SPSS Example: Data Explanations (Data set: class.sav)

# Import the data set
library(haven)
class <- read_sav("Desktop/class.sav")
head(class)
## # A tibble: 6 x 5
##           admit        admit1   gre   gpa  rank
##       <dbl+lbl>     <dbl+lbl> <dbl> <dbl> <dbl>
## 1 0 [not admit] 1 [not admit]   380  3.61     3
## 2 1 [admit]     2 [admit]       660  3.67     3
## 3 1 [admit]     2 [admit]       800  4        1
## 4 1 [admit]     2 [admit]       640  3.19     4
## 5 0 [not admit] 1 [not admit]   520  2.93     4
## 6 1 [admit]     2 [admit]       760  3        2
# Let's do a simple descriptive analysis first
summary(class)
##      admit            admit1           gre             gpa       
##  Min.   :0.0000   Min.   :1.000   Min.   :220.0   Min.   :2.260  
##  1st Qu.:0.0000   1st Qu.:1.000   1st Qu.:520.0   1st Qu.:3.130  
##  Median :0.0000   Median :1.000   Median :580.0   Median :3.395  
##  Mean   :0.3175   Mean   :1.317   Mean   :587.7   Mean   :3.390  
##  3rd Qu.:1.0000   3rd Qu.:2.000   3rd Qu.:660.0   3rd Qu.:3.670  
##  Max.   :1.0000   Max.   :2.000   Max.   :800.0   Max.   :4.000  
##       rank      
##  Min.   :1.000  
##  1st Qu.:2.000  
##  Median :2.000  
##  Mean   :2.485  
##  3rd Qu.:3.000  
##  Max.   :4.000
sapply(class,sd)
##       admit      admit1         gre         gpa        rank 
##   0.4660867   0.4660867 115.5165364   0.3805668   0.9444602
table(class$admit)
## 
##   0   1 
## 273 127
table(class$rank)
## 
##   1   2   3   4 
##  61 151 121  67
# To see the crosstable, we need CrossTable function from gmodels package
library(gmodels)
# Build a crosstable between admit and rank
CrossTable(class$admit, class$rank)
## 
##  
##    Cell Contents
## |-------------------------|
## |                       N |
## | Chi-square contribution |
## |           N / Row Total |
## |           N / Col Total |
## |         N / Table Total |
## |-------------------------|
## 
##  
## Total Observations in Table:  400 
## 
##  
##              | class$rank 
##  class$admit |         1 |         2 |         3 |         4 | Row Total | 
## -------------|-----------|-----------|-----------|-----------|-----------|
##            0 |        28 |        97 |        93 |        55 |       273 | 
##              |     4.464 |     0.356 |     1.314 |     1.880 |           | 
##              |     0.103 |     0.355 |     0.341 |     0.201 |     0.682 | 
##              |     0.459 |     0.642 |     0.769 |     0.821 |           | 
##              |     0.070 |     0.242 |     0.233 |     0.138 |           | 
## -------------|-----------|-----------|-----------|-----------|-----------|
##            1 |        33 |        54 |        28 |        12 |       127 | 
##              |     9.596 |     0.765 |     2.825 |     4.042 |           | 
##              |     0.260 |     0.425 |     0.220 |     0.094 |     0.318 | 
##              |     0.541 |     0.358 |     0.231 |     0.179 |           | 
##              |     0.083 |     0.135 |     0.070 |     0.030 |           | 
## -------------|-----------|-----------|-----------|-----------|-----------|
## Column Total |        61 |       151 |       121 |        67 |       400 | 
##              |     0.152 |     0.378 |     0.302 |     0.168 |           | 
## -------------|-----------|-----------|-----------|-----------|-----------|
## 
## 
# Before we run a binary logistic regression, we need check the previous two-way contingency table of categorical outcome and predictors. We want to make sure there is no zero in any cells.
# In our case, no zero cells could be found. Another simple way to do that is:
xtabs(~admit+rank,data=class)
##      rank
## admit  1  2  3  4
##     0 28 97 93 55
##     1 33 54 28 12
hist(class$rank)

This dataset 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. We can get basic descriptives for the entire data set by using summary. To get the standard deviations, we use sapply to apply the sd function to each variable in the dataset.

