Binomial Data
Many statistical problems involve binary response variables. For example, we could classify individuals as alive/dead, healthy/unwell, employ/unemployed, left/right, right/wrong, … etc. A regression of binary data is possible if at least one of the predictors is continuous (otherwise you would use some kind of categorical test, such as a Chi-squared test).
The response variable contains only 0s and 1s (e.g., dead = 0, alive = 1) in a single vector. R treats such binary data is if each row came from a binomial trial with sample size 1. If the probability that an individual is dead is p, then the probability of obtaining y (where y is either dead or alive, 0 or 1) is given by an abbreviated form of the binomial distribution with n = 1, known as the Bernoulli distribution.
Binomial data can either be modelled at the individual (binary response) or group (proportion) level. If you have unique values of one or more explanatory variables for each and every individual case, then a model with a binary response variable will work. If not, you should aggregate the data until you do have such explanatory variables (e.g., at the level of the household, town, state, population, …).
A. Individual-level: binary response (0/1) = logistic regression
At the individual level we can model each individual in terms of dead or alive (male/female, tall/short, green/red, …), using logistic regression (or logit model). In the logit model, the log odds (logarithm of the odds) of the outcome is modeled as a linear combination of the predictor variables. If p is a probability, then p/(1 − p) is the corresponding odds; the logit of the probability is the logarithm of the odds.
To model a binary response variable, we first need (or need to create) a single vector of 0 and 1s. We then model this vector with glm() and family = 'binomial'. We can compare models using a test for a change in deviance with chi-squared. Overdispersion is not an issue with a binary response variable.
Let’s look at the incidence (presence/absence) of a breeding bird on various islands. Examine the top 6 rows of the isolation dataset.
head(isolation)## incidence area distance
## 1 1 7.928 3.317
## 2 0 1.925 7.554
## 3 1 2.045 5.883
## 4 0 4.781 5.932
## 5 0 1.536 5.308
## 6 1 7.369 4.934The data show the $incidence of the bird (present = 1, absent = 0) on islands of different sizes ($area in km2) and distance ($distance in km) from the mainland.
We can easily imagine how we might be more likely to find a specific bird on larger islands that are closer to the mainland, and less likely to find that bird on smaller islands further away from the mainland.
Now let’s plot the data. First, plot $incidence as a function of $distance.
plot(incidence ~ distance, data = isolation)
Because we only have two options for the y-axis (response variable), 0 or 1, we have two lines of data points at different distances (x-axis). It looks like there are more 1’s than 0’s at lower values of distance and fewer 1’s than 0’s at higher values of distance.
Now make a similar plot for $area.
plot(incidence ~ area, data = isolation)
You can see that area has the opposite effect: as area increases, we see fewer 0’s and more 1’s. Let’s test these relationships using a generalized linear model.
We can begin examining these patterns in more detail by fitting a model of incidence as a function of area and distance. We will first consider the additive model (i.e., area and distance are independent). Use glm() to run this model, and call it m0. Remember to use a formula, set the data = argument, and the family = binomial argument.
m0 <- glm(incidence ~ area + distance, family = binomial, data = isolation)What are we actually stating in this additive model? We are saying that area and distance have the same effect on incidence, but that the intercept may be different between them. If we plotted both on the same graph, they would have the same slope, but maybe different intercepts.
Ok, what if we think that there is an interaction between area and distance, i.e., the slope of each variable varies as well as the intercept? Run the model of an interaction between area and distance, called m2.
m1 <- glm(incidence ~ area * distance, family = binomial, data = isolation)We can compare these two models using ANOVA. The function anova() computes analysis of variance (or deviance) tables for one or more fitted model objects. Run this function on m0 and m1. In the case of binary data, we need to do a Chi-squared test (test = 'Chi').
anova(m0, m1, test = "Chi")## Analysis of Deviance Table
##
## Model 1: incidence ~ area + distance
## Model 2: incidence ~ area * distance
## Resid. Df Resid. Dev Df Deviance Pr(>Chi)
## 1 47 28.402
## 2 46 28.252 1 0.15043 0.6981The simpler model (m0) is not significantly worse (p = 0.6981), so we can go with that one.
