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Estimation

The end-points or bounds l and u are called the lower- and upper-confidence limits, respectively,
and 1 − α is called the confidence coefficient.
√
¯
In our problem situation, because Z = (X − µ)/(σ n) has a standard normal distribution,
we may write
{ ¯ }
X − µ
P −z α/2 ≤ √ ≤ z α/2 = 1 − α.
σ n
Simplifying this expression we obtain next
{ }
σ σ
¯
¯
P X − z α/2 √ ≤ µ ≤ X + z α/2 √ = 1 − α (2.19)
n n
From consideration (2.19), the lower and upper limits of the inequalities in (2.19) are the lower-
and upper-confidence limits L and U, respectively. This leads to the following definition.

Definition 2.1. If ¯x is the sample mean of a random sample of size n from a
2
normal population with known variance σ , a 100(1 − α)% CI on µ is given by
√ √
¯ x − z α/2 σ/ n ≤ µ ≤ ¯x + z α/2 σ/ n (2.20)

where z α/2 is the upper 100α/2 percentage point of the standard normal
distribution. ✓

Example 2.5. ASTM Standard E23 defines standard test methods for notched bar
impact testing of metallic materials. The Charpy V-notch (CVN) technique measures
impact energy and is often used to determine whether or not a material experiences
a ductile-to-brittle transition with decreasing temperature. Ten measurements of
0
impact energy (J) on specimens of A238 steel cut at 60 C are as follows: 64.1, 64.7,
64.5, 64.6, 64.5, 64.3, 64.6, 64.8, 64.2, and 64.3. Assume that impact energy is normally
distributed with σ = 1J. We want to find a 95% CI for µ, the mean impact energy.
The required quantities are z α/2 = z 0.025 = 1.96, n = 10, σ = 1, and ¯x = 64.46. The
resulting 95% CI is found from Equation (2.20) as follows:

σ σ
¯ x − z α/2 √ ≤ µ ≤¯] + z α/2 √ ,
n n
1 1
64.46 − 1.96√ ≤ µ ≤ 64.46 + 1.96√ ,
10 10
63.84 ≤ µ ≤ 65.08

Practical Interpretation: Based on the sample data, a range of highly plausible
0
values for mean impact energy for A238 steel at 60 C is 63.84J ≤ µ ≤ 65.08J. ,

Interpreting a Confidence Interval. How does one interpret a confidence interval? In the
impact energy estimation problem in previous example, the 95% CI is 63.84 ≤ µ ≤ 65.08, so it is
tempting to conclude that µ is within this interval with probability 0.95. However, with a little
reflection, it’s easy to see that this cannot be correct; the true value of µ is unknown and the
statement 63.84 ≤ µ ≤ 65.08 is either correct (true with probability 1) or incorrect (false with
probability 1). The correct interpretation lies in the realization that a CI is a random interval
because in the probability statement defining the end-points of the interval (2.17), L and U are
random variables. Consequently, the correct interpretation of a 100(1 − α)% CI depends on the
relative frequency view of probability. Specifically, if an infinite number of random samples are
collected and a 100(1 − α)% confidence interval for µ is computed from each sample,
100(1 − α)% of these intervals will contain the true value of µ.

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