Welch confidence interval for the difference

◈ 4 cards

(x̄₁ − x̄₂) ± t_{α/2; df} × √(s₁²/n₁ + s₂²/n₂) with the hand df; an interval entirely above zero says μ₁ > μ₂, one that straddles zero says the data do not settle which is larger.

Estimate ± critical value × SE, once more

The interval for $μ_{1} - μ_{2}$ has the shape every interval in this course has had:

$(\overset{x}{ˉ}_{1} - \overset{x}{ˉ}_{2}) \pm t_{α /2; df} \frac{s _{1}^{2}}{n _{1}} + \frac{s _{2}^{2}}{n _{2}}, df = min (n_{1} - 1, n_{2} - 1)$

The SE is L11.3’s Welch SE and the df is the same hand rule. Because the interval is two-sided, the critical value is the $α /2$ column: for 90 % confidence, column 0.05; for 95 %, column 0.025.

Worked example — how much faster?

The test in L11.3 said the new billing system is paid faster. The controller’s next question is the useful one: by how much? A 90 % interval for $μ_{1} - μ_{2}$ (current minus new):

$7.1 \pm t_{0.05; 9} \times 3.485 = 7.1 \pm 1.833 \times 3.485 = 7.1 \pm 6.39$

$(0.71, 13.49) days$

With 90 % confidence, invoices under the current system take between 0.7 and 13.5 days longer to be paid than under the new system. The interval lies entirely above zero, which is the interval’s way of saying $μ_{1} > μ_{2}$ — the same verdict as the test — but it also says how imprecise the estimate is: the saving could be under a day or nearly two weeks. Two samples of 12 and 10 with SDs near 9 cannot do better.

Reading the sign

The interval is for a difference, so its relation to zero carries the message:

Interval for $μ_{1} - μ_{2}$	Reading
entirely positive, e.g. (0.71, 13.49)	$μ_{1} > μ_{2}$ at that confidence
entirely negative, e.g. (−9.2, −1.4)	$μ_{1} < μ_{2}$
straddles zero, e.g. (−2.1, 5.4)	no evidence of a difference at that level

An interval that straddles zero does not show the means are equal — it shows the data cannot tell which is larger. Module 10’s duality carries over: a $(1 - α)$ interval that excludes zero is exactly a two-sided test rejecting at $α$ . Note the pairing of levels and tails, though: the 90 % interval here matches a two-sided test at 10 %, not the one-sided 5 % test of L11.3 — the two happen to agree on this data because the 5 % one-sided critical value and the 90 % two-sided one are the same 1.833.

Order matters, but only for the sign

Subtracting new minus current would give $- 7.1 \pm 6.39 = (- 13.49, - 0.71)$ : the same information with the sign flipped. Say which way round the difference is taken, in words, in the conclusion, and keep it the same as in the hypotheses.