Poisson regression is used to model count variables.

This page uses the following packages. Make sure that you can load
them before trying to run the examples on this page. If you do not have
a package installed, run: install.packages("packagename"), or
if you see the version is out of date, run: update.packages().

require(ggplot2)require(sandwich)require(msm)

Version info: Code for this page was tested in R version 3.1.1 (2014-07-10)
On: 2014-08-11
With: sandwich 2.3-1; boot 1.3-11; knitr 1.6; pscl 1.04.4; vcd 1.3-1; gam 1.09.1; coda 0.16-1; mvtnorm 1.0-0; GGally 0.4.7; plyr 1.8.1; MASS 7.3-33; Hmisc 3.14-4; Formula 1.1-2; survival 2.37-7; psych 1.4.5; reshape2 1.4; msm 1.4; phia 0.1-5; RColorBrewer 1.0-5; effects 3.0-0; colorspace 1.2-4; lattice 0.20-29; pequod 0.0-3; car 2.0-20; ggplot2 1.0.0


Please note: The purpose of this page is to show how to use various data
analysis commands. It does not cover all aspects of the research process which
researchers are expected to do. In particular, it does not cover data
cleaning and checking, verification of assumptions, model diagnostics or
potential follow-up analyses.

Examples of Poisson regression

Example 1. The number of persons killed by mule or horse kicks in the
Prussian army per year. Ladislaus Bortkiewicz collected data from 20 volumes of
Preussischen Statistik. These data were collected on 10 corps of
the Prussian army in the late 1800s over the course of 20 years.

Example 2. The number of people in line in front of you at the grocery store.
Predictors may include the number of items currently offered at a special
discounted price and whether a special event (e.g., a holiday, a big sporting
event) is three or fewer days away.

Example 3. The number of awards earned by students at one high school.
Predictors of the number of awards earned include the type of program in which the
student was enrolled (e.g., vocational, general or academic) and the score on their
final exam in math.

Description of the data

For the purpose of illustration, we have simulated a data set for Example 3 above.
In this example, num_awards is the outcome variable and indicates the
number of awards earned by students at a high school in a year, math is a continuous
predictor variable and represents students’ scores on their math final exam, and prog is a categorical predictor variable with
three levels indicating the type of program in which the students were
enrolled. It is coded as 1 = “General”, 2 = “Academic” and 3 = “Vocational”.
Let’s start with loading the data and looking at some descriptive statistics.

p<-read.csv("https://stats.idre.ucla.edu/stat/data/poisson_sim.csv")p<-within(p, {prog<-factor(prog,levels=1:3,labels=c("General","Academic",                                                     "Vocational"))id<-factor(id)})summary(p)

##        id        num_awards           prog          math     
##  1      :  1   Min.   :0.00   General   : 45   Min.   :33.0  
##  2      :  1   1st Qu.:0.00   Academic  :105   1st Qu.:45.0  
##  3      :  1   Median :0.00   Vocational: 50   Median :52.0  
##  4      :  1   Mean   :0.63                    Mean   :52.6  
##  5      :  1   3rd Qu.:1.00                    3rd Qu.:59.0  
##  6      :  1   Max.   :6.00                    Max.   :75.0  
##  (Other):194

Each variable has 200 valid observations and their distributions seem quite
reasonable. The unconditional mean and variance of our outcome variable
are not extremely different. Our model assumes that these values, conditioned on
the predictor variables, will be equal (or at least roughly so).

We can use the tapply function to display the summary statistics by program
type. The table below shows the average numbers of awards by program type
and seems to suggest that program type is a good candidate for
predicting the number of awards, our outcome variable, because the mean value of
the outcome appears to vary by prog. Additionally, the
means and variances within each level of prog–the conditional
means and variances–are similar. A conditional histogram separated out by
program type is plotted to show the distribution.

with(p,tapply(num_awards, prog,function(x) {sprintf("M (SD) = %1.2f (%1.2f)",mean(x),sd(x))}))

##                General               Academic             Vocational 
## "M (SD) = 0.20 (0.40)" "M (SD) = 1.00 (1.28)" "M (SD) = 0.24 (0.52)"

ggplot(p,aes(num_awards,fill= prog))+geom_histogram(binwidth=.5,position="dodge")

Analysis methods you might consider

Below is a list of some analysis methods you may have
encountered. Some of the methods listed are quite reasonable, while others have
either fallen out of favor or have limitations.

Poisson regression

At this point, we are ready to perform our Poisson model analysis using
the glm function. We fit the model and store it in the object m1
and get a summary of the model at the same time.

summary(m1<-glm(num_awards~prog+math,family="poisson",data=p))

## 
## Call:
## glm(formula = num_awards ~ prog + math, family = "poisson", data = p)
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -2.204  -0.844  -0.511   0.256   2.680  
## 
## Coefficients:
##                Estimate Std. Error z value Pr(>|z|)    
## (Intercept)     -5.2471     0.6585   -7.97  1.6e-15 ***
## progAcademic     1.0839     0.3583    3.03   0.0025 ** 
## progVocational   0.3698     0.4411    0.84   0.4018    
## math             0.0702     0.0106    6.62  3.6e-11 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for poisson family taken to be 1)
## 
##     Null deviance: 287.67  on 199  degrees of freedom
## Residual deviance: 189.45  on 196  degrees of freedom
## AIC: 373.5
## 
## Number of Fisher Scoring iterations: 6

