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Welch's Standard Error for Two Means

Calculate and interpret Welch’s standard error for a difference between two independent sample means, without assuming equal group variances.

Intermediate 9 min read

What You'll Learn

  • Identify the independent-sample setting for Welch’s standard error.
  • Derive the standard error by combining the estimated variances of two sample means.
  • Calculate Welch’s standard error from two sample standard deviations and sample sizes.
  • Distinguish the standard error of a difference in means from either group’s standard deviation.
  • Explain how each group contributes to the estimated uncertainty.
  • Recognize why Welch’s formula does not require equal population variances.

Why Two Means Have a Standard Error

A difference between two sample means varies from sample to sample. Even if the underlying population means stay fixed, a new pair of samples will usually produce a somewhat different difference. The standard error summarizes the estimated sampling variability of that difference. For two independent groups, Welch’s standard error combines information about the spread of observations in each group and how many observations were sampled.

The standard error is not the spread of individual observations in either group. A group’s sample standard deviation describes how much its individual data values vary around that group’s sample mean. The standard error concerns the variability of a difference between sample means. Its size depends on both groups’ standard deviations and both sample sizes.

Definition: For two independent samples, let \(\bar{x}_1\) and \(\bar{x}_2\) be their sample means, \(s_1\) and \(s_2\) their sample standard deviations, and \(n_1\) and \(n_2\) their sample sizes. Welch’s estimated standard error for the difference \(\bar{x}_1-\bar{x}_2\) is
$$ SE(\bar{x}_1-\bar{x}_2) = \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}. $$
The result has the same units as the measured variable. The formula does not require the two population variances to be equal.

Where the Formula Comes From

Suppose the observations in group 1 are independent draws from a population with mean \(\mu_1\) and variance \(\sigma_1^2\), and those in group 2 are independent draws from a population with mean \(\mu_2\) and variance \(\sigma_2^2\). Assume also that the two samples are independent of each other and that the population variances are finite. These are the assumptions for the variance calculation below; they do not say that the two population variances are the same.

The difference in sample means estimates the population mean difference \(\mu_1-\mu_2\). For independent observations within a group, the variance of a sample mean is its population variance divided by its sample size. Independence between the two groups means their sample means have covariance zero. To see how these facts combine, use the identity for the variance of a difference:

$$ \begin{aligned} \operatorname{Var}(\bar{X}_1-\bar{X}_2) &=\operatorname{Var}(\bar{X}_1)+\operatorname{Var}(\bar{X}_2) -2\operatorname{Cov}(\bar{X}_1,\bar{X}_2)\\ &=\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}-2(0)\\ &=\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}. \end{aligned} $$

The population standard deviations \(\sigma_1\) and \(\sigma_2\) are generally unknown. Estimate each population variance with the corresponding sample variance, \(s_1^2\) or \(s_2^2\). Substituting those estimates into the variance expression gives the estimated variance of the difference; taking its square root gives Welch’s standard error:

$$ \widehat{\operatorname{Var}}(\bar{X}_1-\bar{X}_2) =\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2} \quad\Longrightarrow\quad SE(\bar{x}_1-\bar{x}_2) =\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}. $$

Each term, \(s_i^2/n_i\), estimates the variance contributed by one sample mean. Those estimated variances are added, and only then is the square root taken. The result is expressed in the original measurement units rather than squared units.

Worked Example: Two Groups With Different Spreads

In a hypothetical evaluation, two independent groups of students complete the same assessment. Group 1 has \(n_1=25\) students and a sample standard deviation of \(s_1=8\) points. Group 2 has \(n_2=36\) students and a sample standard deviation of \(s_2=12\) points. Calculate Welch’s standard error for the difference in their mean scores.

The groups are described as independent, so the independent-sample formula applies. The group standard deviations and sizes provide the needed inputs; the group means are not needed to calculate this standard error.

First calculate each sample mean’s estimated variance contribution:

$$ \frac{s_1^2}{n_1}=\frac{8^2}{25}=\frac{64}{25}=2.56\ \text{points}^2, \qquad \frac{s_2^2}{n_2}=\frac{12^2}{36}=\frac{144}{36}=4.00\ \text{points}^2. $$

Add these contributions and take the square root:

$$ SE(\bar{x}_1-\bar{x}_2) =\sqrt{2.56+4.00} =\sqrt{6.56} \approx 2.561\ \text{points}. $$

The estimated standard error of the difference is about 2.561 points. It is not 8 or 12 points: those numbers are the standard deviations of individual scores in the separate groups. Nor is the standard error the difference in sample means; it describes how much that estimated difference would vary across repeated sampling under the model.

Why the Standard Deviations Are Not Averaged

The two groups can contribute different amounts to the uncertainty. Their contributions are determined not only by how spread out their observations are, but also by how many observations are in each sample. A larger sample size reduces that group’s contribution, because the variance of its sample mean is \(s_i^2/n_i\). The formula therefore does not average the two standard deviations, and it does not use either group’s standard deviation alone.

The square root matters as well. The quantities \(s_1^2/n_1\) and \(s_2^2/n_2\) are expressed in squared measurement units, so they can be added as variance contributions. The standard error is the square root of their sum, bringing the final result back to the measurement scale.

