Confidence interval for the slope, and testing r

◈ 5 cards

β̂₁ ± t_{α/2; n−2} SE(β̂₁), read in units of y per unit of x; and the test of ρ = 0 via t = r√(n − 2)/√(1 − r²) is the slope test in different clothes — same t, same df, same decision.

The interval, by the grammar

Module 9’s one line — estimate ± critical value × standard error — with the slope filled in:

$\hat{β}_{1} \pm t_{α /2; n - 2} SE (\hat{β}_{1})$

The critical value is $t$ on $n - 2$ df because $s$ estimated $σ$ ; the SE is L13.2’s $s / S S_{xx}$ .

Worked example — how much does a thousand dollars buy?

Northfield Credit: $\hat{β}_{1} = 2.0659$ , $SE = 0.1432$ , $df = 6$ , 95 %: $t_{0.025; 6} = 2.447$ .

$2.0659 \pm 2.447 \times 0.1432 = 2.0659 \pm 0.3504 = (1.72, 2.42)$

With 95 % confidence, each additional $1,000 of annual income is associated with between 1.72 and 2.42 more points of credit score, on average, among Northfield’s clients. The units — points per $1,000 — are the interpretation; an interval quoted without them is a pair of numbers.

The interval excludes zero. That is Module 10’s duality again: a 95 % interval that excludes 0 is the same statement as rejecting $H_{0} : β_{1} = 0$ two-sided at 5 %. It excludes 1 and 3 as well, so a claim that "a thousand dollars is worth three points" is rejected at 5 % too — the interval answers every two-sided test at once.

Testing the correlation: the same test

Some papers ask instead whether the population correlation $ρ$ is zero. The statistic is

$t = \frac{r n - 2}{1 - r ^{2}}, df = n - 2$

For Northfield, with $r$ carried in full ( $0.98589$ ):

$t = \frac{0.98589 \times 6}{1 - 0.97197} = \frac{2.4149}{0.16743} = 14.42$

— identical to the slope $t$ of L13.2, to every decimal, and always will be. $H_{0} : ρ = 0$ and $H_{0} : β_{1} = 0$ are one hypothesis (the slope is $r s_{y} / s_{x}$ , so one is zero exactly when the other is), and the two formulas are algebraic rearrangements of each other. Say so once on the paper and run whichever the given numbers make easier: the slope route when $s$ and $S S_{xx}$ are printed, the $r$ route when only $r$ and $n$ are.

Two cautions. Rounding $r$ to 0.986 first gives $14.48$ — the third-decimal drift Module 4 warned about, harmless for the decision, visible in the number. And IS1 offers a table of critical $r$ values as a third route; the course does not ship it, because the $t$ route needs no new table and the paper favours it.

Reading the interval as a lender

The width, 0.70 points per $1,000, is large relative to the estimate — a consequence of $n = 8$. Doubling the sample would cut $\mathrm{SE}$ by about $\sqrt{2}$ and, with $t$ on 14 df instead of 6 (2.145 instead of 2.447), narrow the interval by nearly half. When a question asks "how could the firm estimate the slope more precisely?", the answer is more clients across a wider income range — both $n $an d$ SS_{xx}$ are in the SE.