Probit regression, also called a probit model, is used to model dichotomous or binary outcome variables. In the probit model, the inverse standard normal distribution of the probability is modeled as a linear combination of the predictors.
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them before trying to run the examples on this page. If you do not have
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if you see the version is out of date, run:
Version info: Code for this page was tested in R version 3.1.1 (2014-07-10)
With: ggplot2 1.0.1; aod 1.3; knitr 1.10.5
Please Note: The purpose of this page is to show how to use various data analysis commands. It does not cover all aspects of the research process which researchers are expected to do. In particular, it does not cover data cleaning and checking, verification of assumptions, model diagnostics and potential follow-up analyses.
Example 1: Suppose that we are interested in the factors that influence whether a political candidate wins an election. The outcome (response) variable is binary (0/1); win or lose. The predictor variables of interest are the amount of money spent on the campaign, the amount of time spent campaigning negatively and whether the candidate is an incumbent.
Example 2: A researcher is interested in how variables, such as GRE (Graduate Record Exam scores), GPA (grade point average) and prestige of the undergraduate institution, effect admission into graduate school. The response variable, admit/don't admit, is a binary variable.
For our data analysis below, we are going to expand on Example 2 about getting into graduate school. We have generated hypothetical data, which can be obtained from our website in R. Note that R requires forward slashes (/) not back slashes (\) when specifying a file location even if the file is on your hard drive.
mydata <- read.csv("http://www.ats.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 2
This data set has a binary response (outcome, dependent) variable called
There are three predictor variables:
rank. We will treat the
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.
## admit gre gpa rank ## Min. :0.000 Min. :220 Min. :2.26 1: 61 ## 1st Qu.:0.000 1st Qu.:520 1st Qu.:3.13 2:151 ## Median :0.000 Median :580 Median :3.40 3:121 ## Mean :0.318 Mean :588 Mean :3.39 4: 67 ## 3rd Qu.:1.000 3rd Qu.:660 3rd Qu.:3.67 ## Max. :1.000 Max. :800 Max. :4.00
xtabs(~ rank + admit, data = mydata)
## admit ## rank 0 1 ## 1 28 33 ## 2 97 54 ## 3 93 28 ## 4 55 12
Below is a list of some analysis methods you may have encountered. Some of the methods listed are quite reasonable while others have either fallen out of favor or have limitations.
The code below estimates a probit regression model using the glm (generalized linear model) function.
Since we stored our model output in the object "myprobit", R will not print anything to the console. We
can use the
summary function to get a summary of the
model and all the estimates.
myprobit <- glm(admit ~ gre + gpa + rank, family=binomial(link="probit"), data=mydata) ## model summary summary(myprobit)
## ## Call: ## glm(formula = admit ~ gre + gpa + rank, family = binomial(link = "probit"), ## data = mydata) ## ## Deviance Residuals: ## Min 1Q Median 3Q Max ## -1.616 -0.871 -0.639 1.156 2.103 ## ## Coefficients: ## Estimate Std. Error z value Pr(>|z|) ## (Intercept) -2.38684 0.67395 -3.54 0.00040 *** ## gre 0.00138 0.00065 2.12 0.03433 * ## gpa 0.47773 0.19720 2.42 0.01541 * ## rank2 -0.41540 0.19498 -2.13 0.03313 * ## rank3 -0.81214 0.20836 -3.90 9.7e-05 *** ## rank4 -0.93590 0.24527 -3.82 0.00014 *** ## --- ## 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: 470.4 ## ## Number of Fisher Scoring iterations: 4
gpa, and the three terms for
rankare statistically significant. The probit regression coefficients give the change in the z-score or probit index for a one unit change in the predictor.
gre, the z-score increases by 0.001.
gpa, the z-score increases by 0.478.
rankhave a slightly different interpretation. For example, having attended an undergraduate institution of
rankof 2, versus an institution with a
rankof 1 (the reference group), decreases the z-score by 0.415.
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.
## 2.5 % 97.5 % ## (Intercept) -3.72011 -1.07633 ## gre 0.00011 0.00266 ## gpa 0.09607 0.86261 ## rank2 -0.79921 -0.03300 ## rank3 -1.22310 -0.40501 ## rank4 -1.42342 -0.45954
We can test for an overall effect of
rank using the
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
We use the
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
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-05
The 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
The first line of code below creates a row matrix
l that defines the test we
want to perform. In this case, we want to test the difference (subtraction) of
the terms for
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 row matrix
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) = 0.018
The 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
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 2.5 1 ## 2 206 2.5 1 ## 3 212 2.5 1 ## 4 218 2.5 1 ## 5 224 2.5 1 ## 6 230 2.5 1
Now 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
## change in deviance with(myprobit, null.deviance - deviance)
##  41.6
## change in degrees of freedom with(myprobit, df.null - df.residual)
##  5
## chi square test p-value with(myprobit, pchisq(null.deviance - deviance, df.null - df.residual, lower.tail = FALSE))
##  7.22e-08
The 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. This is sometimes called a likelihood ratio test (the deviance residual is -2*log likelihood). To see the model's log likelihood, we type:
## 'log Lik.' -229 (df=6)
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