Standard Deviation of the Sample Mean

How Spread in Individual Values Affects Sample Means

In “Mean of the Sampling Distribution of x-bar,” we saw that the sampling distribution of \(\bar{x}\) is centered at the population mean \(\mu\) for appropriate random samples. But sample means do not all have to be the same: different samples can produce different values of \(\bar{x}\). This tutorial measures how much those sample means vary.

Suppose the population has mean \(\mu=50\) and standard deviation \(\sigma=12\). A sample mean is an average of \(n\) individual observations. Averaging reduces variation: unusually high and low observations tend to balance one another, so sample means are less spread out than individual observations. The amount of reduction depends on \(n\).

When the observations are independent, the standard deviation of the sample mean is the population standard deviation divided by the square root of the sample size. Thus, when \(\sigma=12\), the standard deviation of \(\bar{x}\) depends on \(n\), not on the particular sample mean observed.

Why the Square Root of the Sample Size Appears

The formula follows from how the sample mean combines the observations. Think of the \(n\) observations as random variables \(X_1,X_2,\ldots,X_n\). Each has standard deviation \(\sigma\), so each has variance \(\sigma^2\). The sample mean is their sum divided by \(n\).

For independent observations, the variance of a sum is the sum of the variances. Dividing the sum by \(n\) divides its variance by \(n^2\). The variance of the sample mean is therefore \(n\sigma^2/n^2=\sigma^2/n\). Taking the square root gives \(\sigma/\sqrt{n}\).

The square root matters. For example, quadrupling the sample size divides the standard deviation of the sample mean by 2, not by 4. A larger sample makes sample means less variable, but the decrease follows the square root of \(n\).

The formula describes spread, not center. As established in the earlier tutorial “Mean of the Sampling Distribution of x-bar,” the mean of the sampling distribution is \(\mu_{\bar{x}}=\mu\). Knowing both results tells us the sampling distribution is centered at \(\mu\) and has standard deviation \(\sigma/\sqrt{n}\), provided the independence condition is appropriate.

Calculate the Standard Deviation for Several Sample Sizes

Use the same population values \(\mu=50\) and \(\sigma=12\) to compare sample sizes. Because \(\sigma\) is in the original measurement units, every result for \(\sigma_{\bar{x}}\) is in those same units.

The values also show why “larger sample” does not mean “standard deviation divided by the sample size.” For \(n=16\), the standard deviation is \(12/4=3\) units, because \(\sqrt{16}=4\). Dividing 12 by 16 would give 0.75 units, which is not the result of the formula.

Sampling Without Replacement and the 10% Condition

When a simple random sample is selected without replacement from a finite population, the observations are dependent: after one individual is selected, the population available for the next selection has changed. The formula \(\sigma/\sqrt{n}\) treats observations as independent. In AP Statistics, the 10% condition provides a practical basis for using that formula as an approximation: the sample size should be no more than 10% of the population size.

For a finite population sampled without replacement, the exact standard deviation includes a finite-population adjustment. If the population standard deviation \(\sigma\) is calculated using the population size \(N\), the adjustment is \(\sqrt{(N-n)/(N-1)}\). When the sample is a small fraction of the population, this factor is close to 1, so the simpler formula is a close approximation.

Using the Formula in Context

The formula can also be used to find a sample size that produces a specified standard deviation, provided the other assumptions remain appropriate. Start with \(\sigma_{\bar{x}}=\sigma/\sqrt{n}\), then solve for \(n\). This is an algebraic use of the same relationship, not a different measure of spread.

This calculation concerns the standard deviation of sample means across repeated samples. It does not identify the actual sample mean for a particular set of 36 observations, and it does not by itself describe the shape of the sampling distribution.

Common Mistakes and AP Exam Tip

• Dividing by \(n\) instead of \(\sqrt{n}\): The standard deviation is \(\sigma/\sqrt{n}\). For \(\sigma=12\) and \(n=9\), the result is \(12/3=4\), not \(12/9\).
• Confusing the population spread with the sample-mean spread: \(\sigma=12\) units describes individual population values. For \(n=4\), the standard deviation of \(\bar{x}\) is 6 units under the independent-observation model.
• Leaving out units or context: State what the number measures. For example, “For repeated samples of size 9, sample means typically vary around 50 units with a standard deviation of 4 units.”
• Using the formula without checking sampling dependence: For sampling without replacement, check the 10% condition. If it holds, \(\sigma/\sqrt{n}\) is the AP approximation; it is not the exact finite-population value.
• Claiming every sample mean is close to \(\mu\): A standard deviation summarizes the spread of a sampling distribution. It does not guarantee a particular sample mean is within one standard deviation of the population mean.
• Assuming the standard deviation determines the distribution’s shape: The formula gives spread. It does not, by itself, establish that the sampling distribution is normal.

For full-credit communication, name the sampling distribution, show the substitution into \(\sigma/\sqrt{n}\), and report the result with units. If the sample is drawn without replacement, state whether the 10% condition is met and identify the usual formula’s result as approximate. Keep the mean and standard deviation distinct: the mean describes the center, while the standard deviation describes the spread.