7.4Estimating the Difference in Means of Two Normal Populations 253
it follows that, with confidence .90,
σ^2 ∈(7.267× 10 −^6 , 36.875× 10 −^6 )Taking square roots yields that, with confidence .90,
σ∈(2.696× 10 −^3 , 6.072× 10 −^3 ) ■One-sided confidence intervals forσ^2 are obtained by similar reasoning and are
presented in Table 7.1, which sums up the results of this section.
7.4Estimating the Difference in Means of Two Normal Populations
LetX 1 ,X 2 ,...,Xnbe a sample of sizenfrom a normal population having meanμ 1 and
varianceσ 12 and letY 1 ,...,Ymbe a sample of sizemfrom a different normal population
having meanμ 2 and varianceσ 22 and suppose that the two samples are independent of
each other. We are interested in estimatingμ 1 −μ 2.
SinceX=
∑n
i= 1 Xi/nandY=∑m
i= 1 Yi/mare the maximum likelihood estimators of
μ 1 andμ 2 it seems intuitive (and can be proven) thatX−Yis the maximum likelihood
estimator ofμ 1 −μ 2.
To obtain a confidence interval estimator, we need the distribution ofX−Y. Because
X∼N(μ 1 ,σ 12 /n)
Y∼N(μ 2 ,σ 22 /m)it follows from the fact that the sum of independent normal random variables is also
normal, that
X−Y∼N(
μ 1 −μ 2 ,σ 12
n+σ 22
m)Hence, assumingσ 12 andσ 22 are known, we have that
X−Y−(μ 1 −μ 2 )
√
σ 12
n+σ 22
m∼N(0, 1) (7.4.1)and so
P
−zα/2<X−Y−(μ 1 −μ 2 )
√
σ 12
n+σ 22
m<zα/2
= 1 −α