Q31. Interpreting Poisson regression coefficients
[4 marks]
The model in Question 30 produces:
| Term | Estimate |
|---|---|
| Intercept | 1.50 |
distance |
-0.04 |
log(area) |
0.60 |
(a) Interpret the intercept in the context of the study.
(b) Interpret the distance coefficient on the scale of the expected species count.
(c) Compare two lakes at the same distance from the coast, where one has twice the area of the other. What does the model predict about their expected species counts?
(d) Assess this statement: “An extra hectare increases the expected species count by 60%.”
Definitions. Let \widehat\mu(d,A) be the fitted expected count at distance d kilometres and area A>0 hectares. A log link gives a mean ratio \mu_2/\mu_1=\exp(\eta_2-\eta_1), where \eta=\log\mu. For a logged covariate, the coefficient is an elasticity: the change in \log\mu per unit change in \log A. Each comparison below holds the other predictor fixed.
The fitted mean at distance d and area A is \widehat\mu(d,A)=\exp\{1.50-0.04d+0.60\log A\}.
(a) Intercept. At d=0 and \log A=0, so that A=1 hectare, \widehat\mu(0,1)=e^{1.50}\approx4.48. Conclusion: A one-hectare lake at the coast has a fitted expected species count of approximately 4.48. This is a reference prediction; it may involve extrapolation if such lakes are not represented in the data.
(b) Distance coefficient. Holding area fixed, increasing distance by one kilometre gives \frac{\widehat\mu(d+1,A)}{\widehat\mu(d,A)} =\frac{e^{1.50-0.04(d+1)+0.60\log A}} {e^{1.50-0.04d+0.60\log A}} =e^{-0.04}\approx0.9608. Conclusion: Each additional kilometre from the coast is associated with approximately 100(1-e^{-0.04})=3.92\% fewer expected species, holding lake area fixed.
(c) Doubling the area. At the same distance from the coast, \frac{\widehat\mu(d,2A)}{\widehat\mu(d,A)} =e^{0.60\{\log(2A)-\log A\}} =e^{0.60\log2}=2^{0.60}\approx1.516. Conclusion: The lake with twice the area has approximately 51.6\% more expected species.
(d) Assessing the extra-hectare claim. The statement is incorrect: the predictor is \log A, not A. An increase from A to A+1 hectares gives the ratio \frac{\widehat\mu(d,A+1)}{\widehat\mu(d,A)} =\left(\frac{A+1}{A}\right)^{0.60}, which depends on the original area. The coefficient 0.60 is an elasticity: \frac{\partial\log\mu}{\partial\log A}=0.60. For a small percentage increase in area, the expected count increases by approximately 0.60 times that percentage, holding distance fixed.