Probit Model
Priyank Goyal
06/04/2020
loading libraries
library(aod)## Warning: package 'aod' was built under R version 3.6.3library(ggplot2)## Registered S3 methods overwritten by 'ggplot2':
## method from
## [.quosures rlang
## c.quosures rlang
## print.quosures rlangReading Data
mydata <- read.csv("https://stats.idre.ucla.edu/stat/data/binary.csv")
## convert rank to a factor (categorical variable)
mydata$rank <- factor(mydata$rank)
## view first few rows
head(mydata)## admit gre gpa rank
## 1 0 380 3.61 3
## 2 1 660 3.67 3
## 3 1 800 4.00 1
## 4 1 640 3.19 4
## 5 0 520 2.93 4
## 6 1 760 3.00 2This data set has a binary response (outcome, dependent) variable called admit. There are three predictor variables: gre, gpa and rank. We will treat the variables gre and gpa as continuous. The variable rank takes on the values 1 through 4. Institutions with a rank of 1 have the highest prestige, while those with a rank of 4 have the lowest.
summary(mydata)## admit gre gpa rank
## Min. :0.0000 Min. :220.0 Min. :2.260 1: 61
## 1st Qu.:0.0000 1st Qu.:520.0 1st Qu.:3.130 2:151
## Median :0.0000 Median :580.0 Median :3.395 3:121
## Mean :0.3175 Mean :587.7 Mean :3.390 4: 67
## 3rd Qu.:1.0000 3rd Qu.:660.0 3rd Qu.:3.670
## Max. :1.0000 Max. :800.0 Max. :4.000Cross tabulating
xtabs(~rank + admit, data = mydata)## admit
## rank 0 1
## 1 28 33
## 2 97 54
## 3 93 28
## 4 55 12Applying the model
myprobit <- glm(admit ~ gre + gpa + rank, family = binomial(link = "probit"),
data = mydata)getting Summary
summary(myprobit)##
## Call:
## glm(formula = admit ~ gre + gpa + rank, family = binomial(link = "probit"),
## data = mydata)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.6163 -0.8710 -0.6389 1.1560 2.1035
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -2.386836 0.673946 -3.542 0.000398 ***
## gre 0.001376 0.000650 2.116 0.034329 *
## gpa 0.477730 0.197197 2.423 0.015410 *
## rank2 -0.415399 0.194977 -2.131 0.033130 *
## rank3 -0.812138 0.208358 -3.898 9.71e-05 ***
## rank4 -0.935899 0.245272 -3.816 0.000136 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 499.98 on 399 degrees of freedom
## Residual deviance: 458.41 on 394 degrees of freedom
## AIC: 470.41
##
## Number of Fisher Scoring iterations: 4In the output above, the first thing we see is the call, this is R reminding us what the model we ran was, what options we specified, etc.
Next we see the deviance residuals, which are a measure of model fit. This part of output shows the distribution of the deviance residuals for individual cases used in the model. Below we discuss how to use summaries of the deviance statistic to asses model fit.
The next part of the output shows the coefficients, their standard errors, the z-statistic (sometimes called a Wald z-statistic), and the associated p-values. Both gre, gpa, and the three terms for rank are statistically significant. The probit regression coefficients give the change in the z-score or probit index for a one unit change in the predictor.
For a one unit increase in gre, the z-score increases by 0.001.
For each one unit increase in gpa, the z-score increases by 0.478.
The indicator variables for rank have a slightly different interpretation. For example, having attended an undergraduate institution of rank of 2, versus an institution with a rank of 1 (the reference group), decreases the z-score by 0.415.
Below the table of coefficients are fit indices, including the null and deviance residuals and the AIC. Later we show an example of how you can use these values to help assess model fit.
We can use the confint function to obtain confidence intervals for the coefficient estimates. These will be profiled confidence intervals by default, created by profiling the likelihood function. As such, they are not necessarily symmetric.
confint(myprobit)## Waiting for profiling to be done...## 2.5 % 97.5 %
## (Intercept) -3.7201050682 -1.076327713
## gre 0.0001104101 0.002655157
## gpa 0.0960654793 0.862610221
## rank2 -0.7992113929 -0.032995019
## rank3 -1.2230955861 -0.405008112
## rank4 -1.4234218227 -0.459538829We can test for an overall effect of rank using the wald.test function of the aod library. The order in which the coefficients are given in the table of coefficients is the same as the order of the terms in the model. This is important because the wald.test function refers to the coefficients by their order in the model.
