The crabs data set

The crabs data set is derived from Agresti (2007, Table 3.2, pp.76-77). It gives 4 variables for each of 173 female horseshoe crabs.

library(glm2)

data(crabs)
head(crabs) %>% kable()
Satellites Width Dark GoodSpine Rep1 Rep2
8 28.3 no no 2 2
0 22.5 yes no 4 5
9 26.0 no yes 5 6
0 24.8 yes no 6 6
4 26.0 yes no 6 8
0 23.8 no no 8 8
summary(crabs[,1:4])  %>% kable()
Satellites Width Dark GoodSpine
Min. : 0.000 Min. :21.0 no :107 no :121
1st Qu.: 0.000 1st Qu.:24.9 yes: 66 yes: 52
Median : 2.000 Median :26.1 NA NA
Mean : 2.919 Mean :26.3 NA NA
3rd Qu.: 5.000 3rd Qu.:27.7 NA NA
Max. :15.000 Max. :33.5 NA NA

Question

Fit a Poisson regression model with the number of Satellites as the outcome and the width of the female as the covariate.

What is the multiplicative change in the expected number of crabs for each additional centimeter of width?

crabs.pois <- glm2( Satellites ~ Width, data=crabs, family="poisson")

summary(crabs.pois)
## 
## Call:
## glm2(formula = Satellites ~ Width, family = "poisson", data = crabs)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.8526  -1.9884  -0.4933   1.0970   4.9221  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -3.30476    0.54224  -6.095  1.1e-09 ***
## Width        0.16405    0.01997   8.216  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for poisson family taken to be 1)
## 
##     Null deviance: 632.79  on 172  degrees of freedom
## Residual deviance: 567.88  on 171  degrees of freedom
## AIC: 927.18
## 
## Number of Fisher Scoring iterations: 6
exp(-3.30476)*exp(0.164*25)
## [1] 2.214973
plot(crabs$Width, crabs$Satellites,
     pch=16, col="darkred")
points(crabs$Width, crabs.pois$fitted.values, 
     col="green", type="p", lwd=3)

Question

What is the expected number of Satellites for a female of width 22cm?

Given a set of parameters \(\{\beta_0,\beta_1, \ldots, \beta_n\}\) and an input vector \(x\) (i.e \(\{x_1,x_2, \ldots x_n\}\)), the mean of the predicted Poisson distribution is given by \[\operatorname{E}(Y|x)=e^{\beta_0+\beta_1x_1 + \ldots + \beta_nx_n}\, \] In the case of one predictor variable, we can say \[\operatorname{E}(Y|x)=e^{\beta_0+\beta_1x}=e^{\beta_0}\times e^{\beta_1x}\,\]

Remark: The expected value does not have to be an integer.

exp(-3.30476)*exp(0.16405*22)
## [1] 1.35573