Stat 5421 Homework 4

Date Assigned: Wednesday, October 29, 2025

Date Due: Monday, November 24, 2025 at one minute before midnight

Instructions

Do using Rmarkdown. Upload two files to Canvas before the time Canvas says the homework is due

• the R markdown source (by convention this has extension .Rmd) and

• the R markdown output. Use output format HTML or output format PDF (your choice). This must be the result of running Rmarkdown on the source you submit.

Rmarkdown file

• https://www.stat.umn.edu/geyer/3701/repro/baz.Rmd

and the corresponding output

• https://www.stat.umn.edu/geyer/3701/repro/baz.html

• https://www.stat.umn.edu/geyer/3701/repro/baz.pdf

cover the basics of Rmarkdown. For the basics of markdown text formatting, see

• https://bookdown.org/yihui/rmarkdown/markdown-syntax.html which is just one section of a whole book R Markdown: The Definitive Guide.

For further info on Rmarkdown, the source for all of the course notes is always available for seeing how anything in any of the course notes was done.

Assignment

Solve each problem. Explain your reasoning. No credit for answers with no explanation.

4-1

Agresti problem 8.28 except just analyze as a contingency table. Do not treat any variable as the “response.” These data are in R package CatDataAnalysis.

library(CatDataAnalysis)
data(exercise_8.28)
sapply(exercise_8.28, class)

## Satisfaction      Contact    Influence      Housing       counts 
##     "factor"     "factor"     "factor"     "factor"    "numeric"

4-2

Agresti problem 8.28 as stated by Agresti (so this is a redo of the preceding problem treating Satisfaction as the “response”).

4-3

Agresti problem 8.8 except rather than use the analysis in Agresti do your own analysis using the R function polr in the R package MASS. The data can be read into R as follows as follows

u <- url("http://www.stat.umn.edu/geyer/5421/data/table-8.18.txt")
foo <- read.table(u, header = TRUE)
sapply(foo, class)

##      status   seat.belt    location      gender      counts 
## "character" "character" "character" "character"   "integer"

lapply(foo, unique)

## $status
## [1] "not injured"                   "injured, no ER"               
## [3] "injured, ER, not hospitalized" "hospitalized did not die"     
## [5] "died"                         
## 
## $seat.belt
## [1] "No"  "Yes"
## 
## $location
## [1] "Urban" "Rural"
## 
## $gender
## [1] "Female" "Male"  
## 
## $counts
##  [1]  7287   175   720    91    10 11587   126   577    48     8  3246    73
## [13]   710   159    31  6134    94   564    82    17 10381   136   566    96
## [25]    14 10969    83   259    37     1  6123   141   188    45  6693    74
## [37]   353    12

Note that status is not an ordered factor. You have to handle that yourself. The note on the home page about ordered factors may help.

Also note that counts cannot be the “response” variable on the left-hand side of the formula, because the “response” variable is the ordered factor, in this case status (or whatever variable name you call status after you turn it into an ordered factor). So where to you tell R function polr what the counts are? One of the examples on the help for the polr function shows how to use the weights argument for that.

Find the best model you can. R generic function anova function produces \(P\)-values for tests of model comparison. It has a method for objects of class "polr" produced by R function polr.

I’m not sure exactly what Agresti meant for (b). Just make a confidence interval for the regression coefficient for gender.

In part (c) the cumulative odds ratio Agresti asks about is discussed on page 302 in Agresti.

For part (c) interpret all of the regression coefficients. What does it mean that they have the signs they do (if anything). In interpreting an interaction term (like the location by seat belt interaction Agresti asks about), figure out what the size of the effects are for all four classes (for this interaction, rural without seat belt, rural with seat belt, urban without seat belt, urban with seat belt).

It may help to read the details section of the on-line help for the polr function. It may also help to look at the probabilities produced by the method of R generic function predict applied to R objects of class "polr" produced by R function polr. It is weird that this function has no on-line help but has arguments unlike the method of R generic function predict for R objects of class "glm" produced by R function glm. Say

predict(out, type = "probs")

to get predicted class probabilities. The way I figured this out was to look at the source, which is found with

getS3method("predict", "polr")

(shouldn’t have to do that). Maybe Venables and Ripley expect you to look in their book.

4-4

Agresti problem 4.13. The data can be read into R as follows

library(CatDataAnalysis)
data(exercise_4.13)
sapply(exercise_4.13, class)

##      game      made  attempts 
## "integer" "integer" "integer"