Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 5I / Solution

The R command fits a Poisson regression with the default log link:with independent annual counts conditional on the supplied temperature index. The fitted mean is $\hat{\lambda }(T)=\exp (2.26061+0.48870T)$. The intercept gives about $9.59$ expected named storms at index zero. A one-unit increase in the index multiplies the expected count by $e^{0.48870}\simeq 1.63$; a $0.1$ increase multiplies it by about $1.050$.

The reported standard errors estimate the sampling uncertainty of the coefficients. Each displayed $z$ is the estimate divided by its standard error, compared with a standard normal distribution under a zero-coefficient null hypothesis. The slope has $z=2.879$ and two-sided $p=0.00399$, evidence of positive association within this model. The intercept test is against mean one at index zero and is not itself a test of a climatic mechanism. Association does not establish causation.

There are $61$ annual observations and two fitted coefficients, giving $59$ residual degrees of freedom. The residual deviance iswith $0\log 0=0$. It is not large relative to $59$ and gives no obvious overdispersion signal, although temporal dependence and omitted covariates still require substantive assessment.

For 2005 insert the index into the fitted mean:This is an expected count, approximately fourteen storms, rather than a guarantee of an integer outcome. In R one would use predict(Mod, newdata=data.frame(Temp=0.743), type="response"). A predictive distribution must include Poisson variation and, if desired, coefficient-estimation uncertainty; the latter needs the coefficient covariance matrix, not just the two printed standard errors.