Why Use a Welch Interval?
A confidence interval for the difference between two independent population means must account for sampling variation in both samples. When the population variances are known, the variance of the difference can be calculated directly. In practice, those variances are often unknown and estimated from the samples. If the two population variances may differ, the Welch interval uses a separate estimate from each sample rather than imposing a common variance.
This method is useful when comparing two distinct populations, such as hypothetical claim-processing times under two independent workflows. The target is the difference in population means, not the difference between individual paired observations. The method is aligned with the SOA VEE Mathematical Statistics syllabus.
For the usual Welch interval, both population variances are unknown and are not assumed equal. The sample sizes need not be equal. If both populations are normal, the sample means and sample variances have the standard normal-sampling properties used to motivate the method. With nonnormal populations, the interval is an approximation; its reliability depends on the sample sizes and the population shapes. Small samples with strong skewness or influential outliers can make the approximation unreliable.
The Welch Interval
The estimated standard error of \(\bar X-\bar Y\) is obtained by adding the two estimated contributions to variance. The approximate degrees of freedom account for the fact that the two contributions are estimated with different amounts of information.
where \(t_{1-\alpha/2,\,\widehat{\nu}}\) is the \(1-\alpha/2\) quantile of the Student \(t\) distribution with \(\widehat{\nu}\) degrees of freedom, and
The two variance contributions \(s_1^2/n_1\) and \(s_2^2/n_2\) remain separate throughout the calculation. In particular, the formula does not pool \(s_1^2\) and \(s_2^2\) into one common estimate.
Why the Degrees of Freedom Have This Form
Under normal sampling, each sample variance is related to a chi-square random variable: \((n_j-1)s_j^2/\sigma_j^2\) has a chi-square distribution with \(n_j-1\) degrees of freedom, for \(j=1,2\). The two sample variances are independent. Thus, the estimated variance of the difference is a sum of two scaled chi-square quantities, not generally one scaled chi-square quantity.
The Satterthwaite approximation replaces this sum by a single scaled chi-square variable whose mean and variance match those of the sum. To see the resulting degrees of freedom, write \(a_j=s_j^2/n_j\). The estimated variance is \(a_1+a_2\); approximating its distribution by a scaled chi-square gives the moment-matching degrees of freedom \((a_1+a_2)^2/[a_1^2/(n_1-1)+a_2^2/(n_2-1)]\), which is the stated formula. Because the two chi-square contributions generally have different scales, their sum is not exactly a scaled chi-square. Consequently, the Welch \(t\) statistic is generally approximate, even for normal populations.
This distinction matters when the population variances happen to be equal. Under normality and equal population variances, the pooled two-sample \(t\) procedure is exact. The Welch procedure remains an approximation in general; equal population variances alone do not turn its data-dependent degrees of freedom into an exact \(t\) pivot.
Write the parameter as \(\mu_1-\mu_2\), and calculate \(\bar X-\bar Y\) in the same order.
Find \(a_1=s_1^2/n_1\) and \(a_2=s_2^2/n_2\), then calculate \(SE_W=\sqrt{a_1+a_2}\).
Substitute \(a_1,a_2,n_1,n_2\) in the Welch–Satterthwaite formula. Do not replace the result by a pooled degrees-of-freedom count.
For a two-sided confidence level \(1-\alpha\), use \(t_{1-\alpha/2,\widehat{\nu}}\), then add and subtract the margin of error from \(\bar X-\bar Y\).
Worked Calculations
Worked Example: Comparing Two Independent Processing Times
A hypothetical insurer compares two independent claim-file workflows. A random sample of 10 files processed by Workflow A has mean processing time 18 hours and sample standard deviation 3 hours. An independent random sample of 10 files processed by Workflow B has mean 14 hours and sample standard deviation 4 hours. Construct a 95% Welch interval for the difference in population mean processing times, \(\mu_A-\mu_B\). Assume the two population distributions are normal with unknown, potentially unequal variances. No interest rate or mortality assumption is involved.
Method and estimate. Use the Welch interval, keeping the two sample variance estimates separate. The observed difference is
Standard error. The two contributions are \(a_1=3^2/10=9/10=0.9\) and \(a_2=4^2/10=16/10=1.6\). Therefore,
Degrees of freedom. Substituting the two contributions gives
At 95% confidence, the critical value is approximately \(t_{0.975,16.69}=2.113\). The margin of error is \(2.113(1.5811)\approx3.341\) hours. Thus the interval is \(4\pm3.341\), or approximately \((0.66,7.34)\) hours, rounding the endpoints to two decimal places.
Check and common mistake. The midpoint of the reported interval is 4 hours, as it should be, and its half-width is approximately 3.34 hours. A common mistake is to pool the two sample variances because the sample sizes are equal. Equal sample sizes do not establish equal population variances; the Welch calculation uses \(9/10\) and \(16/10\) separately.
Worked Example: Unequal Sample Sizes and a Positive Estimate
In a hypothetical study of two independent methods for reviewing policy applications, Method 1 is applied to 8 randomly selected applications and Method 2 to 12 different randomly selected applications. The sample means are 27.4 and 24.9 minutes, and the sample standard deviations are 2 and 3 minutes, respectively. Construct a 95% Welch interval for \(\mu_1-\mu_2\). Assume independent normal populations with unknown, potentially unequal variances. No interest rate or mortality assumption is involved.
