How to Determine Sample Size for a Normal Mean

Title: How to Determine Sample Size for a Normal Mean Category: statistics

Determining a sample size requires a goal: estimating a population mean to a chosen precision is different from detecting a difference in a hypothesis test. The most common normal-distribution calculation is the first one. It uses the confidence level, desired margin of error, and population standard deviation.

Sample size for estimating a mean

Suppose you will collect $n$ independent observations from a population with mean $\mu$ and known standard deviation $\sigma$. For independent observations from the same population with finite variance, the sample mean has standard error exactly $\sigma/\sqrt{n}$. If the population is normal, the sample mean is normally distributed; for a nonnormal population, its distribution is approximately normal for a sufficiently large sample under suitable conditions by the central limit theorem.

For a normal population, a two-sided confidence interval with confidence level $1-\alpha$ has margin of error

$$E = z_{1-\alpha/2}\frac{\sigma}{\sqrt{n}},$$

where $z_{1-\alpha/2}$ is the corresponding standard normal quantile. For a sufficiently large sample from a nonnormal population, this is an approximate interval under suitable conditions. Solving for $n$ gives

$$n = \left(\frac{z_{1-\alpha/2}\sigma}{E}\right)^2.$$

Round this result up to the next whole number. Rounding down could make the actual margin of error larger than the one you requested.

Common critical values are $1.645$ for a 90% two-sided confidence interval, $1.96$ for 95%, and approximately $2.576$ for 99%. The value of $E$ is the margin of error on either side of the estimate, not the full width of the interval.

Worked example

Suppose a measurement has a known population standard deviation of $12$ units. You want a 95% confidence interval for the population mean with a margin of error no greater than $3$ units.

Use $z_{0.975}=1.96$, $\sigma=12$, and $E=3$:

$$\begin{aligned} n &= \left(\frac{1.96(12)}{3}\right)^2 \\ &= 7.84^2 \\ &= 61.4656. \end{aligned}$$

Round up: you need 62 observations. With 62 observations, the planned margin of error is approximately $1.96(12/\sqrt{62})=2.99$ units, which is just under 3. With 61 observations it would be slightly over 3.

What if the standard deviation is unknown?

In practice, the population standard deviation is often not known before data collection. Use a credible estimate, usually from a prior study, historical data, or a pilot sample, in place of $\sigma$. The resulting sample size is a planning estimate; if the actual variability is greater, the achieved margin of error will also be greater.

When the standard deviation is estimated from the same sample, a confidence interval for the mean generally uses a Student’s $t$ critical value rather than a normal $z$ value. For a proposed sample size $n$, its margin of error is $t_{1-\alpha/2,n-1}s/\sqrt{n}$. Since the critical value depends on $n$, you may need to try integer sample sizes until the planned margin of error meets your target. For small samples, this is more appropriate than simply treating an estimated standard deviation as known. A pilot estimate can also be uncertain, so allow for that uncertainty when it matters.

If the goal is a hypothesis test

A test needs a different set of inputs: significance level, desired power, variability, and the smallest difference $\delta$ worth detecting. For a two-sided one-sample test of a mean, a common normal approximation is

$$n \approx \left(\frac{(z_{1-\alpha/2}+z_{1-\beta})\sigma}{\delta}\right)^2,$$

where $1-\beta$ is the desired power. For example, with a 5% significance level, 80% power, $\sigma=12$, and a target difference of $6$ units, use $z_{0.975}=1.96$ and $z_{0.80}\approx0.842$:

$$n \approx \left(\frac{(1.96+0.842)(12)}{6}\right)^2 \approx 31.4.$$

This approximation suggests 32 observations. An exact calculation may differ, particularly for small samples or when using a $t$ test. For a two-group comparison, paired design, or a one-sided test, use the formula for that specific design rather than applying this one unchanged.

Finite populations and practical adjustments

The basic formula assumes sampling from a very large population, or sampling with replacement. If you sample by simple random sampling without replacement from a finite population of size $N$, a finite-population correction can reduce the required sample size. Starting with the large-population result $n_0$, one common adjustment is

$$n_{\text{finite}} = \frac{n_0}{1+(n_0-1)/N}.$$

Round up after applying the adjustment. This correction is relevant when the sample is a substantial fraction of the population; it does not fix biased selection or dependent observations.

If some selected participants may not provide usable data, increase the number invited. For an expected response fraction $r$, a simple planning adjustment is to invite about $n/r$ people to obtain $n$ completed observations. Use a realistic response-rate estimate, and do not confuse the number invited with the number analyzed.

Common mistakes

• Using the standard deviation as the standard error. The standard error is $\sigma/\sqrt{n}$; it changes with sample size.
• Rounding to the nearest integer. Always round the required sample size up.
• Using the wrong margin of error. If you want a total interval width of 6 units, set $E=3$, not 6.
• Ignoring the study design. Clustering, repeated measurements, unequal sampling probabilities, and nonresponse can change the required sample size.
• Treating normality as the only requirement. The formula also relies on a suitable sampling process and independent observations. A large sample does not remove bias from a poor design.

Before collecting data, state whether your target is precision or power, and record the assumptions used for the calculation. That makes the resulting sample-size choice interpretable and easier to revise if those assumptions change.