To understand how confidence intervals work, we will first illustrate them by the simplest case, in which the observations have a normal distribution with a known variance.

**CONFIDENCE INTERVALS FOR A SINGLE POPULATION MEAN**

To understand how confidence intervals work, we
will first illustrate them by the simplest case, in which the observations have
a normal distribution with a known variance *σ*^{2} and we want to estimate the population mean, *μ*. Then we know that the sample
mean is and its sampling
distribution has mean equal to the population mean μ and a variance *σ*^{ 2}/*n*, where *n* is the number of samples. Thus, *Z* = (– *μ*)/(* σ* /√*n*) has a standard normal distribution. We can therefore state that *P*(–1.96* *≤* Z *≤* *1.96) = 0.95 based on the standard normal
distribution. Substituting* *( – *μ*)/(*σ*/√*n*) for *Z* we obtain *P*(–1.96 ≤ ( – *μ*)/(*σ*/√*n*) ≤ 1.96) = 0.95 or *P*(–1.96*σ/*√*n *≤* *(* *–* **μ*)* *≤* *1.96*σ*/√*n*) = 0.95 or* P*(–1.96(*σ*/√*n*) –* ** *≤* *–* μ *≤* *1.96(*σ*/√*n*) – ) = 0.95. Multiplying
throughout by –1 and reversing the inequality, we find that *P*(1.96(*σ*/√*n*) + ≥ μ ≥ –1.96(*σ*/√*n*) + ) = 0.95. Rearranging
the foregoing formula, we have *P*( – 1.96*σ*/√*n* ≤ μ ≤ *X* + 1.96*σ*/√*n*) = 0.95. The confidence interval is an interpretation of this
probability statement. The confidence interval [– 1.96*σ*/√*n*, *X* + 1.96*σ*/√*n*] is a
random interval determined by the sample value of ,
*σ*, *n*, and the confidence level (e.g., 95%). is the component to this
interval that makes it random. (See Display 8.2.)

The probability statement *P* [ – 1.96(*σ*/√*n*) ≤ μ ≤ + 1.96(*σ*/√*n*)] = 0.95 says only that the probability that this random interval
includes the population mean is 0.95. This probability pertains to the
procedure for generating random confidence intervals. It does not say what will
happen to the parameter on any particular out-come. If, for example, σ is 5 and
*n* = 25 and we obtain from a sample a
sample mean of 5.96, then the outcome for the random interval is [5.96 – 1.96,
5.96 + 1.96] = [4.00, 7.92]. The population mean will either be inside or
outside the interval. If the mean μ = 7, then it is contained in the interval.
On the other hand, if μ = 8, μ is not contained in the interval.

We *cannot*
say that the probability is 0.95 that the single fixed interval [4.00, 7.92]
contains *μ*. It
either does or it does not. Instead, we say that we have 95% confidence that
such an interval would include (or cover) *μ*. This means that the process will tend to include the true value of the
parameter 95% of the time if we

The confidence interval is formed by
the following equation:

[–1.96*σ*/√*n*,* ** *+ 1.96*σ*/√*n*]

where *n* is the sample size.

were to repeat the process many times. That is to
say, if we generated 100 samples of size 25 and for each sample we generated
the confidence interval as described above, approximately 95 of the intervals
would include μ and the remaining ones would not. (See Figure 8.2.)

Why did we choose 95%? There is no strong reason.
The probability 0.95 is high and indicates we have high confidence that the
interval will include the true para-meter. However, in some situations we may
feel comfortable only with a higher confidence level such as 99%. Let “*C*” denote the *Z* value associated with a particu-lar level of confidence that
corresponds to a particular section of the normal curve. To obtain a 99%
confidence interval, we just go to the table of the standard normal
distribution to find the value *C* such
that *P*(–*C* ≤ *Z* ≤ *C*) = 0.99. We find that *C*
= 2.576. This leads to the interval [ – 2.576*σ*/√*n*, + 2.576*σ*/√*n*]. In the
example

*Figure 8.2. **The results of a computer
simulation of 20 samples of size n = 1000. We assumed that the true value of p = 0.5. At the top is
the sampling distribution of pˆ [normal, with mean p and σ =*

above where the sample mean is 5.96, σ is 5, and *n* = 25, the resulting interval would be
[5.96 – 2.576(5)/ √25, 5.96 + 2.576(5)/ √25] = [3.384, 8.536]. Compare
this to the 95% interval [4.00, 7.92].

Notice that for the same standard deviation and
sample size, increasing the con-fidence level increases the length of the
interval and also increases the chance that such intervals generated by this
prescription would contain the parameter *μ*. Note that in this case, if μ = 8, the 95% interval would not have contained
μ but the 99% interval would. This example could have been one of the 5% of
cases where a 95% confidence interval does not contain the mean but the 99%
interval does. The 99% interval has to be wider because it has to capture the
true mean in 4/5ths of the cases where the 99% confidence interval does not.
That is why the 95% interval is con-tained within the 99% interval.

