Practice: Choosing a Regression Form

1. The response. Callouts are a count: non-negative integers, no upper bound, concentrated near zero with a long right tail. Two consequences. Ordinary least squares on the raw count assumes the response is continuous and unbounded in both directions, and that its variance is the same for every observation. Counts satisfy neither. Least squares on $\log (\text{callouts})$ is unavailable because most meters recorded zero and $\log 0$ is undefined. An offset such as $\log (y+1)$ would permit the fit, but the estimate then depends on whether 1 or 0.5 was added, with no principle to settle it. 2. The fitted model. The fit will predict negative callouts for young meters: at age 0 the intercept is whatever least squares chose, and with most observations at zero and a positive slope the line sits below zero somewhere in the lower age range. A negative count is not an imprecise prediction but an impossible one. The reported standard errors are computed from a single pooled residual variance. Because a count's variance rises with its mean, that figure is too large where callouts are rare and too small where they are common, so the uncertainty is misstated in both directions rather than uniformly. The sentence "0.41 more callouts per year of age" is a defensible reading of that fit; the problem is the fit, not the interpretation. 3. The form I would use. Poisson regression, modelling $\log \mathbb{E}[\text{callouts}\mid \text{age}]={\beta }_0+{\beta }_1\text{age}$. Modelling the log of the mean keeps every fitted mean positive, and the Poisson assumption $\mathrm{Var}(y)=\mathbb{E}[y]$ matches the variance structure the data show. The coefficient is reported by exponentiating: if ${\beta }_1=0.12$ then $e^{0.12}=1.127$, so the expected callout rate rises by about 12.7% per year of meter age, or equivalently is multiplied by 1.127. Quoting $0.12$ itself would state a log-scale slope and would be read as an additive change it is not. I would also check for overdispersion by comparing residual variance against fitted means. If variance runs well above the mean, the point estimates stand but the intervals are too narrow, and a model allowing extra dispersion is needed. 4. A curved relationship. Add a quadratic term in age, fitting $\log \mathbb{E}[y]={\beta }_0+{\beta }_{1}a+{\beta }_2{a}^2$. This is still a linear model in the sense that matters: the coefficients enter to the first power and multiply columns, so the estimation is unchanged and only the design matrix grows. Linearity refers to the coefficients, not to the predictors. The degree is a flexibility parameter, so it is chosen by held-out error rather than by residual deviance, which falls with every added column. If the same data must also report performance, the degree search belongs inside a nested scheme so the reported error is not the minimum of the values the search examined. 5. The projection to age 25. The data stop at 14. A prediction at 25 is nearly twice the observed range, and no observation bears on it. Whatever the fitted form produces there is a statement about the form: a Poisson fit with a positive slope keeps multiplying, a quadratic term makes it accelerate or turn depending on the sign of ${\beta }_2$, and a linear least-squares fit rises steadily. These disagree, and the data cannot adjudicate. This is why the answer does not depend on which form was fitted. Choosing the right form fixes points 1 through 4 and changes nothing about point 5. What can be claimed is the rate relationship over ages 0 to 14, reported with that range stated. Deciding about 25-year-old meters needs either observations of meters that old or an engineering model of the failure mechanism, not an extrapolated curve.