回归：直线与它的警察
引入| English | 释义 | 拼音 |
|---|---|---|
| residual/rɪˈsɪdʒuːəl/ | 残差 | cán chà |
| least squares | 最小二乘 | zuì xiǎo èr chéng |
| correlation/ˌkɒrɪˈleɪʃn/ | 相关系数 | xiāng guān xì shù |
Farther, therefore slower?
- Twelve tracked deliveries: distance 1.0–3.75 km, time 22.2–35.1 minutes. The scatter is tight and upward — but how much of the time is distance, and what would change if you intervened? Those are different questions.
- Regression fits the line; the residuals and the design police the interpretation.
The fit
- Least squares 最小二乘 finds the line minimising squared misses: $\hat y = \mathbf{17.4 + 4.83x}$ — each extra kilometre costs 4.83 minutes.
- Correlation 相关系数 $r = 0.989$, $r^2 = \mathbf{0.979}$: 97.9% of the variance in time travels with distance in this sample.
The slope of the fitted line, in minutes per km, to two decimals?
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$b = 4.83$: on average, each extra kilometre travels with about 4.8 more minutes in this sample.
r = 0.989. The r-squared, to three decimals?
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$0.989^2 = 0.978 \approx 0.979$: about 98% of the time-variance moves with distance - association, not accuracy, not causation.
Read the parts properly
- Predictions: 2 km → $17.4 + 9.7 = 27.1$ minutes ✓; 3 km → 31.9 ✓. The intercept 17.4 at 0 km is prep+handover — meaningful only because the story says so.
- Extrapolation beyond 3.75 km is fiction: no data vouched for the line out there. A 7 km "prediction" of 51 minutes is a story, not a statistic.
Fit: y = 17.4 + 4.83x. Predicted minutes at 2 km, to one decimal?
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$17.4 + 4.83 \times 2 = 17.4 + 9.66 = 27.1$. At 3 km: 31.9 - inside the data's range, so a prediction, not a story.
A stakeholder asks for the predicted time at 7 km. The correct answer?
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The arithmetic extends; the evidence does not. No delivery beyond 3.75 km has vouched for the line - say so in the same breath.
What does the intercept 17.4 mean?
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x = 0 is outside the data but inside the story (a zero-km order is prep plus handover). State the reading and its basis - or say the intercept has no meaning here.
The residuals are the report's other half
- Residual 残差 = $y - \hat y$: the part the line missed. Here they run −1.2 to +1.1 minutes with NO pattern ✓ — the straight line is adequate.
- A curved or funnel-shaped residual plot means the fit is wrong somewhere: curvature, or unequal spread. Never report a line without its residual picture.
What does the fit license?
Sort claims a regression output supports from claims it cannot make, and residual patterns from adequate fits.
Residuals within plus or minus 1.2 minutes with no visible pattern mean the straight line is adequate here.
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Small and patternless residuals are exactly what an adequate fit leaves behind. Curvature or a funnel would say otherwise.
Match each regression quantity to its reading.
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Each number answers one question - and none of them answers causation.
The causal police
- $r = 0.989$ is as strong as business data gets — and proves no direction. Traffic density moves both distance and time; route assignment (far orders at off-peak) can correlate them without the fitted causal rate.
- The honest sentence: "distance PREDICTS time in this sample; a causal rate needs route assignment held fixed". Association is about this dataset's joint variation; causation is about what would change under intervention.
Which study would justify "cutting 1 km saves 4.83 minutes"?
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Only breaking the entanglement - by design - turns an association into a causal rate. More data or a bigger r refine the SAME associational claim.
Three regression slips
- Reading $r^2$ as accuracy. 0.979 describes variance shared, not predictions guaranteed.
- Blessing the intercept. x = 0 is often outside meaning; say so or stay quiet.
- Correlation → causation. No mechanism, no experiment, no caution — no claim.
Recap
- $\hat y = a + bx$ by least squares; slope in minutes per km.
- Examine the residuals; they grade the fit.
- $r^2$ is association; only design earns causation.