A ** confidence interval** gives an estimated range of values
which is likely to include an unknown population parameter, the estimated
range being calculated from a given set of sample data.
(

The common notation for the parameter in question is .
Often, this parameter is the population mean , which is
estimated through the

The * level C* of a confidence interval gives the probability
that the interval produced by the method employed includes the true value
of the parameter .

Suppose a student measuring the boiling temperature of a certain liquid observes the readings (in degrees Celsius) 102.5, 101.7, 103.1, 100.9, 100.5, and 102.2 on 6 different samples of the liquid. He calculates the sample mean to be 101.82. If he knows that the standard deviation for this procedure is 1.2 degrees, what is the confidence interval for the population mean at a 95% confidence level?

In other words, the student wishes to estimate the true mean boiling temperature
of the liquid using the results of his measurements. If the
measurements follow a normal distribution, then the sample mean will
have the distribution
* N(,)*. Since the sample size is 6, the standard
deviation of the sample mean is equal to 1.2/sqrt(6) = 0.49.

The selection of a confidence level for an interval determines the probability that the confidence interval produced will contain the true parameter value. Common choices for the confidence level

The value *z ^{*}* representing the point on the standard normal density
curve such that the probability of observing a value greater than

**
Note: This interval is only exact when the population distribution is normal. For
large samples from other population distributions, the interval is approximately correct by
the Central Limit Theorem.**

In the example above, the student calculated the sample mean of the boiling temperatures to be 101.82, with standard deviation 0.49. The critical value for a 95% confidence interval is 1.96, where (1-0.95)/2 = 0.025. A 95% confidence interval for the unknown mean is ((101.82 - (1.96*0.49)), (101.82 + (1.96*0.49))) = (101.82 - 0.96, 101.82 + 0.96) = (100.86, 102.78).

As the level of confidence decreases, the size of the corresponding interval will decrease. Suppose
the student was interested in a 90% confidence interval for the boiling temperature. In this case,
*C* = 0.90, and (1-*C*)/2 = 0.05. The critical value *z ^{*}* for this level
is equal to 1.645, so the 90% confidence interval is ((101.82 - (1.645*0.49)), (101.82 + (1.645*0.49)))
= (101.82 - 0.81, 101.82 + 0.81) = (101.01, 102.63)

An increase in sample size will decrease the length of the confidence interval without reducing the level of confidence. This is because the standard deviation decreases as

Suppose in the example above, the student wishes to have a margin of error equal to 0.5 with
95% confidence. Substituting the appropriate values into the expression for *m* and
solving for *n* gives the calculation *n* = (1.96*1.2/0.5)² = (2.35/0.5)²
= 4.7² = 22.09. To achieve a 95%
confidence interval for the mean boiling point with total length less than 1 degree, the student will
have to take 23 measurements.

**For a population with unknown mean and unknown standard
deviation, a confidence interval for the population mean,
based on a simple random sample (SRS) of size n,
is +
t^{*},
where t^{*} is the upper (1-C)/2 critical value for the t
distribution with n-1 degrees of freedom, t(n-1).**

The dataset "Normal Body Temperature, Gender, and Heart Rate" contains 130 observations of body temperature, along with the gender of each individual and his or her heart rate. Using the MINITAB "DESCRIBE" command provides the following information:

Descriptive Statistics Variable N Mean Median Tr Mean StDev SE Mean TEMP 130 98.249 98.300 98.253 0.733 0.064 Variable Min Max Q1 Q3 TEMP 96.300 100.800 97.800 98.700To find a 95% confidence interval for the mean based on the sample mean 98.249 and sample standard deviation 0.733, first find the 0.025 critical value

For a more precise (and more simply achieved) result, the MINITAB "TINTERVAL" command, written as follows, gives an exact 95% confidence interval for 129 degrees of freedom:

MTB > tinterval 95 c1 Confidence Intervals Variable N Mean StDev SE Mean 95.0 % CI TEMP 130 98.2492 0.7332 0.0643 ( 98.1220, 98.3765)According to these results, the usual assumed normal body temperature of 98.6 degrees Fahrenheit is not within a 95% confidence interval for the mean.

*Data source: Data presented in Mackowiak, P.A., Wasserman, S.S., and Levine, M.M. (1992),
"A Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Normal Body Temperature, and
Other Legacies of Carl Reinhold August Wunderlich," Journal of the American Medical
Association, 268, 1578-1580. Dataset available through the
JSE Dataset Archive.*

For some more definitions and examples, see the confidence interval index in Valerie J. Easton and John H. McColl's