Determining Sample Size for Estimating the Mean

Suppose we want to estimate the average GPA $\mu$ of WMU undergraduates this school year. Historically, the SD of student GPA is known to be $\sigma=.30$. If a random sample of size n=25 yields a sample mean of $\overline{X}=3.05$, then the population mean $\mu$ is estimated as lying within the interval $\overline{X} \pm 2(\sigma/\sqrt{n})= 3.05 \pm 2(.30/ \sqrt{25}) = 3.05 \pm .12$ 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 $\mu$ is within .12 of the sample mean 3.05''. However, .12 is a large number. The value $3.05 \pm .12$ 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: $2(\sigma/\sqrt{n})$. Observe that if sample size increases, the margin of error decreases. Let M denote the desired margin of error. Solving $M=2(\sigma/\sqrt{n})$ 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 $3.05 \pm 2(.30/ \sqrt{144})= 3.05 \pm .05$.

Note that the population SD $\sigma$ 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.