Worked Example: A Smaller Group With More Variability

In a hypothetical study of task completion time, Group 1 has \(n_1=16\) participants and a sample standard deviation of \(s_1=4\) minutes. Group 2 has \(n_2=25\) participants and a sample standard deviation of \(s_2=10\) minutes. Calculate the standard error and compare the two groups’ contributions.

The samples are independent. Group 1 contributes:

$$ \frac{s_1^2}{n_1}=\frac{4^2}{16}=\frac{16}{16}=1.00\ \text{minutes}^2. $$

Group 2 contributes:

$$ \frac{s_2^2}{n_2}=\frac{10^2}{25}=\frac{100}{25}=4.00\ \text{minutes}^2. $$

Thus the standard error is:

$$ SE(\bar{x}_1-\bar{x}_2) =\sqrt{1.00+4.00} =\sqrt{5} \approx 2.236\ \text{minutes}. $$

Although Group 2 has more participants, its much greater sample spread makes its estimated variance contribution four times Group 1’s contribution. Its standard deviation of 10 minutes is not the standard error; the standard error combines both groups’ contributions and is about 2.236 minutes.

Calculating Carefully and Interpreting the Result

A reliable calculation keeps the order of operations visible: square each sample standard deviation, divide by that group’s sample size, add the two resulting quantities, and take one square root. Keep units attached to the values so it is clear that the terms being added are variances and that the final result returns to the original units.

For example, if the outcome is measured in kilograms, each \(s_i\) is in kilograms, each \(s_i^2/n_i\) is in kilograms squared, and the final standard error is in kilograms. If the group order is reversed, the estimated standard error stays the same: exchanging the subscripts simply exchanges the two terms in the sum. The estimated difference changes sign when the order reverses, but its standard error does not.

Welch’s formula uses each group’s own sample variance rather than requiring one shared variance estimate. It is therefore suitable for estimating the standard error when the population variances may differ. This does not mean every other assumption is automatically satisfied: the groups must still be independent for this formula, and the sampling design and data structure must support treating observations within each group as independent. If observations are paired or otherwise dependent across groups, this independent-sample formula is not the appropriate calculation.

Worked Example: Converting Group Summaries Into an SE

For a hypothetical comparison of daily exercise time, a sample of \(n_1=18\) adults in Group 1 has a standard deviation of \(s_1=5.4\) minutes, and an independent sample of \(n_2=12\) adults in Group 2 has a standard deviation of \(s_2=7.2\) minutes. Calculate the standard error for the difference in sample mean exercise times.

The independent groups and their separate sample summaries match the setting for Welch’s standard error. Calculate the two contributions:

$$ \frac{s_1^2}{n_1} =\frac{5.4^2}{18} =\frac{29.16}{18} =1.62\ \text{minutes}^2, $$
$$ \frac{s_2^2}{n_2} =\frac{7.2^2}{12} =\frac{51.84}{12} =4.32\ \text{minutes}^2. $$

Then combine them:

$$ SE(\bar{x}_1-\bar{x}_2) =\sqrt{1.62+4.32} =\sqrt{5.94} \approx 2.437\ \text{minutes}. $$

The estimated standard error is about 2.437 minutes. The sample standard deviations, 5.4 and 7.2 minutes, describe variation among individual adults within the groups. The standard error instead estimates the sampling variation of the difference between the groups’ mean exercise times.

Common Mistakes and Why the Distinction Matters

  • Using a group’s standard deviation as the standard error. This ignores the other group and the sample sizes. Use both terms \(s_1^2/n_1\) and \(s_2^2/n_2\).
  • Averaging the standard deviations. The formula combines estimated variances of sample means, not the two standard deviations. Square, divide by the matching sample size, add, and take the square root.
  • Dividing by the wrong sample size. Each group’s variance must be divided by that same group’s \(n_i\); swapping the denominators can change the answer.
  • Adding the variance terms without taking the square root. Their sum is an estimated variance in squared units. The standard error is its square root in the original units.
  • Treating the formula as a result about the mean difference itself. The standard error describes estimated sampling variability, not the observed difference, and by itself does not establish whether a difference is important or convincing.

The distinction is useful because an observed difference and its uncertainty answer different questions. The difference \(\bar{x}_1-\bar{x}_2\) gives the estimated direction and size of the contrast in the sample. Welch’s standard error describes the estimated sampling variability associated with that contrast under the independent-sample model. Keeping them separate makes later calculations and interpretations less error-prone.

Key takeaway: For two independent sample means, Welch’s standard error is \(\sqrt{s_1^2/n_1+s_2^2/n_2}\). Each group contributes its own estimated variance of the sample mean; the result is not the standard deviation of either group and does not require equal population variances.

Check Your Understanding

Use the formula and the distinction between group spread and sampling variability to answer these questions.

  1. Independent samples have \(s_1=6\), \(n_1=9\), \(s_2=8\), and \(n_2=16\). Calculate Welch’s standard error, showing both variance contributions and the final units.
  2. In your own words, distinguish a group’s sample standard deviation from the standard error of the difference in two sample means.
  3. Why can two groups with different population variances still be used in Welch’s standard-error formula?
  4. If the order of the groups is reversed, what happens to the estimated difference and to its standard error? Explain using the formula.
  5. A researcher has paired measurements on the same people in both conditions. Why should the independent-sample Welch formula not be applied directly to those two groups?