Which statistical model is appropriate when the response is continuous and the predictors are a mix of continuous and categorical? What is the disadvantage in using GLM combined with gaussian family?

Here is my dataset and model in R:

df <- structure(list(as.factor.pred. = structure(c(1L, 1L, 5L, 3L, 
2L, 8L, 3L, 5L, 2L, 2L, 3L, 2L, 4L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 
1L, 2L, 4L, 1L, 1L, 1L, 1L, 3L, 3L, 2L, 2L, 2L, 1L, 7L, 3L, 3L, 
6L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 5L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 
1L, 1L, 1L, 1L, 2L), .Label = c("A", "B", "C", "D", "E", "F", 
"G", "H"), class = "factor"), res = c(33, 33, 37, 32, 32, 26, 
33, 28, 25, 34, 29, 35, 26, 20, 27, 19, 30, 33, 27, 24, 26, 28, 
27, 23, 26, 25, 24, 26, 24, 25, 21, 21, 23, 24, 23, 27, 23, 20, 
21, 22, 22, 22, 22, 23, 23, 21, 22, 21, 21, 23, 23, 18, 20, 18, 
18, 18, 19)), .Names = c("as.factor.pred.", "res"), row.names = c(NA, 
-57L), class = "data.frame")

names(df)[1] <- "pred" ## fix up the names to match formula

model <- glm(res ~ pred, data = df)


glm(formula = res ~ pred, data = df)

Deviance Residuals: 
   Min      1Q  Median      3Q     Max  
-6.625  -3.000  -0.625   2.000  10.000  

              Estimate Std. Error t value Pr(>|t|)    
(Intercept)  2.300e+01  9.080e-01  25.331   <2e-16 ***
predB        2.625e+00  1.471e+00   1.784   0.0806 .  
predC        4.714e+00  1.971e+00   2.391   0.0207 *  
predD        3.500e+00  3.397e+00   1.030   0.3080    
predE        6.333e+00  2.823e+00   2.243   0.0294 *  
predF       -8.283e-15  4.718e+00   0.000   1.0000    
predG        1.000e+00  4.718e+00   0.212   0.8330    
predH        3.000e+00  4.718e+00   0.636   0.5278    
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for gaussian family taken to be 21.43562)

    Null deviance: 1277.6  on 56  degrees of freedom
Residual deviance: 1050.3  on 49  degrees of freedom
AIC: 345.85  
  • $\begingroup$ The choice might also depend on the distribution/nature of the variables or on the shape of the relationship but it sounds like multiple regression would be the usual starting point. $\endgroup$
    – Gala
    Jul 8, 2013 at 15:40
  • $\begingroup$ What have to looked into on the matter? Why would you think there's any disadvantage? What would the advantage or disadvantage be against? Why are you thinking you want to go that way? $\endgroup$
    – John
    Jul 8, 2013 at 15:46

1 Answer 1


If the model you wish to fit is linear in its parameters and the errors are Gaussian with constant variance then a linear model would be a reasonable start, via the lm() function for example in R.

As the linear model is a special case of the GLM there is no real difference, but minor differences may show up in the implementation of the two models due to differences in their algorithms. Fitting that same model in R via glm() should give the same fit (coefficients) up to machine precision or some small differences in the last few decimal places. However, fitting via glm(...., family = gaussian) would be exceedingly inefficient compared to fitting via lm().

Note the lm() function in R fits the so-called General Linear Model, the fusion of "regression" and ANOVA. Hence it is fully capable of dealing with continuous and factor variables.

The above is conditional upon the distribution of the errors and hence the response. You'll need to specify more about the exact problem for a more informed response.


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.

  • $\begingroup$ +1 it's a good point but while it might very well be what the OP had in mind, there is no mention of R in the question… $\endgroup$
    – Gala
    Jul 8, 2013 at 15:46
  • $\begingroup$ I took glm to be an indicator of potential R-ish-ness, but I see you edit that away. That said, what R fits via lm() and glm() is standard statistical, common or garden LMs or GLMs. $\endgroup$ Jul 8, 2013 at 16:19
  • $\begingroup$ Every technical word (“outcome”, “predictor”, “continuous”) was typeset as code. I didn't know what to make of that. $\endgroup$
    – Gala
    Jul 8, 2013 at 16:20
  • $\begingroup$ @GavinSimpson,Thanks. What can I infer here in this test data? pred=c("A","A","E","C","B","H","C","E","B","B","C","B","D","A","A","A","A","B","B","B","A","B","D","A","A","A","A","C","C","B","B","B","A","G","C","C","F","B","A","A","A","A","A","A","E","B","B","A","A","B","A","A","A","A","A","A","B") res = c(33,33,37,32,32,26,33,28,25,34,29,35,26,20,27,19,30,33,27,24,26,28,27,23,26,25,24,26,24,25,21,21,23,24,23,27,23,20,21,22,22,22,22,23,23,21,22,21,21,23,23,18,20,18,18,18,19 ) df = data.frame(pred, res) fmla = as.formula(res ~ pred) model = glm(fmla, data = df) summary(model). $\endgroup$ Jul 8, 2013 at 17:39
  • $\begingroup$ @user1140126 You should post that into your Question. I'm getting strange characters when pasting that into R from the comment, which is causing issues. $\endgroup$ Jul 8, 2013 at 18:03

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