Pooled Versus Unpooled Two-Sample Intervals

Two Ways to Estimate Uncertainty

In “Order of Subtraction and Interval Interpretation,” you kept the order of a difference consistent from the parameter through to the interpretation. Here we focus on a different choice in a two-sample t interval: whether to estimate the two populations’ variability separately or combine the estimates. The standard AP Statistics approach uses separate estimates, an approach called unpooled.

Suppose two independent samples provide sample standard deviations \(s_1\) and \(s_2\). The unpooled standard error for \(\bar{x}_1-\bar{x}_2\) uses a separate variance contribution from each group. A pooled method instead combines the two sample variances to estimate one common population variance. That combination is appropriate only if we assume the two populations have the same variance.

Definition: An unpooled two-sample t interval estimates the variability in \(\bar{x}_1-\bar{x}_2\) from the two samples separately. A pooled interval combines the sample variances and assumes the two populations have a common variance, or equivalently a common population standard deviation.

Pooling does not mean combining the observations into one group or averaging the two sample means. The parameter remains \(\mu_1-\mu_2\), and the center of either interval remains \(\bar{x}_1-\bar{x}_2\). Pooling changes how the standard error and degrees of freedom are calculated.

The Unpooled and Pooled Formulas

As in “Standard Error for a Difference in Means” and “Computing Two-Sample Degrees of Freedom With the Welch Formula,” the unpooled standard error gives each group its own variance contribution. The interval uses Welch degrees of freedom, usually calculated by a calculator.

SE_{\text{unpooled}}= \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}, \qquad \bar{x}_1-\bar{x}_2\ \pm\ t^*SE_{\text{unpooled}}

The pooled method first calculates a weighted average of the two sample variances. The weights are the groups’ degrees of freedom, \(n_1-1\) and \(n_2-1\). Its estimated common standard deviation is \(s_p\).

s_p^2= \frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}, \qquad SE_{\text{pooled}}= s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}

A pooled interval uses \(df=n_1+n_2-2\). In contrast, the unpooled interval uses Welch degrees of freedom, which can be fractional. The formulas can produce different standard errors, critical values, and interval widths, especially when the sample sizes and sample standard deviations differ.

Key distinction: The standard AP two-sample t interval is unpooled. It does not require the two population standard deviations to be equal. Pooling requires an additional common-variance assumption that is not part of the standard AP approach.

Both methods still require the conditions for a two-sample t procedure: independent groups, random sampling or an appropriate randomized process, the 10% condition when sampling without replacement, and suitable distributions within each population. For small samples, check each group’s distribution for strong skewness or outliers; a larger sample can make t procedures more robust to nonnormality. Pooling does not remove these conditions.

Worked Examples

Worked Example: Comparing Approaches for Monthly Study Time

Consider an invented comparison of monthly study time for students at two independent high schools. A random sample of 10 students from each school gives the following summaries. Study time is measured in hours per month.

Group	\(n\)	\(\bar{x}\)	\(s\)
School A	10	42	4
School B	10	36	3

The target is a 95% confidence interval for \(\mu_1-\mu_2\), where \(\mu_1\) is the true mean monthly study time for students at School A and \(\mu_2\) is the true mean monthly study time for students at School B.

State: We want to estimate \(\mu_1-\mu_2\), the difference in true mean monthly study time for students at School A minus students at School B.

Plan: Use an unpooled two-sample t interval. The samples are described as random, and the two groups contain different students, so there is no pairing. Assume each school has at least 100 students, making each sample no more than 10% of its population. Since both sample sizes are small, suppose plots of the two samples show no strong skewness or outliers. These conditions support the interval.

Do: The difference in sample means is \(42-36=6\) hours per month. The unpooled standard error is:

SE_{\text{unpooled}} =\sqrt{\frac{4^2}{10}+\frac{3^2}{10}} =\sqrt{1.6+0.9} =\sqrt{2.5} \approx 1.581

Welch degrees of freedom are:

df= \frac{(1.6+0.9)^2}{\frac{1.6^2}{9}+\frac{0.9^2}{9}} = \frac{6.25}{0.3744} \approx 16.69

For 95% confidence and \(df\approx16.69\), \(t^*\approx2.113\). The margin of error is \(2.113(1.581)\approx3.341\) hours per month. Therefore:

6\pm3.341=(2.659,\ 9.341)\text{ hours per month}

For comparison only, suppose we calculate a pooled interval. Its pooled standard deviation is:

s_p= \sqrt{\frac{9(4^2)+9(3^2)}{18}} =\sqrt{12.5} \approx3.536

The pooled standard error is \(3.536\sqrt{1/10+1/10}\approx1.581\), and the pooled degrees of freedom are \(10+10-2=18\). With \(t^*\approx2.101\), its margin of error is about \(3.322\), giving an interval of approximately \((2.678,\ 9.322)\) hours per month.

Conclude: The unpooled interval gives 95% confidence that the true mean monthly study time at School A minus the true mean monthly study time at School B is between about 2.659 and 9.341 hours per month. Pooling gives a slightly different interval, but it requires assuming equal population variances. The standard AP answer is the unpooled interval; the sample standard deviations alone do not establish that the population variances are equal.

