Multiple Linear Regression

Last updated on 2025-04-15 | Edit this page

Overview

Questions

How do you run a multiple linear regression?
What about conditional effects?

Objectives

Learn how to build and trim a model with multiple independent variabels
Learn how to build a model with interaction terms

Introduction

A linear regression is relatively simple to understand and visualise. We have a dependent variable that we try to predict - model - using an independent variable. In principle we can change the independent variable, and the model will predict the change in the dependent variable. We can visualise this using a relatively simple graph, plotting the two variables against each other, and adding the line representing the model.

But what if we have reason to believe that the dependent variable depends on more than one independent variable?

Let us find some data for illustrating this.

Before anything else, we load the tidyverse package, download the data, and read it using read_csv():

R

library(tidyverse)
download.file("https://raw.githubusercontent.com/KUBDatalab/R-toolbox/main/episodes/data/BONEDEN.csv",
              destfile = "data/BONEDEN.csv", mode = "wb")
boneden <- read_csv("data/BONEDEN.csv")

The data records the bonedensity in different parts of the skeleton of female monozygotic twins, as well as height, age, weight, smoking, alcohol, tea and coffee-habits. We are going to study the second twins, where all variables are suffixed with the digit “2”:

R

density <- boneden %>% select(age, contains("2"))
head(density)

OUTPUT

# A tibble: 6 × 12
    age   ht2   wt2  tea2  cof2  alc2  cur2  men2  pyr2   ls2   fn2   fs2
  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1    27   160    56    42    21     0     0     0  13.8  0.76  0.68  1.04
2    42   159    72    20    21     1     1     0  48    0.89  0.64  1.11
3    59   156    54     7    28     0     0     1  20.5  0.51  0.64  0.86
4    61   162    58    21    35     0     0     1  29.8  0.85  0.69  1.03
5    47   150    58    91     0     0     1     1  25    0.59  0.54  0.96
6    33   158    54    14    14     0     0     0   5    0.83  0.5   1.06

Callout

The package tidyselect is a part of the tidyverse. It contains functions that allow us to choose variables/columns based on patterns in their names rather than specifying the names directly. contains("2") here return all columns with a name containing the string “2”.

More specically we are trying to model the bone density in the femoral shaft (lårbensskaft). As an intial hypothesis, we are going to assume that the density depends on height (ht2), age (age), weight (wt2), tea (tea2) and coffee consumption (cof2).

That is, we assume that there is a linear correlation between the bone density and each of the independent variables.

The simple linear regresion between just the bone density and the height, can be done with:

R

lm(fs2 ~ ht2, data = density) %>% 
  summary()

OUTPUT


Call:
lm(formula = fs2 ~ ht2, data = density)

Residuals:
     Min       1Q   Median       3Q      Max
-0.41406 -0.02810  0.00190  0.07658  0.30722

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept) -1.78744    0.71928  -2.485 0.017351 *
ht2          0.01766    0.00445   3.969 0.000301 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.1315 on 39 degrees of freedom
Multiple R-squared:  0.2877,	Adjusted R-squared:  0.2694
F-statistic: 15.75 on 1 and 39 DF,  p-value: 0.0003006

Where we pipe the output of the regression to the summary function to take a closer look. There is a significant correlation between bone density and height. If we increase the height of a woman by one centimeter, the density of her femoral shaft is expected to increase by 0.0177 \(g/cm^3\). The correlation is significant, that is, we are rather sure that the estimate of this coefficient is different from 0.

Now we would like to add the other independent variables to the model.

The model we are building can be described using math as:

\[ \text{fs2} = w_0 + w_1\text{ht2} + w_2\text{wt2} + w_3\text{tea2} + w_4\text{cof2} + \epsilon \] That is - we would like to figure out which coefficients \(w_1\) to \(w_4\), and which intercept \(w_0\) will give us estimates of fs2, that minimizes the error \(\epsilon\)

This is pretty straight forward. Instead of running lm(fs2~ht2, data = density, we add the additional independent variables using the + sign. And again we pipe the result to the summary function to take a closer look:

R

lm(fs2 ~ ht2 + age + wt2 + tea2 + cof2, data = density) %>% 
summary()

OUTPUT


Call:
lm(formula = fs2 ~ ht2 + age + wt2 + tea2 + cof2, data = density)

Residuals:
      Min        1Q    Median        3Q       Max
-0.247601 -0.061517  0.005634  0.067826  0.166474

Coefficients:
              Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.2800114  0.7774507  -0.360 0.720887
ht2          0.0090956  0.0046455   1.958 0.058246 .
age         -0.0068373  0.0015669  -4.364 0.000108 ***
wt2          0.0038274  0.0014934   2.563 0.014836 *
tea2        -0.0009246  0.0009837  -0.940 0.353677
cof2        -0.0005310  0.0009328  -0.569 0.572778
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.1043 on 35 degrees of freedom
Multiple R-squared:  0.5978,	Adjusted R-squared:  0.5403
F-statistic:  10.4 on 5 and 35 DF,  p-value: 3.566e-06

It is that simple. Interpreting the results on the other hand is a bit more complicated.

