258 Chapter 7: Parameter Estimation
Therefore, when the data result in the valuesX =x,Y =y,Sp=sp, we obtain the
following 100(1−α) percent confidence interval forμ 1 −μ 2 :
(
x−y−tα/2,n+m− 2 sp√
1/n+1/m, x−y+tα/2,n+m− 2 sp√
1/n+1/m)
(7.4.4)One-sided confidence intervals are similarly obtained.
Program 7.4.2 can be used to obtain both one- and two-sided confidence intervals for
the difference in means in two normal populations having unknown but equal variances.
EXAMPLE 7.4b There are two different techniques a given manufacturer can employ to
produce batteries. A random selection of 12 batteries produced by technique I and of 14
produced by technique II resulted in the following capacities (in ampere hours):
Technique I Technique II
140 132 144 134
136 142 132 130
138 150 136 146
150 154 140 128
152 136 128 131
144 142 150 137
130 135Determine a 90 percent level two-sided confidence interval for the difference in means,
assuming a common variance. Also determine a 95 percent upper confidence interval for
μI−μII.
SOLUTION We run Program 7.4.2 to obtain the solution (see Figure 7.5). ■
REMARK
The confidence interval given by Equation 7.4.4 was obtained under the assumption that
the population variances are equal; withσ^2 as their common value, it follows that
X−Y−(μ 1 −μ 2 )
√
σ^2 /n+σ^2 /m=X−Y−(μ 1 −μ 2 )
σ√
1/n+1/mhas a standard normal distribution. However, sinceσ^2 is unknown this result cannot be
immediately applied to obtain a confidence interval;σ^2 must first be estimated. To do so,
note that both sample variances are estimators ofσ^2 ; moreover, sinceS 12 hasn−1 degrees
of freedom andS 22 hasm−1, the appropriate estimator is to take a weighted average of the
two sample variances, with the weights proportional to these degrees of freedom. That is,