Page 94 - 4660

Estimation

One-Sided Confidence Bounds. The confidence interval in (2.20) gives both a lower confidence
bound and an upper confidence bound for µ. Thus, it provides a two-sided CI. It is also possible to
obtain one-sided confidence bounds for µ by setting either l = −∞ or u = ∞ and replacing z α/2
by z α .
A 100(1 − α)% upper-confidence bound for µ is
√
µ ≤ u = ¯x + z α σ/ n (2.22)

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

√
¯ x − z α σ/ n = l ≤ µ. (2.23)

Example 2.7. One-Sided Confidence Bound: the same data for impact testing from
Example 2.5 are used to construct a lower, one-sided 95% confidence interval for
the mean impact energy. Recall that ¯x = 64.46, σ = 1J, and n = 10. The interval is
σ 1
¯ x − z α √ ≤ µ, 64.46 − 1.64 √ ≤ µ, 63.94 ≤ µ,
n 10
Practical Interpretation: The lower limit for the two-sided interval in Example
2.5 was 63.84. Because z α < z α/2 , the lower limit of a one-sided interval is always
greater than the lower limit of a two-sided interval of equal confidence. The
one- sided interval does not bound µ from above so that it still achieves 95%
confidence with a slightly greater lower limit. If our interest is only in the
lower limit for µ, then the one-sided interval is preferred because it provides
equal confidence with a greater lower limit. Similarly, a one-sided upper limit
is always less than a two-sided upper limit of equal confidence. ,

General Method to Derive a Confidence Interval. It is easy to give a general method for finding
a confidence interval for an unknown parameter θ. Let X 1 , X 2 , . . . , X n be a random sample of n
observations. Suppose we can find a statistic g(X 1 , X 2 , . . . , X n ; θ) with the following properties:
1. g(X 1 , X 2 , . . . , X n ; θ) depends on both the sample and µ.
2. The probability distribution of g(X 1 , X 2 , . . . , X n ; θ) does not depend on µ or any other
unknown parameter.
In the case considered in this section, the parameter θ = µ. The random variable
√
¯
g(X 1 , X 2 , . . . , X n ; µ) = (X − µ)/(σ/ n)

and satisfies both conditions above; it depends on the sample and on µ, and it has a standard
normal distribution since σ is known. Now one must find constants C L and C U so that

P[C L ≤ g(X 1 , X 2 , . . . , X n ; θ) ≤ C u ] = 1 − α (2.24)

Because of property 2, CL and CU do not depend on θ. In our example, C L = −z α/2 and C U = z α/2 .
Finally, you must manipulate the inequalities in the probability statement so that

P[L(X 1 , X 2 , . . . , X n ) ≤ θ ≤ U(X 1 , X 2 , . . . , X n )] = 1 − α. (2.25)

This gives L(X 1 , X 2 , . . . , X n ) and U(X 1 , X 2 , . . . , X n ) as the lower and upper confidence limits
defining the 100(1 − α)% confidence interval for θ. The quantity g(X 1 , X 2 , . . . , X n ; θ) is often
called a ”pivotal quantity” because we pivot on this quantity in (2.24) to produce (2.25). In our
√
¯
example, we manipulated the pivotal quantity (X − µ)/(σ n) to obtain
√
√
¯
¯
L(X 1 , X 2 , . . . , X n ) = X − z α/2 σ/ n and U(X 1 , X 2 , . . . , X n ) = X + z α/2 σ/ n.
94