# Using the logit model: The code below estimates a logistic regression model using the glm (generalized linear model) function. First, we convert rank to a factor to indicate that rank should be treated as a categorical variable.
class$rank <- factor(class$rank)
mylogit <- glm(admit ~ gre + gpa + rank, data = class, family = "binomial")
# Check the result of our model
summary(mylogit)
## 
## Call:
## glm(formula = admit ~ gre + gpa + rank, family = "binomial", 
##     data = class)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.6268  -0.8662  -0.6388   1.1490   2.0790  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -3.989979   1.139951  -3.500 0.000465 ***
## gre          0.002264   0.001094   2.070 0.038465 *  
## gpa          0.804038   0.331819   2.423 0.015388 *  
## rank2       -0.675443   0.316490  -2.134 0.032829 *  
## rank3       -1.340204   0.345306  -3.881 0.000104 ***
## rank4       -1.551464   0.417832  -3.713 0.000205 ***
## ---
## 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: 458.52  on 394  degrees of freedom
## AIC: 470.52
## 
## Number of Fisher Scoring iterations: 4

How to do the interpretation?

Why the coefficient value of rank (B) are different with the SPSS outputs? - In R, the glm automatically made the Rank 1 as the references group. However, in our SPSS example, we set the rank 4 as the reference group.

We 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.

# Load the package aod to use the wald.test function
library(aod)
# Run a wald.test
wald.test(b=coef(mylogit),Sigma=vcov(mylogit), Terms=4:6)
## Wald test:
## ----------
## 
## Chi-squared test:
## X2 = 20.9, df = 3, P(> X2) = 0.00011
# The chi-squared test statistic of 20.9, with three degrees of freedom is associated with a p-value of 0.00011 indicating that the overall effect of rank is statistically significant.

We can also test additional hypotheses about the differences in the coefficients for the 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).

# Prepare the comparison code
l <- cbind(0, 0, 0, 1, -1, 0)
# Run wald test to see whether Rank 2 is equal to Rank 3.
wald.test(b = coef(mylogit), Sigma = vcov(mylogit), L = l)
## Wald test:
## ----------
## 
## Chi-squared test:
## X2 = 5.5, df = 1, P(> X2) = 0.019
# 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 rank=2 and the coefficient for rank=3 is statistically significant.

You can also exponentiate the coefficients and interpret them as odds-ratios. R will do this computation for you. To get the exponentiated coefficients, you tell R that you want to exponentiate (exp), and that the object you want to exponentiate is called coefficients and it is part of mylogit (coef(mylogit)). We can use the same logic to get odds ratios and their confidence intervals, by exponentiating the confidence intervals from before. To put it all in one table, we use cbind to bind the coefficients and confidence intervals column-wise.

## odds ratios only
exp(coef(mylogit))
## (Intercept)         gre         gpa       rank2       rank3       rank4 
##   0.0185001   1.0022670   2.2345451   0.5089309   0.2617923   0.2119375
## odds ratios and 95% CI
exp(cbind(OR = coef(mylogit), confint(mylogit)))
## Waiting for profiling to be done...
##                    OR       2.5 %    97.5 %
## (Intercept) 0.0185001 0.001889165 0.1665354
## gre         1.0022670 1.000137602 1.0044457
## gpa         2.2345451 1.173858341 4.3238354
## rank2       0.5089309 0.272289673 0.9448343
## rank3       0.2617923 0.131641715 0.5115181
## rank4       0.2119375 0.090715546 0.4706961
# We will start by calculating the predicted probability of admission at each value of rank, holding gre and gpa at their means. First we create and view the data frame.
newdata1 <- with(class, data.frame(gre = mean(gre), gpa = mean(gpa), rank = factor(1:4)))

## view data frame
newdata1
##     gre    gpa rank
## 1 587.7 3.3899    1
## 2 587.7 3.3899    2
## 3 587.7 3.3899    3
## 4 587.7 3.3899    4

These objects must have the same names as the variables in your logistic regression above (e.g. in this example the mean for gre must be named gre). Now that we have the data frame we want to use to calculate the predicted probabilities, we can tell R to create the predicted probabilities. The first line of code below is quite compact, we will break it apart to discuss what various components do. The newdata1$rankP tells R that we want to create a new variable in the dataset (data frame) newdata1 called rankP, the rest of the command tells R that the values of rankP should be predictions made using the predict( ) function. The options within the parentheses tell R that the predictions should be based on the analysis mylogit with values of the predictor variables coming from newdata1 and that the type of prediction is a predicted probability (type=“response”). The second line of the code lists the values in the data frame newdata1. Although not particularly pretty, this is a table of predicted probabilities.