The bone density, for a newborn woman (0 years old), that is 0 centimeters high, weighing 0 kilogram, who does not drink either tea or coffee, is -0.28 \(g/cm^3\) - which is obviously either impossible, or a discovery that would get you an invitation to Stockholm.

For each centimeter the woman grows, if all other variables are held constant! the bone density is expected to increase by 0.0091 \(g/cm^3\).

For each year the woman grows older, if all other variables are held constant! the bone density is expected to decrease by 0.0068 \(g/cm^3\).

For each kilogram the woman puts on weight, if all other variables are held constant! the bone density is expected to increase by 0.0038 \(g/cm^3\).

The coefficient for the height variable is not significant (if we have decided on a 5% significanse level), and neither consumption of teo nor coffee has a significant influence on bone density. According to this model.

Is this a good model? No. The coefficients for Tea and coffee consumption are not significant, and should not be in the model. Let us remove them.

But let us remove them one by one. First coffee that has the worst p-value:

R

lm(fs2 ~ ht2 + age + wt2 + tea2, data = density) %>% summary()

OUTPUT


Call:
lm(formula = fs2 ~ ht2 + age + wt2 + tea2, data = density)

Residuals:
     Min       1Q   Median       3Q      Max
-0.26036 -0.06261  0.01682  0.06366  0.17393

Coefficients:
              Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.2209640  0.7632342  -0.290   0.7739
ht2          0.0085364  0.0044976   1.898   0.0657 .
age         -0.0068645  0.0015514  -4.425 8.58e-05 ***
wt2          0.0041140  0.0013927   2.954   0.0055 **
tea2        -0.0007497  0.0009257  -0.810   0.4233
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.1034 on 36 degrees of freedom
Multiple R-squared:  0.5941,	Adjusted R-squared:  0.5489
F-statistic: 13.17 on 4 and 36 DF,  p-value: 1.048e-06

Note that not only is the coefficient for tea still not significant, it actually gets “worse”. On the other hand, the coefficient for weight now is larger, ie explains more of the variation.

Callout

Why does this happen?

What the model does is “trying” to explain the variation in fs2, using the independent variables. We have an underlying assumption that we can change those variables independently. However, they are often correlated. In this case there is a relatively large correlation between cof2 and wt2. When we remove cof2 from the model, some of the variation explained by cof2 can now be captured by wt2.

We do not actually explain that much of the variation in fs2. The R² value tells us how much of the variation is explained by the model, and 0.5941 is not that impressive. We will be able to explain everything, including random noise in the data. We just need to add enough independent variables to our model. Some of the random noise will be explained by even a random varible.

This is called overfitting, and one of ways we try to identify that, is by looking at the Adjusted R-squared instead. This adjusts the R² for the number of variables we have in the model. This is the measure of how good the model is, that we prefer.

We continue our “leave-one-out” approach. tea2 is the “worst” performing variable measured by the p-value, and that is the next one we will remove:

R

lm(fs2 ~ ht2 + age + wt2 , data = density) %>% summary()

OUTPUT


Call:
lm(formula = fs2 ~ ht2 + age + wt2, data = density)

Residuals:
      Min        1Q    Median        3Q       Max
-0.273490 -0.058038  0.005241  0.067527  0.169512

Coefficients:
             Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.569569   0.627355  -0.908  0.36981
ht2          0.010524   0.003751   2.806  0.00796 **
age         -0.006489   0.001474  -4.404 8.74e-05 ***
wt2          0.004077   0.001385   2.943  0.00559 **
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.1029 on 37 degrees of freedom
Multiple R-squared:  0.5867,	Adjusted R-squared:  0.5531
F-statistic:  17.5 on 3 and 37 DF,  p-value: 3.078e-07

Now we have a model where all coefficients are significant.

Callout

What about the intercept?

The estimate of the intercept is not significant. Is that a problem?

No, not in this case. The intercept is the estimate of what fs2 would be if all the independent variables are 0. That might be meaningful for age, but definitely not for height or weight.

Also the thing we are interested in is not actually making predictions, in that case we would like a better estimate of the intercept, but in studying the effect of height, age and weight on bonedensity.

Challenge

The dataset contains more variables than we have examined here. Alcohol consumption alc, and how many years the individual women have been smoking pyr.

We also only studied half of the dataset, the variables ending in “2”. Data on the other twins have variables ending in “1”.

Build a model predicting the bonedensity of the femoral shaft fs1 using a selection of the variables age, ht1,wt1,tea1,cof1,alc1andpyr1`.

Make sure that only significant (p<0.05) variables are included.

What is the adjusted R squared value?

Show me the solution

R

dens_exercise <- boneden %>% 
  select(age, contains("1"), -c(cur1, ls1, fn1, men1 ))

model_1 <- lm(fs1~., data = dens_exercise)
model_2 <- lm(fs1 ~ age + ht1 + wt1+ cof1 + alc1 +pyr1, data = dens_exercise)
model_3 <- lm(fs1 ~ age + ht1 + wt1 + alc1 +pyr1, data = dens_exercise)
model_4 <- lm(fs1 ~ age + ht1 + wt1 + pyr1, data = dens_exercise)
model_5 <- lm(fs1 ~ age + ht1 + pyr1, data = dens_exercise)
model_6 <- lm(fs1 ~ age + ht1, data = dens_exercise)

summary(model_6)$adj.r.squared

OUTPUT

[1] 0.2555525

Categorical data

We also have categorical variables in the data. Can we use them?

