Introduction to Probability and Statistics for Engineers and Scientists

252 Chapter 7: Parameter Estimation

Hence whenS^2 =s^2 , a 100(1−α) percent confidence interval forσ^2 is

{

(n−1)s^2

χα^2 /2,n− 1

,

(n−1)s^2

χ 12 −α/2,n− 1

}

EXAMPLE 7.3h A standardized procedure is expected to produce washers with very small
deviation in their thicknesses. Suppose that 10 such washers were chosen and measured.
If the thicknesses of these washers were, in inches,

.123 .133

.124 .125

.126 .128

.120 .124

.130 .126

what is a 90 percent confidence interval for the standard deviation of the thickness of a
washer produced by this procedure?

SOLUTION A computation gives that

S^2 =1.366× 10 −^5

Becauseχ.05,9^2 =16.917 andχ.95,9^2 =3.334, and because

9 ×1.366× 10 −^5

16.917

=7.267× 10 −^6 ,

9 ×1.366× 10 −^5

3.334

=36.875× 10 −^6

TABLE 7.1 100(1−α)Percent Confidence Intervals
X 1 ,...,Xn∼N(μ,σ^2 )

X=

∑n

i= 1

Xi/n, S=

√√

√√∑n

i= 1

(Xi−X)^2 /(n−1)

Assumption Parameter Confidence Interval Lower Interval Upper Interval

σ^2 known μ X±zα/2√σn

(

−∞,X+zα√σn

)(

X+zα√σn,∞

)

σ^2 unknown μ X±tα/2,n− 1 √Sn

(

−∞,X+tα,n− 1 √Sn

)(

X−tα,n− 1 √Sn,∞

)

μunknown σ^2

(

(n−1)S^2

χα^2 /2,n− 1

,(n−1)S

2

χ^21 −α/2,n− 1

)(

0,(n−1)S

2

χ 12 −α,n− 1

)(

(n−1)S^2

χα^2 ,n− 1

,∞

)