Predicting the difference between two groups in R

Question

I'm working on practice exercises with R. From the data, I now have to determine the difference in predicted male compared to female spending while keeping all other predictors constant.

I replaced the sex 0,1 with male, female:

> spending$sex=factor(spending$sex)
> levels(spending$sex)=c("male", "female")
> tail(spending)
    sex status income verbal gamble
42 male     61  15.00      9   69.7
43 male     75   3.00      8   13.3
44 male     66   3.25      9    0.6
45 male     62   4.94      6   38.0
46 male     71   1.50      7   14.4
47 male     71   2.50      9   19.2
> summary(spending)
     sex         status          income           verbal     
 male  :28   Min.   :18.00   Min.   : 0.600   Min.   : 1.00  
 female:19   1st Qu.:28.00   1st Qu.: 2.000   1st Qu.: 6.00  
             Median :43.00   Median : 3.250   Median : 7.00  
             Mean   :45.23   Mean   : 4.642   Mean   : 6.66  
             3rd Qu.:61.50   3rd Qu.: 6.210   3rd Qu.: 8.00  
             Max.   :75.00   Max.   :15.000   Max.   :10.00  
     gamble     
 Min.   :  0.0  
 1st Qu.:  1.1  
 Median :  6.0  
 Mean   : 19.3  
 3rd Qu.: 19.4  
 Max.   :156.0  

Now, this is where I got stumped, how do I set this up with constants and only use the sex variable?

Answer 1

You've got one big problem in your data as posted: there's no "spending" column, so you don't have a response. Once you get that, just run a linear model with sex as your only predictor.

 yourModel <- lm(spending ~ sex, data = spending)
 summary(yourModel)

The base level (intercept only) corresponds to your factor's 0-level, in this case male. The difference between the levels is sex coefficient in the model output.

When you have more than 2 levels, the output is explicit. For example, using the built-in dataset mtcars

head(mtcars)
mtcars$cyl <- factor(mtcars$cyl)
mylm <- lm(mpg ~ cyl, data = mtcars)

> levels(mtcars$cyl)
[1] "4" "6" "8"

> summary(mylm)

Call:
lm(formula = mpg ~ cyl, data = mtcars)

Residuals:
    Min      1Q  Median      3Q     Max 
-5.2636 -1.8357  0.0286  1.3893  7.2364 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  26.6636     0.9718  27.437  < 2e-16 ***
cyl6         -6.9208     1.5583  -4.441 0.000119 ***
cyl8        -11.5636     1.2986  -8.905 8.57e-10 ***
---

Here the factor levels are called out, with the 0-level (4-cylinder) corresponding the the intercept. As in your example, when there are no other predictors, this is as simple as taking the means for each factor level. The 4-cyl mean is 26.66, same as the intercept, and then adding -6.9 we get the 6-cyl mean, and adding -11.56 (to the original 26.66) we get the 8-cyl mean of 15.1.

> with(mtcars, tapply(mpg, cyl, mean)) ## just calculates mean mpg at each level of cyl
       4        6        8 
26.66364 19.74286 15.10000