Solved – Model to predict categorical outcome from continuous and categorical variables

categorical datacontinuous datageneralized linear modelpredictive-modelsself-study

I have to fit a model to test whether Learning (1=learned, 0=failed) depends on lizard sex (M or F), Lizard SVL (snout-vent length), or an interaction of the two.

I am new to both R and this website. Please explain each step fully.

This is random data given to us as part of a zoology/statistics assignment. It is related to lectures 'Generalized Linear Models' and 'Model Selection and Model Averaging'.

Snout-vent length is used as a a measure of size (in millimeters).

Specifically, what R code is required to fit a GLM when a categorical and continuous predictor variable are used to predict a categorical dependent variable? The following is from a previous example with two categorical predictor variables that we have worked through if that helps with what I am trying to ask:

plot(ProportionSurvived ~ Treatment, data=dat)
interaction.plot(dat$Sculpin, dat$Lake, response = dat$NSurvivors/dat$NSticklebackAdded)

glm(cbind(NSurvivors, NSticklebackAdded - NSurvivors) ~ Lake * Sculpin, family = binomial(link = "logit"), data=dat)

Treatment is one of four possible conditions (2 levels of 2 predictors; lake and sculpin present or absent). This example was testing the proportion of sticklebacks that survived (proportionsurvived) in a pond.

Best Answer

Looks to me like you should use binary logistic regression. Your dependent variable is learning (yes/no) and all other variables are independent variables. Binary logistic regression is extremely flexible with regard to scales of measurement used with the predictors.

Update

In light of the OP posting data, we can show the equivalence. In the below, model is as per the OP's question, whilst model2 is the same model fitted via lm() instead of glm()

> anova(model, test = "F")
Analysis of Deviance Table

Model: gaussian, link: identity

Response: res

Terms added sequentially (first to last)


     Df Deviance Resid. Df Resid. Dev      F Pr(>F)
NULL                    56     1277.6              
pred  7   227.23        49     1050.3 1.5144 0.1846
> anova(model2)
Analysis of Variance Table

Response: res
          Df  Sum Sq Mean Sq F value Pr(>F)
pred       7  227.23  32.462  1.5144 0.1846
Residuals 49 1050.35  21.436

Notice the Deviance of the model is the same as the sums of squares in model2 and the rest of the important numbers, F and its p-value are the same. Likewise, the estimated values of rht model coefficients are the same

> coef(model)
  (Intercept)         predB         predC         predD         predE 
 2.300000e+01  2.625000e+00  4.714286e+00  3.500000e+00  6.333333e+00 
        predF         predG         predH 
-8.283393e-15  1.000000e+00  3.000000e+00 
> coef(model2)
  (Intercept)         predB         predC         predD         predE 
 2.300000e+01  2.625000e+00  4.714286e+00  3.500000e+00  6.333333e+00 
        predF         predG         predH 
-8.283393e-15  1.000000e+00  3.000000e+00 
> all.equal(coef(model), coef(model2))
[1] TRUE

There appears to be some small differences between between groups A and C and A and E but overall pred does not explain a significant amount of variance in the response.

Best Answer

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Update

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