Pooled t when the variances are equal

◈ 6 cards

When the question says σ₁ = σ₂, pool the variances: s_p² = ((n₁ − 1)s₁² + (n₂ − 1)s₂²)/(n₁ + n₂ − 2), SE = s_p √(1/n₁ + 1/n₂), df = n₁ + n₂ − 2 — pool variances, never SDs.

One variance, estimated twice

If the two populations are known — or assumed by the question — to have the same variance $σ^{2}$ , then $s_{1}^{2}$ and $s_{2}^{2}$ are two estimates of one number, and it is wasteful to keep them apart. The pooled variance averages them, weighting each by its df:

$s_{p}^{2} = \frac{( n _{1} - 1 ) s _{1}^{2} + ( n _{2} - 1 ) s _{2}^{2}}{n _{1} + n _{2} - 2}$

With equal $n$ it is the plain average of the two variances; otherwise the larger sample counts more. The SE of the difference then factors:

$SE = \frac{s _{p}^{2}}{n _{1}} + \frac{s _{p}^{2}}{n _{2}} = s_{p} \frac{1}{n _{1}} + \frac{1}{n _{2}}$

and the statistic $t = (\overset{x}{ˉ}_{1} - \overset{x}{ˉ}_{2} - δ_{0}) / SE$ has an exact t distribution on $df = n_{1} + n_{2} - 2$ — the two samples’ df added. That is the reward for the assumption: a real df, larger than the min rule’s, with no approximation.

Worked example — basket size at two grocery chains

A retail analyst compares the mean basket size (\$) at two chains, sampling $n = 10$ baskets from each. Chain A: $\overset{x}{ˉ}_{1} = 48.6$ , $s_{1} = 5.2$ . Chain B: $\overset{x}{ˉ}_{2} = 44.9$ , $s_{2} = 4.7$ . Assume the population variances are equal. Do the mean basket sizes differ, at $α = 0.10$ ?

Step 1. $H_{0} : μ_{1} - μ_{2} = 0$ , $H_{a} : μ_{1} - μ_{2} \neq = 0$ . Two-sided, from differ.

Step 2. Variances: $5. 2^{2} = 27.04$ , $4. 7^{2} = 22.09$ .

$s_{p}^{2} = \frac{9 \times 27.04 + 9 \times 22.09}{18} = \frac{243.36 + 198.81}{18} = 24.565, s_{p} = 4.956$

$SE = 4.956 \frac{1}{10} + \frac{1}{10} = 4.956 \times 0.4472 = 2.217, t = \frac{48.6 - 44.9}{2.217} = \frac{3.7}{2.217} = 1.669$

Step 3. $df = 10 + 10 - 2 = 18$ . Two-sided at $α = 0.10$ : column 0.05, $t_{0.05; 18} = 1.734$ . Bracket: 1.669 sits between 1.330 (0.10) and 1.734 (0.05), so one-sided $0.05 < p < 0.10$ , two-sided $0.10 < p < 0.20$ .

Step 4. $1.669 < 1.734$ : do not reject $H_{0}$ at the 10 % level. There is insufficient evidence that mean basket size differs between the chains.

The 95 % interval. Column 0.025, row 18: $t_{0.025; 18} = 2.101$ .

$3.7 \pm 2.101 \times 2.217 = 3.7 \pm 4.66 = (- 0.96, 8.36)$

Straddles zero, as the test said it would (at 5 % two-sided the decision is the same: 1.669 < 2.101).

When to pool

Pool only when the question says the variances are equal — "assume equal population variances", "the two populations have the same SD". Never pool because the two sample SDs look close: 5.2 and 4.7 are close, and that is not an assumption, it is an observation. This course has no test for equal variances; without the statement, use Welch with the min rule. On this data Welch would give $df = 9$ and $t_{0.05; 9} = 1.833$ — the same decision, a wider margin.

Pool variances, never SDs: $(5.2 + 4.7) /2 = 4.95$ happens to be near $s_{p} = 4.956$ here only because the two SDs are nearly equal and the $n$ are equal. With $s_{1} = 2$ and $s_{2} = 10$ the average of the SDs is 6 while $s_{p} = 52 = 7.21$ .