A Worked Pooled T Analysis With Equal Variance

Why Use a Pooled Two-Sample t Analysis?

When a study compares the means of two independent groups, the question may concern the difference between their population means. The pooled two-sample t procedure addresses that question by combining the two sample variances into one estimate of a shared population variance. It can be useful when a common variance is a reasonable model for both populations, but it should not be chosen merely because the two sample standard deviations happen to look similar.

This tutorial works through a pooled test and confidence intervals using sample summaries. The comparison is always stated in a specified order, group 1 minus group 2, so the sign of the difference has a clear interpretation.

Definition: In a pooled two-sample t analysis, two independent samples are modeled as coming from populations with means $\mu_1$ and $\mu_2$ and a common variance $\sigma^2$. The pooled sample variance and pooled standard deviation are $$ s_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}, \qquad s_p=\sqrt{s_p^2}, $$ where $n_1,n_2$ and $s_1,s_2$ are the sample sizes and sample standard deviations. The standard error for the difference in sample means is $$ SE=s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}. $$

Model and Conditions

The pooled method compares exactly two independent samples. The samples must be independent of one another, and observations within each sample must be independent under the sampling design. The data should come from random samples or an appropriate randomized design if the goal is to generalize or make a causal comparison. For small samples, each group’s population should be approximately normal, with no severe outliers. With larger samples, the t procedure is generally more robust to moderate departures from normality, but extreme skewness or influential outliers can still be a concern.

Most importantly, the two population variances are assumed equal: $\sigma_1^2=\sigma_2^2=\sigma^2$. The pooled standard deviation estimates this single common standard deviation. This assumption is part of the model, not a conclusion established just by calculating the pooled value. If unequal variances are plausible, the Welch procedure is the alternative developed in the earlier material on Welch inference.

For a test of equal means, the usual null and two-sided alternative are $H_0:\mu_1-\mu_2=0$ and $H_a:\mu_1-\mu_2\ne0$. The pooled statistic is compared with a t distribution having $df=n_1+n_2-2$ degrees of freedom. For a confidence interval, the same standard error and degrees of freedom are used.

How the Pooling Formula Works

Each sample variance estimates the common variance, but the estimates may differ because of sampling variation. The pooled variance combines them with weights $n_i-1$, the degrees of freedom associated with each sample variance. Under independent sampling and the common-variance model, each sample variance has expected value $\sigma^2$. Therefore,

E(s_p^2) = \frac{(n_1-1)E(s_1^2)+(n_2-1)E(s_2^2)} {n_1+n_2-2} = \frac{(n_1-1)\sigma^2+(n_2-1)\sigma^2} {n_1+n_2-2} =\sigma^2.

Thus, the pooled variance is a weighted average of the two sample variances and is an unbiased estimate of the common variance under the model. The denominator for the pooled standard error follows from the variance of a difference between independent sample means:

\operatorname{Var}(\bar{x}_1-\bar{x}_2) = \frac{\sigma^2}{n_1}+\frac{\sigma^2}{n_2} = \sigma^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right).

Replacing $\sigma$ with its pooled estimate $s_p$ gives the stated $SE$. With independent normal samples and equal population variances, dividing the difference in sample means minus its hypothesized value by this estimated standard error has a t distribution with $n_1+n_2-2$ degrees of freedom. That result supplies the reference distribution for both tests and intervals.

Worked Pooled Confidence Interval

Worked Example: Comparing Two Process Means

Suppose two independent samples measure the output, in points, of two processes. The hypothetical summaries are $n_1=10$, $\bar{x}_1=41.2$, $s_1=4.60$, and $n_2=12$, $\bar{x}_2=38.5$, $s_2=4.8299$. Assume the observations were independently collected, the group distributions have no severe outliers and are reasonably compatible with normal populations, and a common population variance is appropriate. Find a 95% pooled confidence interval for $\mu_1-\mu_2$.

First calculate the pooled variance and standard deviation:

s_p^2 = \frac{9(4.60^2)+11(4.8299^2)}{20} = \frac{190.44+256.6073}{20} \approx22.3524, \qquad s_p\approx4.7278\ \text{points}.

There are $10+12-2=20$ degrees of freedom. The standard error is

SE = \sqrt{22.3524}\sqrt{\frac{1}{10}+\frac{1}{12}} = \sqrt{22.3524}\sqrt{0.183333} \approx2.0243\ \text{points}.

For 95% confidence with $df=20$, the critical value is approximately $t^*=2.086$. The estimated difference is $41.2-38.5=2.7$ points, and the margin of error is $2.086(2.0243)\approx4.223$ points. Therefore,

(\bar{x}_1-\bar{x}_2)\pm t^*SE = 2.7\pm4.223 = (-1.523,\ 6.923)\ \text{points}.

