Generalized Linear Models, Part I: The Logistic Model

R

An applied short approach to Generalized Linear Models.

Author

Mauricio “Pachá” Vargas S.

Published

July 18, 2022

Updated on 2023-01-16: Explains reasoning in confidence intervals to conclude that some parties are similar.

Updated on 2022-08-03: Corrects chances increase in the text for the 2.775 value.

Context

Let’s say we are interested in predicting the gender of a candidate for the British General Elections in 1992 by using the Political Parties as a predictor. We have the next data:

# A tibble: 5 × 3
party women men
<chr> <dbl> <dbl>
1 Tories 57 577
2 Labour 126 508
3 LibDem 136 496
4 Green 60 192
5 Other 135 546

Being the dependent variable a categorical one, we need to propose a Logistic Model.

Let \(Y_i \mid \pi_i \sim Bin(n, \pi_i)\). If \(n=1\), \(Y_i\) indicates that a candidate is a woman or man.

In this case the Generalized Linear Model matches the probability of success (i.e., the probability of the candidate being a woman if we define that \(Y_i=1\) in that case and zero otherwise).

A good reference for all the mathematical details is McCullag and Nelder, 1983.

# A tibble: 10 × 4
party gender candidates gender_bin
<fct> <fct> <dbl> <int>
1 Tories women 57 1
2 Tories men 577 0
3 Labour women 126 1
4 Labour men 508 0
5 LibDem women 136 1
6 LibDem men 496 0
7 Green women 60 1
8 Green men 192 0
9 Other women 135 1
10 Other men 546 0

To specify a Generalized Linear Model that considers Gender (i.e., 1: female, 0: male) as the response and the Political Party as the predictor, we fit the proposed model in R.

fit <-glm(gender_bin ~ party,weights = candidates,family =binomial(link ="logit"),data = elections_long)summary(fit)

Call:
glm(formula = gender_bin ~ party, family = binomial(link = "logit"),
data = elections_long, weights = candidates)
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -1.1632 0.1479 -7.864 3.71e-15 ***
partyLabour -0.2310 0.1783 -1.296 0.195
partyLibDem -0.1308 0.1768 -0.740 0.459
partyOther -0.2342 0.1764 -1.328 0.184
partyTories -1.1516 0.2028 -5.678 1.37e-08 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 2683.2 on 9 degrees of freedom
Residual deviance: 2628.7 on 5 degrees of freedom
AIC: 2638.7
Number of Fisher Scoring iterations: 5

Odds Ratios

We can obtain the odds ratio of a women candidate moving from Tories to Liberal Democrats. This corresponds to

From the model, we have \(\text{logit}(\pi) = \beta_0 + \beta_1 x\), this is the same as \[
\log \left[ \frac{\pi}{1 - \pi} \right] = \beta_0 + \beta_1 x
\implies \frac{\pi}{1 - \pi} = \exp[\beta_0 + \beta_1 x].
\]

In R, we obtain the odds ratio as a substraction of the estimated coefficients No. 5 and No. 3 for this case. This is, \(\exp[\beta_5 - \beta_3]\).

exp(coef(fit)[5] -coef(fit)[3])

partyTories
0.3602814

Which means that the chances of having a women candidate drop around 65% by moving from Tories to Liberal Democrats.

The opposite exercise would tell us that the chances increase around 177% by moving from Liberal Democrats to Tories.

exp(coef(fit)[3] -coef(fit)[5])

partyLibDem
2.775608

Hypothesis Testing

Consider the following hypothesis:

\(H_0: \beta = 0\)

\(H_0: \beta_{labour} = \beta_{libdem}\)

\(H_0: \beta_{labour} = \beta_{green}\)

To test these hypothesis we can estimate the constrats, followed by their exponentials and the respective confidence intervals. The function to use in this case corresponds to the General Linear Hypotheses.

For \(H_0: \beta = 0\) we have

library(multcomp)summary(glht(fit, mcp(party ="Tukey")), test =Chisqtest())

General Linear Hypotheses
Multiple Comparisons of Means: Tukey Contrasts
Linear Hypotheses:
Estimate
Labour - Green == 0 -0.231049
LibDem - Green == 0 -0.130770
Other - Green == 0 -0.234193
Tories - Green == 0 -1.151640
LibDem - Labour == 0 0.100278
Other - Labour == 0 -0.003145
Tories - Labour == 0 -0.920591
Other - LibDem == 0 -0.103423
Tories - LibDem == 0 -1.020870
Tories - Other == 0 -0.917447
Global Test:
Chisq DF Pr(>Chisq)
1 45.75 4 2.773e-09

The global test returns \(p_{calculated} < p_{critical}\) (\(p_{critical} = 0.05\)), therefore we reject this hypothesis.

For the other hypothesis we have

summary(glht(fit, mcp(party ="Tukey")))

Simultaneous Tests for General Linear Hypotheses
Multiple Comparisons of Means: Tukey Contrasts
Fit: glm(formula = gender_bin ~ party, family = binomial(link = "logit"),
data = elections_long, weights = candidates)
Linear Hypotheses:
Estimate Std. Error z value Pr(>|z|)
Labour - Green == 0 -0.231049 0.178269 -1.296 0.689
LibDem - Green == 0 -0.130770 0.176760 -0.740 0.946
Other - Green == 0 -0.234193 0.176391 -1.328 0.669
Tories - Green == 0 -1.151640 0.202841 -5.678 <1e-04 ***
LibDem - Labour == 0 0.100278 0.138831 0.722 0.950
Other - Labour == 0 -0.003145 0.138361 -0.023 1.000
Tories - Labour == 0 -0.920591 0.170806 -5.390 <1e-04 ***
Other - LibDem == 0 -0.103423 0.136411 -0.758 0.941
Tories - LibDem == 0 -1.020870 0.169230 -6.032 <1e-04 ***
Tories - Other == 0 -0.917447 0.168845 -5.434 <1e-04 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Adjusted p values reported -- single-step method)

Estimate lwr upr
Labour - Green 0.7937008 0.4888709 1.2886038
LibDem - Green 0.8774194 0.5426587 1.4186905
Other - Green 0.7912088 0.4898305 1.2780163
Tories - Green 0.3161179 0.1821282 0.5486822
LibDem - Labour 1.1054788 0.7579634 1.6123249
Other - Labour 0.9968603 0.6843630 1.4520516
Tories - Labour 0.3982834 0.2503463 0.6336411
Other - LibDem 0.9017453 0.6223559 1.3065588
Tories - LibDem 0.3602814 0.2274320 0.5707320
Tories - Other 0.3995379 0.2524772 0.6322571
attr(,"conf.level")
[1] 0.95
attr(,"calpha")
[1] 2.718402

Here, the differences that are not statistically significant reveal that some parties are similar to each other (in the gender dimension), which is the case for Green vs Labour and Labour vs LibDem but not for Greens vs Tories.

In this case we cannot ignore the exponential transformation to the confidence interval. What happens here is that the “zero is not round”. My apologies for the non-technical explanation, but here if the number one is contained in the confidence interval (i.e., one is the “non-round zero”), then the difference between parties is not statistically significant, and therefore the two parties are similar.

Changing the Reference Factor

Consider that Green is the reference factor in the previous model. To change the reference, we can use the Tories or any other party.