# How to do a hypothesis test for the ratio of two population variances (in R)

See all solutions.

Let’s say we want to compare the variability of two populations. We take two samples of data, $x_1, x_2, x_3, \ldots, x_k$ from population 1 and $x’_1, x’_2, x’_3, \ldots, x’_k$ from population 2. What hypothesis tests can help us compare the population variances?

## Solution

We’ll use R’s dataset EuStockMarkets to do an example. This dataset has information on the daily closing prices of 4 European stock indices. We’re going to compare the variability of Germany’s DAX and France’s CAC closing prices.

Let’s load the dataset. (See how to quickly load some sample data.) If using your own data, place it into the sample1 and sample2 variables instead of using the code below.

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# install.packages("datasets") # If you have not already done so
library(datasets)

# Load the dataset and convert it to a data frame, then extract two columns
EuStockMarkets <- data.frame(EuStockMarkets)
sample.1 <- EuStockMarkets$DAX sample.2 <- EuStockMarkets$CAC


# Two-tailed test

For all tests below, we will use $\alpha=0.05$ as our Type I Error Rate, but any value between 0.0 and 1.0 can be used.

### Two-tailed test

We can use a two-tailed test to test whether the two population variances are equal. Specifically, the null hypothesis will be:

$H_0: \frac{\sigma_1^2}{\sigma_2^2} = 1$
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sample.1.df <- length(sample.1) - 1            # degrees of freedom
sample.2.df <- length(sample.2) - 1            # degrees of freedom
test.statistic <- var(sample.1)/var(sample.2)  # test statistic
2*pf(test.statistic, df1=sample.1.df, df2=sample.2.df, lower.tail=FALSE) # p-value

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[1] 7.729079e-151


Our $p$-value is smaller than our chosen alpha, so we have sufficient evidence to reject the null hypothesis. The ratio of the variance of the closing prices on Germany’s DAX and France’s CAC is significantly different than 1, so the variances are not equal.

### Right-tailed test

In a right-tailed test, the null hypothesis is that the ratio is less than or equal to 1. This is equivalent to asking if $\sigma_1^2 \le \sigma_2^2$.

$H_0: \frac{\sigma_1^2}{\sigma_2^2} \le 1$

We repeat below some of the code above to make each example easy to copy and paste.

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sample.1.df <- length(sample.1) - 1            # degrees of freedom
sample.2.df <- length(sample.2) - 1            # degrees of freedom
test.statistic <- var(sample.1)/var(sample.2)  # test statistic
pf(test.statistic, df1=sample.1.df, df2=sample.2.df, lower.tail=FALSE) # p-value

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[1] 3.86454e-151


Our $p$-value is smaller than our chosen alpha, so we have sufficient evidence to reject the null hypothesis. The ratio of the variance of the closing prices on Germany’s DAX and France’s CAC is significantly greater than 1, so the variance of closing prices on Germany’s DAX is greater than that of closing prices on France’s CAC.

To test whether $\sigma_1^2 \ge \sigma_2^2$, simply swap the roles of the two data columns in the above code.