Yes!

First way - categorical values as independent variables

Let us add information about menopausal status, men2:

R

density <- density %>% 
  mutate(men2 = factor(men2))

cat_model <- lm(fs2 ~ ht2 + age + wt2 + men2, data = density)
summary(cat_model)

OUTPUT


Call:
lm(formula = fs2 ~ ht2 + age + wt2 + men2, data = density)

Residuals:
      Min        1Q    Median        3Q       Max
-0.282979 -0.049356 -0.006728  0.066710  0.171457

Coefficients:
             Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.532553   0.643852  -0.827 0.413764
ht2          0.010650   0.003898   2.732 0.009799 **
age         -0.008232   0.002098  -3.924 0.000389 ***
wt2          0.004191   0.001414   2.963 0.005450 **
men21        0.053401   0.051460   1.038 0.306514
men22       -0.026286   0.059796  -0.440 0.662939
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.1032 on 35 degrees of freedom
Multiple R-squared:  0.6067,	Adjusted R-squared:  0.5505
F-statistic:  10.8 on 5 and 35 DF,  p-value: 2.462e-06

There are three different categorical values in men2. 0: premenopausal, 1: Postmenopausal and 2: Unknown or hysterectomy.

If we want to add categorical data to the model, we need to make sure they are encoded as categorical, otherwise the model will treat them as continous values. We do that by running factor() on the relevant variable.

Again this is relatively simple, the interpretation on the other hand is a bit more complex. The coefficients for ht2, age and wt2 now tells us how the value of fs2 changes for the base case 0, premenopausal. The coefficient of factor(men2)1 tells us that there is an added “bonus” to the bonedensity to become postmenopausal. And factor(men2)2 that there is a “penalty” of -0.026286 to have the “unknown” status. The coefficients are not significant, so this might be a bad example.

The second way - interaction

Maybe the influence of menopause status is not direct, but rather in the form that the coefficient for wt2 depends on menopausal status. Ie increasing weight might lead to increasing bonedensity. But to different degrees depending on menopausal status.

We can add this interaction term to the model by multiplying the categorical variable to the variabel where we suspect that there might be an interaction:

R

interact_model <- lm(fs2 ~ ht2 + age + wt2*men2, data = density)
summary(interact_model)

OUTPUT


Call:
lm(formula = fs2 ~ ht2 + age + wt2 * men2, data = density)

Residuals:
      Min        1Q    Median        3Q       Max
-0.264590 -0.053534 -0.000568  0.057129  0.168441

Coefficients:
             Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.355334   0.648878  -0.548 0.587642
ht2          0.009842   0.004014   2.452 0.019661 *
age         -0.007966   0.002084  -3.823 0.000555 ***
wt2          0.003272   0.002281   1.434 0.160913
men21       -0.166698   0.196497  -0.848 0.402355
men22        0.149021   0.264316   0.564 0.576703
wt2:men21    0.003455   0.003070   1.126 0.268439
wt2:men22   -0.002592   0.004025  -0.644 0.524081
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.1021 on 33 degrees of freedom
Multiple R-squared:  0.637,	Adjusted R-squared:  0.5601
F-statistic: 8.274 on 7 and 33 DF,  p-value: 8.211e-06

Once again it is “simple” to do, more complicated to interpret.

The (estimated) coefficients for ht2and age should be understood as before.

But now we have a good deal more coefficients. factor(men2)1 and factor(men2)2 should also be understood as before.

wt2 is also still the same as before. But we have two new. If the specific woman is postmenopausal, there is an additional term, wt2:men21 that needs to be added to the prediction.

Challenge

What is the prediction for fs2 for a 52 year old, 164 centimeter high, postmenopausal woman weighing 62 kg?

Show me the solution

We find it by adding all the terms:

intercept: -0.355334
52 × age: 52 × -0.007966 = -0.414232
164 × ht2: 164 × 0.009842 = 1.613488
62 × wt2: 62 × 0.003272 = 0.202864
factor(men2)1 = -0.166698 (because men2 is = 1)
65 × wt2:men21 = 62 × 0.003455 = 0.21421

The result is: 1.094298

An alternative way would be to construct a dataframe containing the values for which we want to make the prediction, and use the predict function:

R

new_data <- data.frame(
  ht2 = 164,
  age = 52,
  wt2 = 62,
  men2 = factor(1, levels = levels(density$men2))
)

predict(interact_model, newdata = new_data)

OUTPUT

       1
1.094928

Key Points

We can do linear regression on multiple independent variables
Be careful not to overfit - only retain variables that are significant (and sensible)
We can fit just as well on categorical variables - but make sure they are categorical
Interpreting linear models with multiple variables are not trivial _ Interpreting linear models with interaction terms are even less trivial