Fitting and interpreting the model
Let’s look at the summary of model m0.
summary(m0)##
## Call:
## glm(formula = incidence ~ area + distance, family = binomial,
## data = isolation)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.8189 -0.3089 0.0490 0.3635 2.1192
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 6.6417 2.9218 2.273 0.02302 *
## area 0.5807 0.2478 2.344 0.01909 *
## distance -1.3719 0.4769 -2.877 0.00401 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 68.029 on 49 degrees of freedom
## Residual deviance: 28.402 on 47 degrees of freedom
## AIC: 34.402
##
## Number of Fisher Scoring iterations: 6As normal, we have the formula, the deviance residuals (a measure of model fit), then the coefficient estimates, their standard errors, the z-statistic (sometimes called a Wald z-statistic), and p-values. What exactly do the coefficients represent?
The coefficients as presented in the model summary are the logits. We can then transform these logits to odds ratios if we want.
Logit/log odds
The logistic regression coefficients give the change in the log odds (i.e., the logarithm of the odds, called the logit).
For continuous predictors, the coefficients give the change in the log odds of the outcome for a one unit increase in the predictor variable. For example, for every 1km^2 increase in area, the log odds of the bird being present (versus absent) increases by 0.58 and the log odds of the bird being present decreases by 1.37 (i.e., -1.37) for every km of distance from the mainland.
For categorical predictors, the coefficient estimate describes the change in the log odds for each level compared to the base level.
As usual, confidence intervals around the coefficient estimates can be calculated with the function confint(). Calculate the CIs for m0.
confint(m0)## Waiting for profiling to be done...## 2.5 % 97.5 %
## (Intercept) 1.8390115 13.735103
## area 0.1636582 1.176611
## distance -2.5953657 -0.639140Logits (log odds) are just one way of describing the model. We can easily convert the logits either to odds ratios or predicted probabilities.
Odds ratios
Odds ratios are a relative measure of effect, which allows the comparison of two groups. It is a popular way of presenting results in medicine, especially of controlled trials. Here, the odds of one group (usually the control) is divided by the odds of another (the treatment) to give a ratio. In this case, if the odds of both groups are the same, the ratio will be 1, suggesting there is no difference. If the OR is > 1 the control is better than the other level. If the OR is < 1 the treatment is better than the control.
However, in R, we are presented with differences or the change in log odds between groups (for categorical variables) or the change in log odds between x and x + 1 (for continuous variables). The directionality is the opposite to the example above because R effectively compares the ‘treatment’ to the ‘control’.
Thus, for categorical predictors, an odds ratio of > 1 suggests a positive difference between levels and the base level (the exponential of a positive number is greater than 1), i.e., the base level has a lower odds and the level has a higher odds. An odds ratio of < 1 suggests a negative difference (the exponential of a negative number lies between 0 and 1).
For continuous predictors, the odds ratio is the odds of x + 1 over x (i.e., the odds ratio for a unit increase in the continuous x predictor variable), and the interpretation is the same as for categorical.
To obtain these odds ratios in R, we need to exponentiate the coefficient estimates. Remember that these estimates are the log of the odds, so the exponent is the actual odds. The function exp() can do this for us.
Remember that we can extract the coefficient estimates from any model with the function coef(). Put exp() and coef() together to obtain the odds ratios from m0.
exp(coef(m0))## (Intercept) area distance
## 766.3669575 1.7873322 0.2536142Since both our predictors are continuous, they are the odds ratios for x+1 to x. The odds ratio for area is 1.79, indicating that for a 1km^2 increase in area we expect a 79% increase in the odds of finding the bird on the island. For a 1km increase in distance of the island from the mainland, we expect a 25% decrease in the odds of finding the bird.
A more detailed discussion of these different representations can be found here: https://stats.idre.ucla.edu/other/mult-pkg/faq/general/faq-how-do-i-interpret-odds-ratios-in-logistic-regression/
Predicted probabilities
We can generate predicted probabilities for both continuous and categorical variables. We can generate predicted probabilities for each row in the original dataset, just by calling the predict() function.
Setting type = 'link' returns the logits; setting type = 'response' returns the predicted probabilities.
Add an extra column to the dataset for the predicted logits, called ‘logits’. Remember that to add a column we just need to call the data object, and add the new column name with a dollar sign.
isolation$logit <- predict(m0, type = 'link')Add another column for the predicted probabilities, called ‘p’.
isolation$p <- predict(m0, type = 'response')However, to better plot these relationships, we should input a specific range of x’s that we predict y’s for.
Plotting binary response data
To plot the fitted line of a generalized linear model, we generally have to use the predict() function, because the model, although linear, is not a straight line that can be plotted with either an intercept or slope, as we used for lm() and abline().
