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Confidence interval on the mean of a normal distribution, variance known

Now in practice, we obtain only one random sample and compute one confidence interval.
Since this interval either will or will not contain the true value of µ, it is not reasonable to attach a
probability level to this specific event. The appropriate statement is that the observed interval
[l, u] brackets the true value of µ with confidence 100(1 − α). This statement has a frequency
interpretation; that is, we don’t know if the statement is true for this specific sample, but the
method used to obtain the interval [l, u] yields correct statements 100(1 − α)% of the time.
The length of a confidence interval is a measure of the precision of estimation. From the
preceeding discussion, we see that precision is inversely related to the confidence level. It is
desirable to obtain a confidence interval that is short enough for decision-making purposes and
that also has adequate confidence. One way to achieve this is by choosing the sample size n to
be large enough to give a CI of specified length or precision with prescribed confidence.

√
Choice of Sample Size. The precision of the confidence interval in Equation (2.20) is 2z α/2 σ/ n.
√
Thismeansthatinusing ¯xtoestimateµ,theerrorE = |¯x−µ|islessthanorequaltoz α/2 σ/ nwith
confidence 100(1 − α). In situations where the sample size can be controlled, we can choose n so
thatweare100(1−α)percentconfidentthattheerrorinestimatingµislessthanaspecifiedbound
√
on the error E. The appropriate sample size is found by choosing n such that z α/2 σ/ n = E.
Solving this equation gives the following formula for n.

Sample Size for Specified Error on the Mean. If ¯x is used as an estimate of µ, we can be 100(1−
α)% confident that the error |¯x − µ| will not exceed a specified amount E when the sample size is

z α/2 σ
( ) 2
n = . (2.21)
E
If the right-hand side of (2.21) is not an integer, it must be rounded up. This will ensure that the
level of confidence does not fall below 100(1 − α)%. Notice that 2E is the length of the resulting
confidence interval.

Example 2.6. To illustrate the use of this procedure, consider the CVN test
described in Example (2.5), and suppose that we wanted to determine how many
specimens must be tested to ensure that the 95% CI on µ for A238 steel cut at
0
60 C has a length of at most 1.0J. Since the bound on error in estimation E is
one-half of the length of the CI, to determine n we use (2.21) with E = 0.5, σ = 1,
and z α/2 = 1.96. The required sample size is 16,

z α/2 σ 1.96 · 1
( ) 2
2
n = = [ ] = 15.37
E 0.5
and because n must be an integer, the required sample size is n = 16. ,

Notice the general relationship between sample size, desired length of the confidence interval
2E, confidence level 100(1 − α), and standard deviation σ :

• As the desired length of the interval 2E decreases, the required sample size n increases for
a fixed value of σ and specified confidence.

• As σ 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 σ.

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