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8-2 CONFIDENCE INTERVAL ON THE MEAN OF A NORMAL DISTRIBUTION, VARIANCE KNOWN 253

As increases, the required sample size n increases for a fixed desired length 2E and
specified confidence.
As the level of confidence increases, the required sample size n increases for fixed
desired length 2E and standard deviation .

8-2.3 One-Sided Confidence Bounds

The confidence interval in Equation 8-7 gives both a lower confidence bound and an upper
confidence bound for . Thus it provides a two-sided CI. It is also possible to obtain one-sided
confidence bounds for by setting either l
or u
and replacing z 2 by z .

Definition
A 100(1 )% upper-confidence bound for is

u x z 1n (8-9)

and a 100(1 )% lower-confidence bound for is

x z 1n l (8-10)

8-2.4 General Method to Derive a Confidence Interval

It is easy to give a general method for finding a confidence interval for an unknown parame-
ter . Let X , X , p , X be a random sample of n observations. Suppose we can find a statistic
1
n
2
g(X , X , p , X ; ) with the following properties:
2
n
1
1. g(X , X , p , X ; ) depends on both the sample and .
n
1
2
2. The probability distribution of g(X , X , p , X ; ) does not depend on or any other
1
n
2
unknown parameter.
In the case considered in this section, the parameter . The random variable g(X , X 2 , p ,
1
; ) 1X 2 1 1n2 and satisfies both conditions above; it depends on the sample and
X n
on , and it has a standard normal distribution since is known. Now one must find constants
C and C so that
U
L
P3C L g 1X 1 , X 2 , p , X n ; 2 C U 4 1 (8-11)
and
U
L
Because of property 2, C and C do not depend on . In our example, C L z 2
C U z 2 . Finally, you must manipulate the inequalities in the probability statement so that
P3L1X , X , p , X 2 U1X , X , p , X 24 1 (8-12)
n
n
2
1
1
2
This gives L(X , X , p , X ) and U(X , X , p , X ) as the lower and upper confidence limits
n
n
2
1
2
1
defining the 100(1 )% confidence interval for . The quantity g(X , X , p , X ; ) is
2
n
1
often called a “pivotal quantity’’ because we pivot on this quantity in Equation 8-11 to pro-
duce Equation 8-12. In our example, we manipulated the pivotal quantity 1X 2 1 1n2
to obtain L 1X , X , p , X 2 X z 2 1n and U 1X , X , p , X 2 X z 2 1n.
n
2
1
n
2
1