The function predict() is a general wrapper for more specific functions, in this case predict.glm(). We have to specify the object (i.e., the fitted model), the newdata (i.e., the values of x for which we want predictions, named identically to the predictors in the object, as a dataframe or list), and the type of prediction required (either on the scale of the linear predictors (‘link’) or on the scale of the response variable (‘response’).
There are, as usual, at least two ways we could do this. First, we will make two separate individual logistic regressions and make predictions from those.
Make a model for incidence as a function of area, called ma.
ma <- glm(incidence ~ area, family = binomial, data = isolation)We then need to make a sequence of x values (e.g., $area) from which to predict the y (predicted probability of $incidence). To do this, first generate a sequence of x values from 0 to 9, in steps of 0.1. Call this vector xv.
xv <- seq(from = 0, to = 9, by = 0.1)Now generate the vector of predicted y values of the log odds (call it yv). Remember to include the model object (ma), the new data (yv, as a dataframe), and the type (response).
yv <- predict(object = ma, newdata = data.frame(area = xv), type = "response")Now, we have two vectors, xv and yv, that we can use to plot the fitted (predicted) line.
Add this fitted line to the plot of incidence on area, using the lines() or points() function.
plot(incidence ~ area, data = isolation)
lines(xv, yv)
Now let’s repeat this process for distance. First, make the model using glm() and call it md.
md <- glm(incidence ~ distance, family = binomial, data = isolation)Then make the sequence of x values for distance. Because in this case, $area and $distance have identical ranges, we can use the same code as before. Generate a sequence of x values from 0 to 9, in steps of 0.1, called xv.
xv <- seq(from = 0, to = 9, by = 0.1)Now generate the vector of predicted y values of the log odds (call it yv). Remember to include the model object (md), the new data (yv, as a dataframe, remember to use ‘distance’ this time, not ‘area’), and the type (response).
yv <- predict(object = md, newdata = data.frame(distance = xv), type = "response")Now we have the data, let’s plot! Make the plot of incidence on distance again …
plot(incidence ~ distance, data = isolation)
And, add the line. This time make the line blue, dashed, and four times as wide.
plot(incidence ~ distance, data = isolation)
lines(xv, yv, col = 'blue', lty = 2, lwd = 4)
So far we have been predicting and plotting lines based solely on the values of a single predictor, using the models ma or md. What if we wanted to take area into account when we display the model for distance? We can go back to our additive model, m0.
To generate predicted values for distance accounting for area, we need to decide what values of area to use. A sensible option might be the mean of area (i.e., the mean size of island). We could then plot the fitted line showing how the effect of distance varies for an average-sized island. Calculate the mean size of island in the dataset.
mean(isolation$area)## [1] 4.31916Now repeat this value to generate a vector the same length as our xv vector (the x-values of distance from which we predict incidence).
rep(mean(isolation$area), times = length(xv))## [1] 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916
## [9] 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916
## [17] 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916
## [25] 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916
## [33] 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916
## [41] 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916
## [49] 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916
## [57] 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916
## [65] 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916
## [73] 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916
## [81] 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916 4.31916
## [89] 4.31916 4.31916 4.31916Ok, so now we have our values for distance (xv), and the vector of mean values of area (unnamed). To put both of these into the predict() function, we need to stick them in a dataframe. Make a data frame of our new values for area and distance. Call this dataframe xv2
xv2 <- data.frame(area = rep(mean(isolation$area), times = length(xv)), distance = xv)Now, put the model (m0), the new data (xv2), and the type of predicted values we want (‘response’), in a call to predict(), and add these value to the datafram xv2, as a new column calledy`.
xv2$y <- predict(m0, newdata = xv2, type = 'response')Add this new line to the graph.
plot(incidence ~ distance, data = isolation)
lines(xv, yv, col = 'blue', lty = 2, lwd = 4)
lines(xv2$distance, xv2$y)
B. Group-level: Proportion data
At the group level, we look at the proportion of successes and failures (or female/male, short/tall, rep/dem, …) in groups or populations or sites, etc. In all these cases, we know both numbers—those of successes and those of failures, in contrast to Poisson regression (covered later) where we just have counts of successes.
In the past, the way proportion data was modelled was to use a single percentage as the response variable. However, this causes problems because: (i) The errors are not normally distributed, (ii) the variance is not constant, (iii) the response is bounded (by 1 above and by 0 below), and (iv) we lose information on the size of the sample, n, from which the proportion was estimated.
