Polytomous Data: nested dichotomies

##       partic hincome children   region
## 3   not.work      45  present  Ontario
## 56  not.work      17   absent Atlantic
## 94  parttime       9  present  Ontario
## 98  fulltime      15   absent  Ontario
## 108 not.work      19  present  Ontario
## 171 fulltime       7  present  Prairie
## 177 not.work      15  present  Ontario
## 252 not.work      23   absent   Quebec

Recode to create dichotomies

Create binary variables working (working or not working), and fulltime (only for those working).

Womenlf <- Womenlf |>
  mutate(working = ifelse(partic=="not.work", 0, 1)) |>
  mutate(fulltime = case_when(
    working & partic == "fulltime" ~ 1,
    working & partic == "parttime" ~ 0)
  )
some(Womenlf, 8)

##       partic hincome children   region working fulltime
## 4   not.work      23  present  Ontario       0       NA
## 9   not.work      15  present  Ontario       0       NA
## 30  fulltime      14  present  Prairie       1        1
## 114 not.work       1  present Atlantic       0       NA
## 136 fulltime      11  present  Ontario       1        1
## 154 parttime      28  present       BC       1        0
## 177 not.work      15  present  Ontario       0       NA
## 180 fulltime      13   absent       BC       1        1

Fit models for each dichotomy

Womenlf <- within(Womenlf, contrasts(children)<- 'contr.treatment')
mod.working <- glm(working ~ hincome + children, family=binomial, data=Womenlf)
mod.fulltime <- glm(fulltime ~ hincome + children, family=binomial, data=Womenlf)

summary(mod.working)

## 
## Call:
## glm(formula = working ~ hincome + children, family = binomial, 
##     data = Womenlf)
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.677  -0.865  -0.777   0.929   1.997  
## 
## Coefficients:
##                 Estimate Std. Error z value Pr(>|z|)    
## (Intercept)       1.3358     0.3838    3.48   0.0005 ***
## hincome          -0.0423     0.0198   -2.14   0.0324 *  
## childrenpresent  -1.5756     0.2923   -5.39    7e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 356.15  on 262  degrees of freedom
## Residual deviance: 319.73  on 260  degrees of freedom
## AIC: 325.7
## 
## Number of Fisher Scoring iterations: 4

summary(mod.fulltime)

## 
## Call:
## glm(formula = fulltime ~ hincome + children, family = binomial, 
##     data = Womenlf)
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -2.405  -0.868   0.395   0.621   1.764  
## 
## Coefficients:
##                 Estimate Std. Error z value Pr(>|z|)    
## (Intercept)       3.4778     0.7671    4.53  5.8e-06 ***
## hincome          -0.1073     0.0392   -2.74   0.0061 ** 
## childrenpresent  -2.6515     0.5411   -4.90  9.6e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 144.34  on 107  degrees of freedom
## Residual deviance: 104.49  on 105  degrees of freedom
##   (155 observations deleted due to missingness)
## AIC: 110.5
## 
## Number of Fisher Scoring iterations: 5

lmtest::coeftest(mod.working)

## 
## z test of coefficients:
## 
##                 Estimate Std. Error z value Pr(>|z|)    
## (Intercept)       1.3358     0.3838    3.48   0.0005 ***
## hincome          -0.0423     0.0198   -2.14   0.0324 *  
## childrenpresent  -1.5756     0.2923   -5.39    7e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

lmtest::coeftest(mod.fulltime)

## 
## z test of coefficients:
## 
##                 Estimate Std. Error z value Pr(>|z|)    
## (Intercept)       3.4778     0.7671    4.53  5.8e-06 ***
## hincome          -0.1073     0.0392   -2.74   0.0061 ** 
## childrenpresent  -2.6515     0.5411   -4.90  9.6e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Anova(mod.working)  

## Analysis of Deviance Table (Type II tests)
## 
## Response: working
##          LR Chisq Df Pr(>Chisq)    
## hincome      4.83  1      0.028 *  
## children    31.32  1    2.2e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Anova(mod.fulltime)

## Analysis of Deviance Table (Type II tests)
## 
## Response: fulltime
##          LR Chisq Df Pr(>Chisq)    
## hincome       9.0  1     0.0027 ** 
## children     32.1  1    1.4e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Prepare for plotting

Create a grid of values of hincome and children. Generate predicted values from the models

attach(Womenlf)
predictors <- expand.grid(hincome=1:45, children=c('absent', 'present'))
# get fitted values for both sub-models
p.work     <- predict(mod.working, predictors, type='response')
p.fulltime <- predict(mod.fulltime, predictors, type='response')

calculate unconditional probs for the three response categories

p.full <- p.work * p.fulltime
p.part <- p.work * (1 - p.fulltime)
p.not <- 1 - p.work

NB: the plot looks best if the plot window is made wider than tall

op <- par(mfrow=c(1,2))
# children absent panel
plot(c(1,45), c(0,1), 
    type='n', xlab="Husband's Income", ylab='Fitted Probability',
    main="Children absent")
lines(1:45, p.not[1:45],  lty=1, lwd=3, col="black")
lines(1:45, p.part[1:45], lty=2, lwd=3, col="blue")
lines(1:45, p.full[1:45], lty=3, lwd=3, col="red")

legend(5, 0.97, lty=1:3, lwd=3, col=c("black", "blue", "red"),
    legend=c('not working', 'part-time', 'full-time'))  

# children present panel
plot(c(1,45), c(0,1), 
    type='n', xlab="Husband's Income", ylab='Fitted Probability',
    main="Children present")
lines(1:45, p.not[46:90],  lty=1, lwd=3, col="black")
lines(1:45, p.part[46:90], lty=2, lwd=3, col="blue")
lines(1:45, p.full[46:90], lty=3, lwd=3, col="red")

par(op)

A more general way to make the plot

op <- par(mfrow=c(1,2))
plotdata <- data.frame(predictors, p.full, p.part, p.not)
Hinc <- 1:max(plotdata$hincome)
for ( kids in c("absent", "present") ) {
    data <- subset(plotdata, children==kids)
    plot( range(data$hincome), c(0,1), type="n",
        xlab="Husband's Income", ylab='Fitted Probability',
        main = paste("Children", kids),
        cex.lab = 1.3)
    lines(Hinc, data$p.not,  lwd=3, col="black", lty=1)
    lines(Hinc, data$p.part, lwd=3, col="blue",  lty=2)
    lines(Hinc, data$p.full, lwd=3, col="red",   lty=3)
  if (kids=="absent") {
  legend(15, 0.97, lty=1:3, lwd=3, col=c("black", "blue", "red"),
    legend=c('not working', 'part-time', 'full-time'))
    }
}

par(op)


detach(Womenlf)