Why Pool the Variances?
Suppose two independent populations are normal, and the goal is to estimate the difference between their means. If their variances are unknown but are assumed to be equal, the two samples can be combined to estimate that shared variance. The resulting pooled two-sample t interval uses information from both groups while accounting for their possibly different sample sizes.
The equal-variance assumption is substantive: it says the populations have the same variance, not that the observed sample variances must be identical. The procedure is appropriate when independent samples come from normal populations with a common variance. It is not the pooled procedure when the populations have different variances, or when the observations are paired rather than independent.
For each sample, let \(\bar X\) and \(\bar Y\) be the sample means, and let \(S_1^2\) and \(S_2^2\) be the usual unbiased sample variances, with denominators \(n_1-1\) and \(n_2-1\), respectively. The pooled variance weights each sample variance by its degrees of freedom:
The weights are not generally equal. A sample with more observations contributes more within-sample degrees of freedom and therefore receives more weight. The pooled standard error for the difference of sample means is
Write \(t_{q,\nu}\) for the \(q\)-quantile of a Student t distribution with \(\nu\) degrees of freedom. For a two-sided confidence level \(1-\alpha\), the pooled interval is
The degrees of freedom are \(n_1+n_2-2\): each sample variance uses one degree of freedom for its estimated sample mean. Keep the order of subtraction consistent throughout. This formula estimates \(\mu_1-\mu_2\); reversing the groups gives an interval for \(\mu_2-\mu_1\), which is the original interval with both signs reversed.
Why the Pooled Standard Error Has This Form
The standard error follows from the variance of a difference of independent sample means. Under the stated model, \(\operatorname{Var}(\bar X)=\sigma^2/n_1\) and \(\operatorname{Var}(\bar Y)=\sigma^2/n_2\). Independence means the covariance between these sample means is zero, so
If \(\sigma^2\) were known, its square root would give the standard error. Since it is unknown, estimate it using the within-sample variation. The unbiased sample variances satisfy \(E[S_1^2]=E[S_2^2]=\sigma^2\). Consequently, a degrees-of-freedom-weighted average also estimates \(\sigma^2\) without bias:
This explains both the weights and the substitution of \(S_p\) for the unknown \(\sigma\) in the standard error. Under the normal, independent, equal-variance model, the corresponding studentized difference uses a t critical value with \(n_1+n_2-2\) degrees of freedom. The confidence interval is centered at the observed difference; its margin of error is the critical value times the estimated standard error.
Constructing the Interval
A reliable calculation keeps the variance pooling, standard error, and t critical value as separate steps. First compute the pooled variance from the two within-sample variances. Then calculate its square root times the square root of the reciprocal sample sizes’ sum. Finally, use the confidence level and pooled degrees of freedom to select the critical value.
Decide whether the target is \(\mu_1-\mu_2\) or \(\mu_2-\mu_1\), and calculate the sample mean difference in that same order.
Multiply each sample variance by its degrees of freedom, add the results, and divide by \(n_1+n_2-2\).
Use \(\nu=n_1+n_2-2\), \(t_{1-\alpha/2,\nu}\), and \(S_p\sqrt{1/n_1+1/n_2}\).
Add and subtract the margin of error from the sample mean difference. State the interval in the original units and in the chosen group order.
Worked Example: Comparing Two Independent Process Times
A hypothetical study compares processing times, in minutes, for two independent teams. Team 1 has \(n_1=8\), sample mean \(\bar X=42.6\), and sample variance \(S_1^2=16\). Team 2 has \(n_2=10\), sample mean \(\bar Y=39.1\), and sample variance \(S_2^2=25\). Assume independent normal populations with a common unknown variance. Construct a 95% confidence interval for \(\mu_1-\mu_2\).
Method and group order. Use the pooled two-sample t interval and keep Team 1 minus Team 2. The observed difference is \(42.6-39.1=3.5\) minutes.
Pool the sample variances. The degrees-of-freedom weights are \(8-1=7\) and \(10-1=9\):
Thus \(S_p=\sqrt{21.0625}\approx4.5894\), and the degrees of freedom are \(8+10-2=16\). For a two-sided 95% interval, \(t_{0.975,16}\approx2.1199\).
Calculate the standard error and margin.
The margin of error is \(2.1199(2.1769)\approx4.6149\) minutes. Therefore the unrounded endpoints are \(3.5-4.6149=-1.1149\) and \(3.5+4.6149=8.1149\). Rounding only now to two decimal places gives the 95% interval \((-1.11,\,8.11)\) minutes.
Check and common mistake. The interval is centered at \(3.5\), since \((-1.1149+8.1149)/2=3.5\). A common error is to average the two sample variances equally; that would ignore the different degrees of freedom. The correct pooled numerator is \(7(16)+9(25)\), not \(16+25\).
Worked Example: Unequal Sample Sizes and a 95% Interval
In a hypothetical study of independent normal populations with a common unknown variance, Group A has \(n_A=6\), mean \(18.4\), and variance \(9\). Group B has \(n_B=9\), mean \(15.7\), and variance \(4\). Find a 95% interval for \(\mu_A-\mu_B\).
