Software 2.6.1 (A two-variable regression). The savings data of Example 1.3.5, with the savings rate regressed on the share of the population under 15:

> fit <- lm(sr ~ pop15, data = LifeCycleSavings)
> summary(fit)

Call:
lm(formula = sr ~ pop15, data = LifeCycleSavings)

Residuals:
   Min     1Q Median     3Q    Max
-8.637 -2.374  0.349  2.022 11.155

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept) 17.49660    2.27972   7.675 6.85e-10 ***
pop15       -0.22302    0.06291  -3.545 0.000887 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 4.03 on 48 degrees of freedom
Multiple R-squared:  0.2075,    Adjusted R-squared:  0.191
F-statistic: 12.57 on 1 and 48 DF,  p-value: 0.0008866

Reading the output, line by line against the theory of this section.

• Residuals: the five-number summary of the \(e_i\). Their mean is zero by Theorem 2.2.2, so it is not printed; a median near zero and roughly symmetric quartiles are what (A5) would lead one to expect.
• Estimate: \(a = 17.497\) and \(b = -0.223\). A country whose share of under-15s is one percentage point higher saves, on average, \(0.22\) percentage points less of its income.
• Std. Error: \(\operatorname {se}(b) = s/\sqrt {\sum x_i^2}\). Here \(\sum x_i^2 = 4103.95\) and \(s = 4.03\), so \(4.03/64.06 = 0.0629\).
• t value: the estimate divided by its standard error, the statistic for \(H_0\colon \beta = 0\). Pr(>|t|) is its two-sided \(p\)-value from \(t_{48}\). The stars are a convention of R’s, not a substitute for reading the \(p\)-value.
• Residual standard error: \(s = \sqrt {\text {RSS}/(n-2)}\), with \(48 = 50 - 2\) degrees of freedom.
• Multiple R-squared: \(R^2 = 0.2075 = (-0.4555)^2\), the squared correlation of Section 1.5 (Theorem 2.2.4). The adjusted \(R^2\) is explained with the \(k\)-variable model.
• F-statistic: \(12.57 = (-3.545)^2\), with the same \(p\)-value as the \(t\) test, by Theorem 2.4.2.

Stata: regress sr pop15. Stata prints the ANOVA table above the coefficients, with confidence intervals in place of \(p\)-value stars. EViews: ls sr c pop15, where c is EViews’ name for the constant.

Software 2.6.2 (The analysis of variance and confidence intervals). The decomposition of Theorem 2.2.3, and intervals for the coefficients:

> anova(fit)
Analysis of Variance Table

Response: sr
          Df Sum Sq Mean Sq F value    Pr(>F)
pop15      1 204.12  204.12  12.569 0.0008866 ***
Residuals 48 779.51   16.24
> confint(fit)
                 2.5 %      97.5 %
(Intercept) 12.9129180 22.08027686
pop15       -0.3494978 -0.09653739

Reading the output. ESS \(= 204.12\) on 1 degree of freedom and RSS \(= 779.51\) on 48. Their sum is TSS \(= 983.63\), and \(R^2 = 204.12/983.63 = 0.2075\). The residual mean square \(16.24 = s^2 = 4.03^2\). The interval for the slope is \(-0.2230 \pm t_{0.025,48}(0.0629)\) with \(t_{0.025,48} = 2.011\). It excludes zero, consistent with the \(t\) test.

Stata: the ANOVA table is printed by regress itself. EViews: the confidence intervals are under View \(\to \) Coefficient Diagnostics \(\to \) Confidence Intervals.

Software 2.6.3 (Predicting the mean and an individual). Predictions at pop15 \(= 25\) and \(45\), near the two ends of the data:

> new <- data.frame(pop15 = c(25, 45))
> predict(fit, new, interval = "confidence")
        fit       lwr       upr
1 11.921158 10.206061 13.636255
2  7.460807  5.762507  9.159107
> predict(fit, new, interval = "prediction")
        fit        lwr      upr
1 11.921158  3.6390423 20.20327
2  7.460807 -0.8178472 15.73946

Reading the output. The point predictions are the same for both questions; only the intervals differ. The mean savings rate of countries with 25% under 15 is estimated to lie between \(10.2\%\) and \(13.6\%\). A single such country could have a savings rate anywhere from \(3.6\%\) to \(20.2\%\). The prediction interval is about five times as wide, because \(s = 4.03\) is large compared with the precision of the line. Its lower limit at 45% under 15 is negative, an impossible savings rate. That is a reminder that the normal disturbance of (A5) is an approximation, and a poor one near a natural boundary.

Stata: after regress, predict yhat, xb; predict sem, stdp gives the standard error for the mean and predict sef, stdf that for an individual. EViews: Proc \(\to \) Forecast in the equation window, which produces the forecast and its standard error (the individual one).