In the GLM framework, we can model proportion data directly. R carries out weighted regression, using the individual sample sizes as weights, and the logit link function to ensure linearity.
Overdispersion
Overdispersion occurs in regression of proportion data when the residual deviance is larger than the residual degrees of freedom. There are two possibilities: either the model is misspecified, or the probability of success, p, is not constant within a given treatment level.
To deal with overdispersion, we can either respecify the model, or set family = quasibinomial rather than family = binomial.
Fitting the model
As above, we use glm() with family = 'binomial'. In contrast to the binomial response, in the case of proportion data our response data is a matrix of two columns, one column of successes and one of failures. This matrix must be a single object (i.e., you could use cbind() to stick two columns together), which in most cases we will call y.
In this case, we have the number of males and females in a range of populations. Does sex ratio vary with population size? Look at the head of the sexratio data.
head(sexratio)## total females males
## 1 1 1 0
## 2 4 3 1
## 3 10 7 3
## 4 22 18 4
## 5 55 22 33
## 6 121 41 80We have columns for the total number of organisms, followed by the number of males and females. We might be interested in how the sex ratio varies as a function of population size.
First calculate the actual proportion of males in each population. Make a new column for this vector, called $p
sexratio$p <- sexratio$males/(sexratio$males + sexratio$females)To model proportion data, R requires a unique form of the response variable—a matrix of two columns of successes and failures, in this case males and females. Use cbind() to create such a variable, y.
y <- cbind(sexratio$males, sexratio$females)Now that we have the y variable, we can construct the model. We use glm() as before, setting y as the response, total as the predictor and family = binomial. Call it m2. A final point to consider is that to generate the predicted values, we need the column names in the original data and new data to match; yet another reason to use the formula approach with a data = argument.
m2 <- glm(y ~ total, data = sexratio, family = binomial)Look at the summary.
summary(m2)##
## Call:
## glm(formula = y ~ total, family = binomial, data = sexratio)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -3.4619 -1.2760 -0.9911 0.5742 1.8795
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.0807368 0.1550376 0.521 0.603
## total 0.0035101 0.0005116 6.862 6.81e-12 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 71.159 on 7 degrees of freedom
## Residual deviance: 22.091 on 6 degrees of freedom
## AIC: 54.618
##
## Number of Fisher Scoring iterations: 4The output is familar. For this continuous predictor, we get an intercept and slope, and the p-values indicate whether they are significantly different from 0. For categorical variables, we get, as usual, the base level mean, then differences between means. Furthermore, the coefficient estimates are in logits (ln(p/(1 - p)). To calculate a proportion (p) from a logit (x), you need to calculate 1/(1 + 1/(exp(x))).
However, we have substantial overdispersion (residual deviance = 22.091 is much greater than residual d.f. = 6). One reason for overdispersion is if the model is misspecified. Let’s try log(total) and see what happens. Try running a new model (called m3) with that.
m3 <- glm(y ~ log(total), data = sexratio, family = binomial)There is no evidence of overdispersion (residual deviance = 5.67 on 6 d.f.) in this new model. We can therefore conclude that the proportion of males increases significantly with increasing density, and that the logistic model is linearized by logarithmic transformation of the explanatory variable (population density).
Plotting proportion data
We can plot the proportion of males ($p) as a function of the log(total) as normal, using plot().
plot(p ~ log(total), data = sexratio)
Now we need to generate the predicted values in order to plot the fitted line. First generate the x values (xv). Remember that we are working with log(total)—look at the x-axis of the plot— so we need values from 0 to 6, in steps of 0.1.
xv <- seq(from = 0, to = 6, by = 0.1)Now for the y values. Here we have to take the exponent of xv, because our model (m3) is of log(total), not log(log(total)). Call this new vector yv as before.
yv <- predict(m3, data.frame(total = exp(xv)), type = "response")Now add the fitted line to the plot, in red and twice the width.
plot(p ~ log(total), data = sexratio)
lines(xv, yv, col = 'red', lwd = 2)
Great! You should be able to analyse the two main types of binomial data now, and produce graphs of the data and fitted lines.
However, there are some kinds of proportion data that are better analysed differently:
Percentage cover data is best analysed using conventional linear models (assuming normal errors and constant variance) following arcsine transformation.
Percentage change in some continuous measurement (e.g., weight, tree diameter) is best modelled either as an ANCOVA of the final measurement with the initial measurement as a covariate, or by specifying the response as a relative change: log(final/initial), both of which should be linear models with normal errors.
2017-11-26