Predicting at x* and the limits of the line

Two questions at the same x*

Module 4 predicted $\hat{y}(75)=544.18+2.066\times 75=699.1$ for a client earning $75,000 and left it as a point. Two different questions hide behind that number:

• What is the average credit score of all clients earning $75,000? — the mean response $\mu_{y \mid 75}$, a parameter on the population line.
• What score will this one new client have? — a single future observation, the line’s value plus that client’s own $\epsilon$.

Both intervals are centred on $\hat{y}(75)$ and both use $t_{\alpha /2;n-2}$; they differ only in the standard error.

$$\text{mean response:}\hat{y}\pm ts\sqrt{\frac{1}{n}+\frac{(x^{\ast }-\bar{x}{)}^2}{S{S}_{xx}}}$$

$$\text{prediction:}\hat{y}\pm ts\sqrt{1+\frac{1}{n}+\frac{(x^{\ast }-\bar{x}{)}^2}{S{S}_{xx}}}$$

The extra 1 is the whole difference. The first root measures how uncertain the line is at $x^{\ast }$ (the intercept and slope were estimated); the second adds the scatter of an individual around the line — one full $s^2$ — because a single client does not sit on the line even when the line is known exactly.

Worked example — Northfield at $75,000

$s=9.642$, $n=8$, $\bar{x}=69.375$, $S{S}_{xx}=4,531.875$, $t_{0.025;6}=2.447$, $\hat{y}=699.12$.

The shared piece: $\frac{1}{8}+\frac{(75-69.375{)}^2}{4,531.875}=0.125+\frac{31.64}{4,531.875}=0.125+0.00698=0.13198$.

Mean response. $\mathrm{SE}=9.642\sqrt{0.13198}=9.642\times 0.3633=3.503$.

$$699.12\pm 2.447\times 3.503=699.12\pm 8.57=(690.55,707.69)$$

With 95 % confidence, the average score of Northfield clients earning $75,000 is between 690.6 and 707.7.

Prediction. $\mathrm{SE}=9.642\sqrt{1.13198}=9.642\times 1.0639=10.259$.

$$699.12\pm 2.447\times 10.259=699.12\pm 25.10=(674.02,724.22)$$

A new client earning $75,000 will have a score between 674 and 724, with 95 % confidence. Three times wider — the individual’s own scatter dominates the line’s uncertainty here, as it does whenever $n$ is small and $x^*$ is near $\bar{x}$.

Both widen away from x̄

The term $(x^{\ast }-\bar{x}{)}^2/S{S}_{xx}$ is zero at $\bar{x}$ and grows quadratically away from it: the line is pinned best at the centre of the data and swings more at the ends, so both intervals are narrowest at $\bar{x}=69.375$ and flare outward. At $x^{\ast }=110$, the edge of the data, the mean-response half-width is already $2.447\times 9.642\sqrt{0.125+0.364}=16.5$ instead of 8.6.

The limits of the line

The intervals are honest only inside the data. Three reasons a lender should not set credit limits from $\widehat{\text{score}}=544+2.07\times \text{income}$:

1. $n=8$. Every interval above is wide because the line rests on eight points; the slope itself is only known to ±0.35.
2. Extrapolation. Incomes ran from $35,000 to $110,000. Outside that range the intervals are formulas, not evidence — nothing in the data says the relationship stays linear.
3. One predictor, observational data. Income is not the only driver of a credit score — age, existing debt, payment history — and the line cannot separate income’s association from theirs. A significant slope (L13.2) is association, not a mechanism to lend against.