# Generalized Linear Models, Part I: The Logistic Model

Mon, Jul 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:

```
library(dplyr)
library(tidyr)
elections <- tibble(
party = c("Tories", "Labour", "LibDem", "Green", "Other"),
women = c(57,126,136,60,135),
men = c(577,508,496,192,546)
)
elections
```

```
## # 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.

# Model Specification

Before proceeding, we need to reshape the data.

```
elections_long <- elections %>%
pivot_longer(-party, names_to = "gender", values_to = "candidates") %>%
mutate(
gender_bin = case_when(
gender == "women" ~ 1L,
TRUE ~ 0L
)
) %>%
mutate_if(is.character, as.factor)
elections_long
```

```
## # 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)
##
## Deviance Residuals:
## 1 2 3 4 5 6 7 8 9 10
## 16.57 -10.43 20.18 -15.00 20.44 -15.50 13.12 -10.22 20.90 -15.53
##
## 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

\[ \frac{\frac{\pi(x = \text{Tories})}{1 - \pi(x = \text{Tories})}}{\frac{\pi(x = \text{LibDem})}{1 - \pi(x = \text{LibDem})}} \]

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)
```

`exp(confint(glht(fit, mcp(party = "Tukey")))[[10]])`

```
## Estimate lwr upr
## Labour - Green 0.7937008 0.4888272 1.2887190
## LibDem - Green 0.8774194 0.5426106 1.4188163
## Other - Green 0.7912088 0.4897871 1.2781294
## Tories - Green 0.3161179 0.1821097 0.5487381
## LibDem - Labour 1.1054788 0.7579106 1.6124372
## Other - Labour 0.9968603 0.6843155 1.4521524
## Tories - Labour 0.3982834 0.2503248 0.6336954
## Other - LibDem 0.9017453 0.6223133 1.3066482
## Tories - LibDem 0.3602814 0.2274127 0.5707804
## Tories - Other 0.3995379 0.2524558 0.6323107
## attr(,"conf.level")
## [1] 0.95
## attr(,"calpha")
## [1] 2.718904
```

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.

```
library(forcats)
elections_long$party <- fct_relevel(elections_long$party, "Tories", after = 0L)
fit <- glm(gender_bin ~ party,
weights = candidates,
family = binomial(link = "logit"),
data = elections_long)
fit
```

```
##
## Call: glm(formula = gender_bin ~ party, family = binomial(link = "logit"),
## data = elections_long, weights = candidates)
##
## Coefficients:
## (Intercept) partyGreen partyLabour partyLibDem partyOther
## -2.3148 1.1516 0.9206 1.0209 0.9174
##
## Degrees of Freedom: 9 Total (i.e. Null); 5 Residual
## Null Deviance: 2683
## Residual Deviance: 2629 AIC: 2639
```

Now we can compute the differences again

`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|)
## Green - Tories == 0 1.151640 0.202841 5.678 <1e-04 ***
## Labour - Tories == 0 0.920591 0.170806 5.390 <1e-04 ***
## LibDem - Tories == 0 1.020870 0.169230 6.032 <1e-04 ***
## Other - Tories == 0 0.917447 0.168845 5.434 <1e-04 ***
## 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
## LibDem - Labour == 0 0.100278 0.138831 0.722 0.950
## Other - Labour == 0 -0.003145 0.138361 -0.023 1.000
## Other - LibDem == 0 -0.103423 0.136411 -0.758 0.941
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## (Adjusted p values reported -- single-step method)
```

Green vs Tories are still different, but the sign is reversed!