Under the stated model, this interval gives plausible values for the difference in population mean output, process 1 minus process 2. Because it includes zero, the data are compatible with equal population means at the corresponding two-sided 5% test level. The interval does not establish that the means are equal.

Worked Pooled Test

Worked Example: Testing for a Difference in Means

Use the same sample sizes and standard deviations, but suppose a separate hypothetical comparison has $\bar{x}_1=41.1984$ points and $\bar{x}_2=38.5000$ points. Test $H_0:\mu_1-\mu_2=0$ against $H_a:\mu_1-\mu_2\ne0$ at $\alpha=0.05$. Assume the independence, distributional, and equal-variance conditions described above are reasonable.

The pooled standard deviation and standard error are the same as in the preceding calculation: $s_p\approx4.7278$ points and $SE\approx2.0243$ points. The observed difference is $41.1984-38.5000=2.6984$ points. Under the null hypothesis, the pooled t statistic is

t = \frac{(\bar{x}_1-\bar{x}_2)-0}{SE} = \frac{2.6984}{2.0243} \approx1.333, \qquad df=20.

For a two-sided test, the p-value is the probability, assuming $H_0$ and the pooled model, of getting a t statistic at least as far from zero as the observed value:

p\text{-value}=2P(T_{20}\ge |1.333|)\approx0.1975.

Since $0.1975>0.05$, we do not reject $H_0$. The sample does not provide statistically significant evidence at the 5% level that the two population means differ. This does not show that the means are equal; it says that the observed difference is not sufficiently unusual under the null model to meet this decision threshold.

Another Confidence Interval

Worked Example: A 90% Interval With Different Sample Sizes

In a hypothetical comparison of two independent groups, suppose $n_1=6$, $\bar{x}_1=74$, $s_1=5.00$, and $n_2=9$, $\bar{x}_2=69$, $s_2=4.7697$, with measurements in units. Assume the independent-sample, normality, and equal-variance conditions are reasonable. Construct a 90% pooled interval for $\mu_1-\mu_2$.

The pooled variance and standard deviation are

s_p^2 = \frac{5(5.00^2)+8(4.7697^2)}{13} \approx23.6154, \qquad s_p\approx4.8596\ \text{units}.

The degrees of freedom are $6+9-2=13$. The standard error and critical value are

SE=4.8596\sqrt{\frac{1}{6}+\frac{1}{9}} =4.8596\sqrt{0.277778} \approx2.5612\ \text{units}, \qquad t^*_{13}\approx1.771.

The margin of error is $1.771(2.5612)\approx4.536$ units. Since the sample difference is $74-69=5$ units, the interval is

5\pm4.536=(0.464,\ 9.536)\ \text{units}.

The interval estimates the population mean difference, group 1 minus group 2. Its positive endpoints suggest group 1 has the greater mean, but the interval’s coverage interpretation depends on the model assumptions being reasonable.

Common Mistake: Pooling Without a Common-Variance Model

A common mistake is to use the pooled procedure because the two observed standard deviations happen to be close, or simply because pooling gives a familiar formula. Sample standard deviations vary from sample to sample, so their similarity alone does not establish equal population variances. Conversely, a moderate difference between observed standard deviations is not by itself proof that the population variances differ.

The correction is to decide whether the common-variance assumption is scientifically and statistically defensible, considering the study design, group distributions, sample sizes, and graphical diagnostics. If the assumption is not credible, use the Welch procedure rather than forcing the data into the pooled model. Also keep the group order explicit: an interval for $\mu_1-\mu_2$ reverses sign if the groups are switched, while the substantive comparison remains the same.

Key takeaway: Pooling estimates one common variance by weighting each sample variance by its degrees of freedom. Under independent samples, suitable distributional conditions, and equal population variances, use $s_p$, $SE=s_p\sqrt{1/n_1+1/n_2}$, and $df=n_1+n_2-2$ for the test or interval for $\mu_1-\mu_2$.

Check Your Understanding

Use these questions to check whether you can set up and interpret a pooled analysis.

Why are the sample variances weighted by $n_1-1$ and $n_2-1$ in the pooled variance formula?
For independent samples with sizes 8 and 11, what are the degrees of freedom for a pooled t procedure?
A 95% pooled confidence interval for $\mu_1-\mu_2$ includes zero. What does that indicate about a two-sided test at the 5% level?
Why does a pooled analysis require a common population variance assumption?
If the group order changes from group 1 minus group 2 to group 2 minus group 1, what happens to the estimated difference and the interval endpoints?