254 Chapter 7: Parameter Estimation
or, equivalently,
P
X−Y−zα/2√
σ 12
n+σ 22
m<μ 1 −μ 2 <X−Y+zα/2√
σ 12
n+σ 22
m
=^1 −αHence, ifXandYare observed to equalxandy, respectively, then a 100(1−α) two-
sided confidence interval estimate forμ 1 −μ 2 is
μ 1 −μ 2 ∈
x−y−zα/2√
σ 12
n+σ 22
m,x−y+zα/2√
σ 12
n+σ 22
m
One-sided confidence intervals forμ 1 −μ 2 are obtained in a similar fashion, and we
leave it for the reader to verify that a 100(1−α) percent one-sided interval is given by
μ 1 −μ 2 ∈(
−∞,x−y+zα√
σ 12 /n+σ 22 /m)Program 7.4.1 will compute both one- and two-sided confidence intervals forμ 1 −μ 2.EXAMPLE 7.4a Two different types of electrical cable insulation have recently been tested
to determine the voltage level at which failures tend to occur. When specimens were
subjected to an increasing voltage stress in a laboratory experiment, failures for the two
types of cable insulation occurred at the following voltages:
Type A Type B
36 54 52 60
44 52 64 44
41 37 38 48
53 51 68 46
38 44 66 70
36 35 52 62
34 44Suppose that it is known that the amount of voltage that cables having type A insulation can
withstand is normally distributed with unknown meanμAand known varianceσA^2 =40,
whereas the corresponding distribution for type B insulation is normal with unknown
meanμBand known varianceσB^2 =100. Determine a 95 percent confidence interval
forμA−μB. Determine a value that we can assert, with 95 percent confidence, exceeds
μA−μB.
SOLUTION We run Program 7.4.1 to obtain the solution (see Figure 7.4). ■