the
dataX 1 ,...,XNconstitute a sample from a normal population, and thus Equation 10.2 is
equal to (N−1)S^2 /σ^2 , whereS^2 is the sample variance. In other words, the conclusion
in this case is just the well-known result (see Section 6.5.2) that (N−1)S^2 /σ^2 is a
chi-square random variable withN−1 degrees of freedom.

10.3 One-Way Analysis of Variance

Considermindependent samples, each of sizen, where the members of theith sample —
Xi 1 ,Xi 2 ,...,Xin— are normal random variables with unknown meanμiand unknown
varianceσ^2. That is,

Xij∼N(μi,σ^2 ), i=1,...,m, j=1,...,n

We will be interested in testing

H 0 :μ 1 =μ 2 = ··· =μm

versus

H 1 :not all the means are equal

That is, we will be testing the null hypothesis that all the population means are equal
against the alternative that at least two of them differ. One way of thinking about this is
to imagine that we havemdifferent treatments, where the result of applying treatment
ion an item is a normal random variable with meanμiand varianceσ 2. We are then
interested in testing the hypothesis that all treatments have the same effect, by applying
each treatment to a (different) sample ofnitems and then analyzing the result.