Worked Example: Unequal Sample Sizes and Different Variability

An invented environmental comparison measures the mean mass of collected litter per shoreline section, in grams. A random sample of 8 sections from site A has \(s_1=2\) grams; a random sample of 20 sections from site B has \(s_2=6\) grams. Let \(\bar{x}_1-\bar{x}_2=4\) grams. Assume the sites’ sections are independent, each sample is no more than 10% of its site’s sections, and plots show no strong skewness or outliers.

The unpooled standard error is:

SE_{\text{unpooled}} =\sqrt{\frac{2^2}{8}+\frac{6^2}{20}} =\sqrt{0.5+1.8} =\sqrt{2.3} \approx1.517\text{ grams}

Welch degrees of freedom are approximately:

df= \frac{(0.5+1.8)^2}{\frac{0.5^2}{7}+\frac{1.8^2}{19}} = \frac{5.29}{0.2062} \approx25.65

For a 90% interval, \(t^*\approx1.706\), so the margin of error is \(1.706(1.517)\approx2.587\) grams. The unpooled interval is \(4\pm2.587=(1.413,\ 6.587)\) grams.

Now calculate the pooled estimate for comparison. Its variance is \([7(2^2)+19(6^2)]/26=712/26\approx27.385\), so \(s_p\approx5.233\) grams. Then:

SE_{\text{pooled}} =5.233\sqrt{\frac{1}{8}+\frac{1}{20}} =5.233\sqrt{0.175} \approx2.189\text{ grams}

With pooled \(df=8+20-2=26\), the 90% critical value is about 1.706. The pooled margin of error is \(1.706(2.189)\approx3.734\) grams, giving an interval of about \((0.266,\ 7.734)\) grams.

The pooled interval is wider here because combining the variances and applying the pooled formula produces a larger standard error. The key point is not that pooling always widens an interval; it does not. The result depends on the sample sizes and variability. The unpooled method accommodates separate variance estimates without requiring the equal-variance assumption.

Worked Example: Equal Sample Sizes Can Hide a Difference

Suppose two independent samples each have \(n=12\). Their sample standard deviations are \(s_1=5\) and \(s_2=5.4\), and their sample means differ by 3 units. Assume the randomization, independence, 10% condition, and distribution conditions are satisfied. Compare the standard errors; this example shows that equal sample sizes can make them match even when the sample standard deviations do not.

The unpooled standard error is:

SE_{\text{unpooled}} =\sqrt{\frac{25}{12}+\frac{29.16}{12}} =\sqrt{4.5133} \approx2.124

The pooled variance is \([11(25)+11(29.16)]/22=27.08\), so \(s_p=\sqrt{27.08}\approx5.204\). The pooled standard error is:

SE_{\text{pooled}} =5.204\sqrt{\frac{1}{12}+\frac{1}{12}} =5.204\sqrt{\frac{1}{6}} \approx2.124

With equal sample sizes, these standard errors are the same: the pooled variance is the average of the two sample variances, and its standard-error formula gives the same result as keeping the two contributions separate. The degrees of freedom may still differ: pooled \(df=22\), while Welch degrees of freedom are about 21.9. Thus the critical values and endpoints can differ slightly even when the standard errors match.

Common Mistakes and AP Exam Tips

Assuming the sample standard deviations must match: The unpooled method allows \(s_1\) and \(s_2\) to differ. Do not replace it with a pooled method just because the sample standard deviations look similar.
Treating pooling as averaging the sample means: Pooling combines variance estimates, not the groups’ means or observations. The center remains \(\bar{x}_1-\bar{x}_2\).
Using pooled degrees of freedom with an unpooled standard error: Match the degrees of freedom to the method. Welch degrees of freedom go with the unpooled standard error; \(n_1+n_2-2\) goes with the pooled method.
Claiming pooling is justified because the sample sizes are equal: Equal sample sizes can make the two standard-error calculations equal, but they do not prove that the population variances are equal.
Forgetting what pooling assumes: A pooled calculation relies on a common population variance. The unpooled AP method does not require that assumption.
Skipping conditions because the formulas are familiar: Identify the independent groups, check random sampling or random assignment as appropriate, apply the 10% condition when needed, and assess the distributions or sample sizes in context.

For a standard AP solution, state that you are using a two-sample t interval for independent means, check its conditions, calculate the unpooled standard error, and use Welch degrees of freedom. Then interpret the interval for \(\mu_1-\mu_2\) in context and in the stated subtraction order. If a question specifically asks about pooling, make the equal-variance assumption explicit rather than treating it as automatic.

Key takeaway: The standard AP two-sample t interval is unpooled: it estimates each group’s variance contribution separately and uses Welch degrees of freedom. A pooled interval combines the variance estimates and requires assuming a common population variance, an assumption the standard AP approach does not make.

Check Your Understanding

Use the distinction between the variance calculation and its assumption to answer these questions.

What does a pooled two-sample t interval assume about the two populations’ variances?
Two independent samples have \(n_1=15\), \(s_1=4\), \(n_2=15\), and \(s_2=7\). Does having equal sample sizes establish that pooling is appropriate? Explain.
Which degrees of freedom are used with the unpooled method, and which are used with the pooled method?
In a standard AP two-sample t interval, why are the sample variances kept separate?
If two groups have unequal sample sizes and noticeably different sample standard deviations, which method avoids assuming a common population variance?

Pooled Versus Unpooled Two-Sample Intervals

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