# Add a predicted Rank column
newdata1$rankP <- predict(mylogit, newdata = newdata1, type = "response")
newdata1
##     gre    gpa rank     rankP
## 1 587.7 3.3899    1 0.5166016
## 2 587.7 3.3899    2 0.3522846
## 3 587.7 3.3899    3 0.2186120
## 4 587.7 3.3899    4 0.1846684
# In the above output we see that the predicted probability of being accepted into a graduate program is 0.52 for students from the highest prestige undergraduate institutions (rank=1), and 0.18 for students from the lowest ranked institutions (rank=4), holding gre and gpa at their means. 
# Now, we are going to do something that do not exist in our SPSS section
# We can do something very similar to create a table of predicted probabilities varying the value of gre and rank. 
# We are going to plot these, so we will create 100 values of gre between 200 and 800, at each value of rank (i.e., 1, 2, 3, and 4).
newdata2 <- with(class, data.frame(gre = rep(seq(from = 200, to = 800, length.out = 100),
    4), gpa = mean(gpa), rank = factor(rep(1:4, each = 100))))
head(newdata2,10)
##         gre    gpa rank
## 1  200.0000 3.3899    1
## 2  206.0606 3.3899    1
## 3  212.1212 3.3899    1
## 4  218.1818 3.3899    1
## 5  224.2424 3.3899    1
## 6  230.3030 3.3899    1
## 7  236.3636 3.3899    1
## 8  242.4242 3.3899    1
## 9  248.4848 3.3899    1
## 10 254.5455 3.3899    1

The code to generate the predicted probabilities (the first line below) is the same as before, except we are also going to ask for standard errors so we can plot a confidence interval. We get the estimates on the link scale and back transform both the predicted values and confidence limits into probabilities.

newdata3 <- cbind(newdata2, predict(mylogit, newdata = newdata2, type = "link",
    se = TRUE))
newdata3 <- within(newdata3, {
    PredictedProb <- plogis(fit)
    LL <- plogis(fit - (1.96 * se.fit))
    UL <- plogis(fit + (1.96 * se.fit))
})

## view first few rows of final dataset
head(newdata3)
##        gre    gpa rank        fit    se.fit residual.scale        UL
## 1 200.0000 3.3899    1 -0.8114870 0.5147714              1 0.5492064
## 2 206.0606 3.3899    1 -0.7977632 0.5090986              1 0.5498513
## 3 212.1212 3.3899    1 -0.7840394 0.5034491              1 0.5505074
## 4 218.1818 3.3899    1 -0.7703156 0.4978239              1 0.5511750
## 5 224.2424 3.3899    1 -0.7565918 0.4922237              1 0.5518545
## 6 230.3030 3.3899    1 -0.7428680 0.4866494              1 0.5525464
##          LL PredictedProb
## 1 0.1393812     0.3075737
## 2 0.1423880     0.3105042
## 3 0.1454429     0.3134500
## 4 0.1485460     0.3164108
## 5 0.1516973     0.3193867
## 6 0.1548967     0.3223773

It can also be helpful to use graphs of predicted probabilities to understand and/or present the model. We will use the ggplot2 package for graphing.

# Load the ggplot2 for the plot
library(ggplot2)
# we make a plot with the predicted probabilities, and 95% confidence intervals.
ggplot(newdata3, aes(x = gre, y = PredictedProb)) + geom_ribbon(aes(ymin = LL,
    ymax = UL, fill = rank), alpha = 0.2) + geom_line(aes(colour = rank),
    size = 1)

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 use the command:

# The test statistic is the difference between the residual deviance for the model with predictors and the null model.
with(mylogit, null.deviance - deviance)
## [1] 41.45903
# The degrees of freedom for the difference between the two models is equal to the number of predictor variables in the mode, and can be obtained using:
with(mylogit, df.null - df.residual)
## [1] 5
# Finally, the p-value can be obtained using:
with(mylogit, pchisq(null.deviance - deviance, df.null - df.residual, lower.tail = FALSE))
## [1] 7.578189e-08

Things to consider:

Note: This Course’s guide could be found on UCLA’s website: https://stats.idre.ucla.edu/r/dae/logit-regression/