We pay a price for the higher confidence in a much
wider interval. For example, by establishing an extremely wide confidence interval,
we are increasingly certain that it contains *μ*. Thus, for example, we could say with extremely high confidence that
the confidence interval for the mean age of the U.S. population is between 0
and 120 years. However, this interval would not be helpful, as we would like to
have a more precise estimate of *μ*.

If we were willing to accept a lower confidence
level such as 90%, we would obtain a value of 1.645 for *C*, where *P*(–*C* ≤ *Z* ≤ *C*) =
0.90. In that case, for the example we are considering the interval would be
[5.96 – 1.645, 5.96 + 1.645] = [4.315, 7.505]. This is a much tighter interval
that is contained within the 95% in-terval. Here we gain a tighter interval at
the price of lower confidence.

Another important point to note is the gain in
precision of the estimate with increase in sample size. This point can be
illustrated by the narrowing of the width of the confidence interval. Let us
consider the 95% confidence interval for the mean that we obtained with a
sample of size 25 and an estimated mean of 5.96. Suppose we increase the sample
size to 100 (a factor of 4 increase) and assume that we still get a sample mean
of 5.96. The 95% interval (assuming the population standard de-viation is known
to be 5) is then [5.96 – 1.96 (5/√100),
5.96 + 1.96 (5/√100)] = [5.96 – 0.98, 5.96 + 0.98] = [4.98, 6.94].

This interval is much narrower and is contained
inside the previous one. The in-terval width is 6.94 – 4.98 = 1.96 as compared
to 7.92 – 4.00 = 3.92. Notice this in-terval is exactly half the width of the
other interval. That is, if the confidence level is left unchanged and the
sample size *n* is increased by a
factor of 4, √*n* is increased by a factor of 2; because the interval width is 2(1.96)/ √*n*, the interval width is reduced by a factor of 2. Exhibit 8.1 summarizes
the critical values of the standard normal distribution for calculating
confidence intervals at various levels of confidence.

If the population standard deviation is unknown and
we want to estimate the mean, we must use the *t* distribution instead of the normal distribution. So we cal-culate
the sample standard deviation *S* and
construct the *t* score ( – *μ*)/(*S*/√*n*). Recall that this quantity has Student’s *t* distribution with *n* – 1
degrees of freedom. Note that this distribution does not depend on the unknown
parameters μ and *σ*, but it does depend on the sample size *n* through the degrees of freedom. This dif*n*.

For standard normal distributions, we
have the following critical points for two-sided confidence intervals:

*C*_{0.90}* *= 1.645

*C*_{0.95}* *= 1.960

*C*_{0.99}* *= 2.576

For a 95% confidence interval, we need to determine
*C* so that *P*(–*C* ≤ *t* ≤ *C*) =
0.95. For *n* = 25, again assume that
the sample mean is 5.96, the sample standard deviation is 5, and *n* = 25. Then the degrees of freedom are
24, and from the table for the *t*
distribution we see that *C* = 2.064.
The statement *P*(–*C* ≤ *t* ≤ *C*) = 0.95 is equivalent to *P*[ – *C*(*S*/√*n*) ≤ μ ≤ *X* + *C*(*S*/√*n*)] =
0.95. So the interval is [ –
*C*(*S*/√*n*),* X *+* C*(*S*/√n)]. Using* C *= 2.064,* S *= 5,* n *= 25, and a
sample mean of* *5.96, we find [5.96 –
2.064, 5.96 + 2.064] or [3.896, 8.024]. Display 8.3 summa-rizes the procedure
for calculating a 95% confidence interval for a population mean when the
population variance is unknown.

You should note that the interval is wider than in
the case in which we knew the variance and used the normal distribution. This
result occurs because there is extra variability in the *t* statistic due to the fact that the random quantity s is used in
place of a fixed quantity *σ*. Remember that the *t* with 24
degrees of freedom has heavier tails than the standard normal distribution;
this fact is reflected in the quantity *C*
= 2.064 in place of *C* = 1.96 for the
standard normal distribution.

Suppose we obtained the same estimates for the
sample mean and the sample
standard deviation *S* but the sample
size was increased to 100; the interval width again would decrease by a factor
of 2, because the width of the interval is 2*C*(*S*/√*n*) and
only *n* is changing.

The confidence interval is given by
the formula

where *n* is the sample size, *C*
is the 97.5 percentile of Student’s *t*
distribution with *n* – 1 degrees of
freedom, and σ is the sample standard deviation.

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