How to perform an analysis of covariance (ANCOVA)

Description

Recall that covariates are variables that may be related to the outcome but are unaffected by treatment assignment. In a randomized experiment with one or more observed covariates, an analysis of covariance (ANCOVA) addresses this question: How would the mean outcome in each treatment group change if all groups were equal with respect to the covariate? The goal is to remove any variability in the outcome associated with the covariate from the unexplained variability used to determine statistical significance.

Related tasks:

• How to do a one-way analysis of variance (ANOVA)
• How to compare two nested linear models
• How to conduct a mixed designs ANOVA
• How to conduct a repeated measures ANOVA

Using pingouin, in Python

View this solution alone.

The solution below uses an example dataset about car design and fuel consumption from a 1974 Motor Trend magazine. (See how to quickly load some sample data.)

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from rdatasets import data
df = data('mtcars')

Let’s use ANCOVA to check the effect of the engine type (0 = V-shaped, 1 = straight, in the variable vs) on the miles per gallon when considering the weight of the car as a covariate. We will use the ancova function from the pingouin package to conduct the test.

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from pingouin import ancova
ancova(data=df, dv='mpg', covar='wt', between='vs')

The $p$-value for each variable is in the p-unc column.

The $p$-value for the wt variable tests the null hypothesis, “The quantities wt and mpg are not related.” Since it is below 0.05, we reject the null hypothesis, and conclude that wt is significant in predicting mpg.

The $p$-value for the vs variable tests the null hypothesis, “The quantities vs and mpg are not related if we hold wt constant.” Since it is below 0.05, we reject the null hypothesis, and conclude that vs is significant in predicting mpg even among cars with equal weight (wt).

Note: Unfortunately, a two-factor ANCOVA is not possible in pingouin. However, a model with more than one covariate is possible, as you can provide a list as the covar parameter when calling ancova.

Content last modified on 24 July 2023.

See a problem? Tell us or edit the source.

Solution, in R

View this solution alone.

The solution below uses an example dataset about car design and fuel consumption from a 1974 Motor Trend magazine. (See how to quickly load some sample data.)

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df <- mtcars
df$vs <- as.factor(df$vs)

Let’s use ANCOVA to check the effect of the engine type (0 = V-shaped, 1 = straight, in the variable vs) on the miles per gallon when considering the weight of the car as a covariate. We will use the ancova function from the pingouin package to conduct the test.

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cov.model <- lm(mpg ~ wt + vs, data = df)
anova(cov.model)

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          Df Sum Sq    Mean Sq    F value    Pr(>F)      
wt         1 847.72525 847.725250 109.704168 2.284396e-11
vs         1  54.22806  54.228061   7.017656 1.292580e-02
Residuals 29 224.09388   7.727375         NA           NA

The $p$-value for each variable can be found in the final column of the output, called Pr(>F).

The $p$-value for the wt variable tests the null hypothesis, “The quantities wt and mpg are not related.” Since it is below 0.05, we reject the null hypothesis, and conclude that wt is significant in predicting mpg.

The $p$-value for the vs variable tests the null hypothesis, “The quantities vs and mpg are not related if we hold wt constant.” Since it is below 0.05, we reject the null hypothesis, and conclude that vs is significant in predicting mpg even among cars with equal weight (wt).

If we wish to create a 2-factor ANCOVA model, we can test to see if the engine type (0 = V-shaped, 1 = straight) and transmission type (0 = automatic, 1 = manual) have an effect on the Miles/gallon per car when considering the weight of the car as a covariate.

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cov.model.2 <- lm(mpg ~ wt + vs + am, data = df)
anova(cov.model.2)

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          Df Sum Sq     Mean Sq    F value    Pr(>F)      
wt         1 847.725250 847.725250 109.729918 3.420018e-11
vs         1  54.228061  54.228061   7.019303 1.310627e-02
am         1   7.778149   7.778149   1.006807 3.242621e-01
Residuals 28 216.315728   7.725562         NA           NA

The $p$-values are again in the final column of output. They show that at the 5% significance level, we would conclude that engine type (vs) significantly impacts the Miles/gallon per car while accounting for the weight of the car (wt) but the transmission type (am) does not.

Content last modified on 24 July 2023.

See a problem? Tell us or edit the source.

Topics that include this task

• Bentley University MA214
• Bentley University MA255

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