We use the wald.test function. b supplies the coefficients, while Sigma supplies the variance covariance matrix of the error terms, finally Terms tells R which terms in the model are to be tested, in this case, terms 4, 5, and 6, are the three terms for the levels of rank.
wald.test(b = coef(myprobit), Sigma = vcov(myprobit), Terms = 4:6)## Wald test:
## ----------
##
## Chi-squared test:
## X2 = 21.4, df = 3, P(> X2) = 8.9e-05The chi-squared test statistic of 21.4 with three degrees of freedom is associated with a p-value of less than 0.001 indicating that the overall effect of rank is statistically significant.
We can also test additional hypotheses about the differences in the coefficients for different levels of rank. Below we test that the coefficient for rank=2 is equal to the coefficient for rank=3. The first line of code below creates a vector l that defines the test we want to perform. In this case, we want to test the difference (subtraction) of the terms for rank=2 and rank=3 (i.e. the 4th and 5th terms in the model). To contrast these two terms, we multiply one of them by 1, and the other by -1. The other terms in the model are not involved in the test, so they are multiplied by 0. The second line of code below uses L=l to tell R that we wish to base the test on the vector l (rather than using the Terms option as we did above).
l <- cbind(0, 0, 0, 1, -1, 0)
wald.test(b = coef(myprobit), Sigma = vcov(myprobit), L = l)## Wald test:
## ----------
##
## Chi-squared test:
## X2 = 5.6, df = 1, P(> X2) = 0.018The chi-squared test statistic of 5.5 with 1 degree of freedom is associated with a p-value of 0.019, indicating that the difference between the coefficient for rank=2 and the coefficient for rank=3 is statistically significant.
You can also use predicted probabilities to help you understand the model. To do this, we first create a data frame containing the values we want for the independent variables.
newdata <- data.frame(gre = rep(seq(from = 200, to = 800, length.out = 100),
4 * 4), gpa = rep(c(2.5, 3, 3.5, 4), each = 100 * 4), rank = factor(rep(rep(1:4,
each = 100), 4)))
head(newdata)## gre gpa rank
## 1 200.0000 2.5 1
## 2 206.0606 2.5 1
## 3 212.1212 2.5 1
## 4 218.1818 2.5 1
## 5 224.2424 2.5 1
## 6 230.3030 2.5 1Now we can predict the probabilities for our input data as well as their standard errors. These are stored as new variable in the data frame with the original data, so we can plot the predicted probabilities for different gre scores. We create four plots, one for each level of gpa we used (2.5, 3, 3.5, 4) with the colour of the lines indicating the rank the predicted probabilities were for.
newdata[, c("p", "se")] <- predict(myprobit, newdata, type = "response", se.fit = TRUE)[-3]
ggplot(newdata, aes(x = gre, y = p, colour = rank)) + geom_line() + facet_wrap(~gpa)
We may also wish to see measures of how well our model fits. This can be particularly useful when comparing competing models. The output produced by summary(mylogit) included indices of fit (shown below the coefficients), including the null and deviance residuals and the AIC. One measure of model fit is the significance of the overall model. This test asks whether the model with predictors fits significantly better than a model with just an intercept (i.e. a null model). The test statistic is the difference between the residual deviance for the model with predictors and the null model. The test statistic is distributed chi-squared with degrees of freedom equal to the differences in degrees of freedom between the current and the null model (i.e. the number of predictor variables in the model). To find the difference in deviance for the two models (i.e. the test statistic) we can compute the change in deviance, and test it using a chi square test—the change in deviance distributed as chi square on the change in degrees of freedom.
## change in deviance
with(myprobit, null.deviance - deviance)## [1] 41.56335## change in degrees of freedom
with(myprobit, df.null - df.residual)## [1] 5## chi square test p-value
with(myprobit, pchisq(null.deviance - deviance, df.null - df.residual, lower.tail = FALSE))## [1] 7.218932e-08The chi-square of 41.56 with 5 degrees of freedom and an associated p-value of less than 0.001 tells us that our model as a whole fits significantly better than an empty model.