Method and direction. The observed difference in the requested order is \(18.4-15.7=2.7\). The pooled degrees of freedom are \(6+9-2=13\).
The pooled variance is
The standard error is
For a 95% two-sided interval with 13 degrees of freedom, \(t_{0.975,13}\approx2.1604\). The margin is \(2.1604(1.2827)\approx2.7711\). The endpoints are \(2.7-2.7711=-0.0711\) and \(2.7+2.7711=5.4711\). Rounded to two decimal places, the interval is \((-0.07,\,5.47)\).
Check and common mistake. The midpoint of the unrounded interval is \(2.7\). Do not use \(n_A+n_B=15\) as the degrees of freedom: the pooled degrees of freedom are \(15-2=13\), not 15.
Worked Example: A 90% Interval With a Different Sample-Variance Pattern
A hypothetical time study takes independent samples from two normal populations assumed to have the same unknown variance. Sample 1 has \(n_1=5\), mean \(31.2\), and variance \(4\). Sample 2 has \(n_2=7\), mean \(28.0\), and variance \(9\). Construct a 90% confidence interval for \(\mu_1-\mu_2\).
Method and observed difference. In the requested order, \(\bar X-\bar Y=31.2-28.0=3.2\). The pooled degrees of freedom are \(5+7-2=10\).
The pooled variance is
For a two-sided 90% interval, each tail has probability \(0.05\), so the critical value is \(t_{0.95,10}\approx1.8125\). The standard error and margin are
The endpoints are \(3.2-2.8079=0.3921\) and \(3.2+2.8079=6.0079\). Rounded to two decimal places, the 90% interval is \((0.39,\,6.01)\).
Check and common mistake. The midpoint is \(3.2\), and the interval uses \(t_{0.95,10}\), not \(t_{0.975,10}\). Using the latter would construct a 95% interval rather than the requested 90% interval.
Assumptions and Common Traps
The pooled interval is useful when the equal-variance model is justified, but the calculation cannot establish that assumption. A close match between observed sample variances is not proof of equal population variances, and different sample variances do not by themselves invalidate the model. The assumption concerns the populations; it should be assessed using the context and the analysis plan.
- Pooling without the equal-variance assumption. The formula estimates one common variance. If the population variances differ, this pooled standard error is not the appropriate one.
- Using the wrong weights. Weight each sample variance by \(n_j-1\), not by \(n_j\), and do not take a simple average unless the degrees of freedom happen to match.
- Using the wrong degrees of freedom. The pooled t distribution has \(n_1+n_2-2\) degrees of freedom.
- Mixing group order. If the interval is for \(\mu_1-\mu_2\), its center must be \(\bar X-\bar Y\). Reversing the order changes both endpoints’ signs.
- Confusing confidence level and tail area. For a two-sided \(1-\alpha\) interval, use \(t_{1-\alpha/2,\nu}\), not \(t_{1-\alpha,\nu}\).
- Rounding too early. Keep extra digits in the pooled variance, standard error, and margin; round the interval endpoints only at the end.
Check Your Understanding
For each problem, use the pooled interval only when the stated equal-variance assumptions apply. Round interval endpoints to two decimal places.
- Two independent normal populations have a common unknown variance. Sample 1 has \(n_1=5\), mean \(24.5\), and sample variance \(4\); Sample 2 has \(n_2=8\), mean \(22.0\), and sample variance \(9\). Find a 95% interval for \(\mu_1-\mu_2\). (Answer: \(S_p^2=79/11\approx7.18\), \(df=11\), interval approximately \((-0.86,\,5.86)\).)
- Independent normal samples with equal population variances have \(n_1=7\), mean \(10.2\), variance \(16\), and \(n_2=9\), mean \(9.0\), variance \(4\). Find a 90% interval for \(\mu_1-\mu_2\). (Answer: \(S_p^2=64/7\approx9.14\), \(df=14\), interval approximately \((-1.48,\,3.88)\).)
- For independent normal populations assumed to have a common unknown variance, Sample 1 has \(n_1=6\), mean \(5.4\), and variance \(2\); Sample 2 has \(n_2=10\), mean \(7.1\), and variance \(8\). Find a 95% interval for \(\mu_1-\mu_2\). (Answer: \(S_p^2=41/7\approx5.86\), \(df=14\), interval approximately \((-4.38,\,0.98)\).)
- Two independent normal populations are assumed to have equal unknown variances. Group B has \(n_B=8\), mean \(42\), and variance \(6\); Group A has \(n_A=12\), mean \(38\), and variance \(3\). Find a 90% interval for \(\mu_B-\mu_A\). (Answer: \(S_p^2=25/6\approx4.17\), \(df=18\), interval approximately \((2.38,\,5.62)\).)
- A hypothetical study compares two independent normal populations under a common unknown variance assumption. Sample 1 has \(n_1=9\), mean \(100.2\), and variance \(25\); Sample 2 has \(n_2=11\), mean \(96.0\), and variance \(16\). Find a 95% interval for \(\mu_1-\mu_2\). (Answer: \(S_p^2=20\), \(df=18\), interval approximately \((-0.02,\,8.42)\).)