Cameron and Trivedi (2009) recommended using robust standard errors for the
parameter estimates to control for mild violation of the distribution
assumption that the variance equals the mean.
We use R package sandwich below to obtain the robust standard errors and
calculated the p-values accordingly. Together with the p-values, we have also
calculated the 95% confidence interval using the parameter estimates and their
robust standard errors.

cov.m1<-vcovHC(m1,type="HC0")std.err<-sqrt(diag(cov.m1))r.est<-cbind(Estimate=coef(m1),"Robust SE"= std.err,"Pr(>|z|)"=2*pnorm(abs(coef(m1)/std.err),lower.tail=FALSE),LL=coef(m1)-1.96*std.err,UL=coef(m1)+1.96*std.err)r.est

##                Estimate Robust SE  Pr(>|z|)      LL       UL
## (Intercept)    -5.24712   0.64600 4.567e-16 -6.5133 -3.98097
## progAcademic    1.08386   0.32105 7.355e-04  0.4546  1.71311
## progVocational  0.36981   0.40042 3.557e-01 -0.4150  1.15463
## math            0.07015   0.01044 1.784e-11  0.0497  0.09061

Now let’s look at the output of function glm more closely.

with(m1,cbind(res.deviance= deviance,df= df.residual,p=pchisq(deviance, df.residual,lower.tail=FALSE)))

##      res.deviance  df      p
## [1,]        189.4 196 0.6182

We can also test the overall effect of prog by comparing the deviance
of the full model with the deviance of the model excluding prog.
The two degree-of-freedom chi-square test indicates that prog, taken
together, is a statistically significant predictor of num_awards.

## update m1 model dropping progm2<-update(m1, .~.-prog)## test model differences with chi square testanova(m2, m1,test="Chisq")

## Analysis of Deviance Table
## 
## Model 1: num_awards ~ math
## Model 2: num_awards ~ prog + math
##   Resid. Df Resid. Dev Df Deviance Pr(>Chi)    
## 1       198        204                         
## 2       196        189  2     14.6  0.00069 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Sometimes, we might want to present the regression results as incident rate
ratios and their standard errors, together with the confidence interval. To
compute the standard error for the incident rate ratios, we will use the
Delta method. To this end, we make use the function deltamethod
implemented in R package msm.

s<-deltamethod(list(~exp(x1),~exp(x2),~exp(x3),~exp(x4)),coef(m1), cov.m1)## exponentiate old estimates dropping the p valuesrexp.est<-exp(r.est[,-3])## replace SEs with estimates for exponentiated coefficientsrexp.est[,"Robust SE"]<-srexp.est

##                Estimate Robust SE       LL      UL
## (Intercept)    0.005263   0.00340 0.001484 0.01867
## progAcademic   2.956065   0.94904 1.575551 5.54620
## progVocational 1.447458   0.57959 0.660335 3.17284
## math           1.072672   0.01119 1.050955 1.09484

The output above indicates that the incident rate for prog = “Academic” is 2.96
times the incident rate for the reference group (prog = “General”). Likewise,
the incident rate for prog = “Vocational” is 1.45 times the incident rate for the
reference group holding the other variables at constant. The percent change in the incident rate of
num_awards is by 7% for every unit increase in math.
For additional information on the various metrics in which the results can be
presented, and the interpretation of such, please see Regression Models for
Categorical Dependent Variables Using Stata, Second Edition by J. Scott Long
and Jeremy Freese (2006).

Sometimes, we might want to look at the expected marginal means. For
example, what are the expected counts for each program type holding math
score at its overall mean? To answer this question, we can make use of
the predict function. First off, we will make a small data set
to apply the predict function to it.

(s1<-data.frame(math=mean(p$math),prog=factor(1:3,levels=1:3,labels=levels(p$prog))))

##    math       prog
## 1 52.65    General
## 2 52.65   Academic
## 3 52.65 Vocational

predict(m1, s1,type="response",se.fit=TRUE)

## $fit
##      1      2      3 
## 0.2114 0.6249 0.3060 
## 
## $se.fit
##       1       2       3 
## 0.07050 0.08628 0.08834 
## 
## $residual.scale
## [1] 1

In the output above, we see that the predicted number of events for level 1
of prog is about .21, holding math at its mean. The predicted
number of events for level 2 of prog is higher at .62, and the
predicted number of events for level 3 of prog is about .31. The ratios
of these predicted counts ((frac{.625}{.211} = 2.96), (frac{.306}{.211} = 1.45)) match
what we saw looking at the IRR.

We can also graph the predicted number of events with the commands below.
The graph indicates that the most awards are predicted for those in the academic
program (prog = 2), especially if the student has a high math score. The
lowest number of predicted awards is for those students in the general program (prog
= 1). The graph overlays the lines of expected values onto the actual points,
although a small amount of random noise was added vertically to lessen
overplotting.

## calculate and store predicted valuesp$phat<-predict(m1,type="response")## order by program and then by mathp<-p[with(p,order(prog, math)), ]## create the plotggplot(p,aes(x= math,y= phat,colour= prog))+geom_point(aes(y= num_awards),alpha=.5,position=position_jitter(h=.2))+geom_line(size=1)+labs(x="Math Score",y="Expected number of awards")

Things to consider

References