Simple linear regression

Twenty days of takings at a coffee shop, and the number of customers on each day: does business scale with foot traffic, and how well can one predict the other? Assignment 2 answered with lm(). This chapter rebuilds everything that call produced.

§5.1Coursework A2 Q3

Fit, intervals and diagnostics

The model $Y_i=\beta_0+\beta_1x_i+\varepsilon_i$ , with independent $\varepsilon_i\sim N(0,\sigma^2)$ , is fitted by least squares:

b_1=\frac{\sum(x_i-\bar x)(y_i-\bar y)}{\sum(x_i-\bar x)^2},

b_0=\bar y-b_1\bar x,

s^2=\frac{\sum e_i^2}{n-2}.

At a new value $x_0$ the mean response and a single new response share the point estimate $\hat y_0=b_0+b_1x_0$ but not the uncertainty:

\text{mean: }\ \hat y_0 \pm t_{n-2}\,s\sqrt{\tfrac1n+\tfrac{(x_0-\bar x)^2}{S_{xx}}},

\text{new day: }\ \hat y_0 \pm t_{n-2}\,s\sqrt{1+\tfrac1n+\tfrac{(x_0-\bar x)^2}{S_{xx}}}.

Data (x, y per line)

20 rows · lines starting with # are ignored

Predict at x₀100

Confidence level95%

Default data are simulated coffee-shop days (sales ≈ −30 + 6.4 × customers + noise) — not the assignment file.

Slope b₁

6.3378

95% CI (5.039, 7.636)

Intercept b₀

-18.13

CI (-142.6, 106.3)

Mean at x₀ = 100

615.7

CI (590.8, 640.5)

New day at x₀ = 100

615.7

PI (507.0, 724.3)

Observed days
Least-squares line
95% confidence band for the mean
95% prediction band for a new day

Figure 5.1.The confidence band describes where the average sales line could be; the prediction band also allows for one day’s scatter around it, so it never shrinks below ±t·s. Both are narrowest at x̄ = 94.1.

Show the numbers as a table

#	x	y	fitted	residual	leverage
1	95	582.8	583.96	-1.16	0.050
2	123	658.2	761.42	-103.22	0.176
3	83	574.3	507.91	66.39	0.069
4	73	423.1	444.53	-21.43	0.117
5	108	731	666.36	64.64	0.079
6	81	504.7	495.23	9.47	0.076
7	103	648	634.67	13.33	0.062
8	103	611.1	634.67	-23.57	0.062
9	68	341.7	412.84	-71.14	0.153
10	102	716	628.33	87.67	0.059
11	74	530.7	450.87	79.83	0.111
12	89	490.7	545.94	-55.24	0.054
13	80	472.5	488.90	-16.40	0.080
14	77	478.3	469.88	8.42	0.094
15	97	566.1	596.64	-30.54	0.051
16	122	749.7	755.08	-5.38	0.167
17	124	782.5	767.76	14.74	0.185
18	87	541	533.26	7.74	0.058
19	122	766.2	755.08	11.12	0.167
20	71	396.6	431.86	-35.26	0.130

R console, recreated

> summary(lm(sales ~ customers))

Residuals:
     Min       1Q   Median       3Q      Max 
-103.222  -25.310    3.287   13.685   87.671 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) -18.1271    59.2481  -0.306    0.763    
customers     6.3378     0.6182  10.253 6.07e-09 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 50.36 on 18 degrees of freedom
Multiple R-squared:  0.8538,	Adjusted R-squared:  0.8457 
F-statistic: 105.1 on 1 and 18 DF,  p-value: 6.074e-09

Formatted with R’s own printing rules (formatReal, printCoefmat, format.pval); the test suite checks the text against R 4.6 character for character, for the 2020 data and 15 synthetic datasets.

Observations (largest three numbered)
LOWESS smooth, as in plot(lm)
Reference

Residuals vs fitted

Normal Q–Q

Scale–location

Residuals vs leverage

Figure 5.2.The four panels R draws for plot(mod). Look for curvature (top left), a straight Q–Q line (top right), constant spread (bottom left) and no single point with both high leverage and a large residual (bottom right, Cook’s distance contours at 0.5 and 1).

The four panels are read by eye. Formal tests of the same assumptions (Shapiro–Wilk, Breusch–Pagan, Durbin–Watson and a Bonferroni outlier test) run on this kind of data in Appendix C.

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