Method and estimate. The sample mean difference is \(27.4-24.9=2.5\) minutes. The separate estimated variance contributions are \(a_1=2^2/8=0.5\) and \(a_2=3^2/12=0.75\). Hence
The degrees of freedom are
The 95% critical value is approximately \(t_{0.975,17.99}=2.101\). The margin of error is \(2.101(1.1180)\approx2.349\) minutes. The interval is \(2.5\pm2.349\), which gives approximately \((0.15,4.85)\) minutes, to two decimal places.
Check and common mistake. The estimate is positive and the interval is centered at 2.5. It includes only positive values at the displayed precision. Do not use \(n_1+n_2-2=18\) as the Welch degrees of freedom merely because that is the familiar pooled count; here the Welch calculation happens to give approximately 17.99, but that is a result of the inputs, not a rule.
Worked Example: A Small Sample with Different Variability
A hypothetical laboratory compares measurement times for two independent procedures. Six randomly selected measurements using Procedure 1 have mean 31 minutes and sample standard deviation 5 minutes. Nine different measurements using Procedure 2 have mean 26 minutes and sample standard deviation 2.5 minutes. Construct a 90% Welch interval for \(\mu_1-\mu_2\). Assume independent normal populations with unknown, potentially unequal variances. No interest rate or mortality assumption is involved.
Method and estimate. The estimated difference is \(31-26=5\) minutes. The variance contributions are
Therefore, \(SE_W=\sqrt{4.166667+0.694444}=\sqrt{4.861111}\approx2.2048\) minutes. The approximate degrees of freedom are
For a 90% two-sided interval, the critical value is approximately \(t_{0.95,6.69}=1.910\). The margin of error is \(1.910(2.2048)\approx4.211\) minutes. Thus the interval is \(5\pm4.211\), or approximately \((0.79,9.21)\) minutes, with endpoints rounded to two decimal places.
Check and common mistake. The estimate of 5 is the midpoint, and the interval is wider than the point estimate’s standard error because it uses a \(t\) critical value. A common error is to round the degrees of freedom down to 6 before selecting the critical value without stating that convention. Using the unrounded approximation, or a stated conservative table convention, is preferable; the calculation above uses approximately 6.69 degrees of freedom.
Interpretation and Exam Traps
A 95% Welch confidence procedure has 95% long-run coverage under its assumptions and approximation. It does not mean that the fixed parameter \(\mu_1-\mu_2\) has a 95% probability of lying in the particular interval after the data have been observed. A confidence interval reports a range of parameter values compatible with the observed difference and its estimated uncertainty.
The interval’s direction follows the order chosen for the difference. If the parameter is \(\mu_1-\mu_2\), reversing the order reverses the signs of the estimate and both endpoints. A positive interval indicates that the first population mean exceeds the second under that stated order; an interval containing zero does not establish a nonzero difference at the corresponding two-sided confidence level.
| Common error | Correction |
|---|---|
| Using a pooled variance despite possible unequal population variances | Use \(s_1^2/n_1+s_2^2/n_2\) and the Welch degrees-of-freedom formula. |
| Using \(n_1+n_2-2\) degrees of freedom by habit | That count belongs to the pooled procedure; calculate \(\widehat{\nu}\) for Welch. |
| Putting \(s_j/n_j\) into the standard error | Use the sample variance contribution \(s_j^2/n_j\), then take the square root of their sum. |
| Using \(t_{1-\alpha}\) for a two-sided interval | Split the excluded probability between both tails and use \(t_{1-\alpha/2,\widehat{\nu}}\). |
| Claiming the Welch interval is exact for equal population variances | The pooled \(t\) interval is exact under normality and equal variances; Welch’s procedure is generally approximate. |
Keep sufficient precision in the standard error and degrees of freedom until the critical value and margin of error have been found. Rounding an intermediate value can move an endpoint, especially in a small-sample calculation. State the order of the means, confidence level, approximate degrees of freedom, and units in a complete answer.
Check Your Understanding
For each problem, state the interval for the first population mean minus the second. Unless a problem says otherwise, assume independent normal populations with unknown, potentially unequal variances. No interest rate or mortality assumption is involved.
- Two independent samples each have size 10. Their sample means are 12 and 9, and their sample standard deviations are both 2. Find a 95% Welch interval. (Answer: approximately \((1.12,4.88)\).)
- For two independent samples, \(n_1=8,\ n_2=8,\ \bar X=15.5,\ \bar Y=14.0,\ s_1=2,\ s_2=4\). Find a 90% Welch interval. (Answer: approximately \((-1.36,4.36)\).)
- Two independent samples have \(n_1=n_2=12\), sample means 44 and 40, and sample standard deviations 3 and 4. Find a 95% Welch interval. (Answer: approximately \((0.99,7.01)\).)
- Two independent samples have \(n_1=6,\ n_2=10,\ \bar X=8,\ \bar Y=6,\ s_1=2,\ s_2=3\). Find a 99% Welch interval. (Answer: approximately \((-1.74,5.74)\).)
- A review compares two independent populations. Sample 1 has \(n_1=5,\ \bar X=42,\ s_1=1\); Sample 2 has \(n_2=20,\ \bar Y=39,\ s_2=5\). Find a 90% Welch interval and state whether it contains zero. (Answer: approximately \((0.94,5.06)\); it does not contain zero.)