Determining Sample Size for Estimating the Mean

Suppose we want to estimate the average GPA
of WMU undergraduates
this school year. Historically, the SD of student GPA is known to be
.
If a random sample of size *n*=25 yields a sample mean
of
,
then the population mean
is estimated as
lying within the interval
with 95% confidence.
The plus-or-minus quantity .12 is called the
*margin of error*
of the sample mean
associated with a 95% confidence level. It is also correct to say
``we are 95% confident that
is within .12 of the sample mean 3.05''.
However, .12 is a large number. The value
can be as low as 2.93 or as high as 3.17.
Is there some way to reduce the margin of error? Yes, and quite easily.
Look at the formula for the margin of
error:
.
Observe that *if sample size increases,
the margin of error decreases*. Let *M* denote the desired margin of error.
Solving
for *n* gives the following:

Example:Suppose we want to reduce the margin of error for estimating mean GPA from .12 to .05. The sample size we need is:n=4 (.30)^{2}/(.05)^{2}=144. To verify that this is the correct sample size, the 95% confidence interval would be computed (if the sample mean remains at 3.05) as .

Note that the population SD is needed in the sample size formula. This typically unknown value may be (i) estimated from historical data, or (ii) from a pilot sample (like the initial sample of size 25 mentioned in the example above). The SD from the pilot sample is then substituted into the sample size formula and the required sample size is calculated.