FS2 7.3 - Two-sample t-tests with pooled variance | Edexcel A-Level Further Mathematics

Two independent samples can estimate one common variance while comparing different population means. You will pool their within-sample variation, test a mean difference and construct its confidence interval under the normal and equal-variance assumptions.

Separate populations, one common unknown variance

If every measurement in group A rises by $100$ units, should the pooled variance rise too?

No. Pooling here combines variation within the groups. Their means may be far apart without changing that within-group spread. The difference between the means belongs in the numerator of the comparison, rather than being mixed into the variance estimate.

You already know

Unknown-variance inference for one normal mean uses a t statistic and unbiased sample variance. An F-test can investigate equality of two normal population variances, but non-rejection does not prove equality. Paired observations require analysis of their differences.

For this procedure, take two independent random samples from normal populations. Assume the population variances are equal but unknown:

X\sim\mathrm N(\mu_A,\sigma^2), \qquad Y\sim\mathrm N(\mu_B,\sigma^2)

The means $\mu_A,\mu_B$ need not be equal. Equality of the variances is a modelling condition that permits a common spread estimate.

Each sample contains independent observations, and the two samples are independent of one another. Measurements repeated on the same people do not satisfy that independent-sample design just because they appear in separate table rows.

An F-test may provide evidence against equal variances. If the common-variance model is unsuitable, the pooled t procedure below is not justified. Do not quietly replace its assumptions with an unrelated z calculation.

Key idea

Pool the within-group variance information, then compare population means. Independence, normal populations and equal unknown variances are distinct requirements.

Check the whole model, not only one convenient condition.

Pool centred sums, not raw measurements

The unbiased variance $s_A^2$ summarises a centred squared total $(n_A-1)s_A^2$ . Add the two groups' centred totals, then divide by their combined degrees of freedom.

Pooled estimate of the common variance

For independent normal samples with a common unknown population variance, and $n_A,n_B\ge2$ ,

s_p^2= \frac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}

The estimated standard error of $\bar X-\bar Y$ is

s_p\sqrt{\frac1{n_A}+\frac1{n_B}} =\sqrt{s_p^2\left(\frac1{n_A}+\frac1{n_B}\right)}

The pooled variance and its use in the two-sample t denominator are supplied. In the formula booklet.

The weights reflect the information contributing to each variance estimate. Only when the sample sizes are equal does pooling reduce to their ordinary arithmetic average.

Worked example 1: unequal sample sizes

Question. Independent samples from normal populations with equal unknown variances have $n_A=6,n_B=4$ , with unbiased variances $s_A^2=9\,\mathrm{mm^2}$ and $s_B^2=25\,\mathrm{mm^2}$ . Find the pooled variance and estimated standard error of their mean difference.

Plan. Pool the within-group centred totals, then take the square root of the estimated mean-difference variance.

\begin{aligned} (n_A-1)s_A^2&=5(9)=45,\qquad(n_B-1)s_B^2=3(25)=75\\ s_p^2&=\frac{45+75}{6+4-2}=\frac{120}{8}=15\\ \widehat{\operatorname{se}}(\bar X-\bar Y) &=\sqrt{15\left(\frac16+\frac14\right)} =\sqrt{\frac{25}{4}}=2.5 \end{aligned}

Why this step? Pooling combines centred squared information with $5+3=8$ degrees of freedom; it does not combine the raw measurements around a single grand mean.

Answer. The pooled variance is $15\,\mathrm{mm^2}$ , and the estimated standard error is $2.5\,\mathrm{mm}$ .

Check. The pooled variance lies between $9$ and $25$ , closer to $9$ because that group contributes more degrees of freedom.

Worked example 2 changes only group B's sample variance from $25$ to $9$ .

Worked example 2: both estimates now agree

Question. Recalculate the pooled variance and standard error after this change.

Plan. Retain the weights and replace the second centred total.

\begin{aligned} s_p^2&=\frac{5(9)+3(9)}8=9\\ \widehat{\operatorname{se}}(\bar X-\bar Y) &=\sqrt{9\left(\frac16+\frac14\right)} =\frac{\sqrt{15}}2=1.93649\ldots \end{aligned}

Why this step? Equal sample-variance estimates remain equal after weighting, but the mean-difference standard error still depends on both sample sizes.

Answer. The pooled variance is $9\,\mathrm{mm^2}$ ; the estimated standard error is $1.94\,\mathrm{mm}$ , to three significant figures.

Check. Lower estimated spread produces a lower estimated standard error.

Adding $100\,\mathrm{mm}$ to every A measurement adds the same amount to its sample mean. Every deviation $x_i-\bar x$ stays unchanged, so $s_A^2$ , $s_p^2$ and the estimated standard error stay unchanged. This answers the opening question.

Watch out

Wrong: use $(9+25)/2=17$ , or divide a difference of means by $s_p^2$ . Right: weight by $n_A-1,n_B-1$ , then use the square root of the estimated mean-difference variance.

Change only group B's sample variance again.

Test a population mean difference

Let $\Delta=\mu_A-\mu_B$ . A zero null difference tests equality of the means; a non-zero null difference assesses a specified gap. Keep the order of the means consistent with the alternative.

Two-sample t statistic with equal unknown variances

For the independent common-variance normal model, under $H_0:\mu_A-\mu_B=\Delta_0$ ,

T= \frac{\bar X-\bar Y-\Delta_0} {S_p\sqrt{1/n_A+1/n_B}} \sim t_{n_A+n_B-2}

Use observed $s_p$ for the numerical statistic. In the formula booklet.

Two estimated group centres cost two degrees of freedom in total. The hypotheses concern population means, not their observed sample estimates.

Worked example 3: a two-sided comparison

Question. Independent component-length samples from normal populations with equal unknown variances give

n_A=6,\quad n_B=4,\quad \bar x=55\,\mathrm{mm},\quad\bar y=50\,\mathrm{mm}, \quad s_A^2=9\,\mathrm{mm^2},\quad s_B^2=25\,\mathrm{mm^2}

Test at $5\%$ whether the population means differ.

Plan. Use the pooled estimate from Worked example 1 and a two-sided $t_8$ comparison.

\begin{aligned} H_0&:\mu_A-\mu_B=0,\qquad H_1:\mu_A-\mu_B\ne0\\ s_p^2&=\frac{5(9)+3(25)}8=15\\ t&=\frac{55-50}{\sqrt{15(1/6+1/4)}}=\frac5{2.5}=2\\ |2|&<2.306 &&8\text{ degrees of freedom, }2.5\%\text{ each tail} \end{aligned}

Why this step? Both sample means contribute uncertainty, and both variance estimates contribute to the common denominator.

Answer. Do not reject $H_0$ . There is insufficient evidence at $5\%$ that the population mean component lengths differ.

Check. A sample difference of $5\,\mathrm{mm}$ need not be significant when its estimated standard error is $2.5\,\mathrm{mm}$ .

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FS2 7.3 - Two-sample t-tests with pooled variance

Separate populations, one common unknown variance

Pool centred sums, not raw measurements

Test a population mean difference

Continue this lesson.

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