# How to do a hypothesis test for population variance (in R)

## Task

Assume we want to estimate the variability of a quantity across a population, starting from a sample of data, $x_1, x_2, x_3, \ldots x_k$. How might we test whether the population variance is equal to, greater than, or less than a hypothesized value?

Related tasks:

- How to compute a confidence interval for the population proportion
- How to do a hypothesis test for a mean difference (matched pairs)
- How to do a hypothesis test for a population proportion
- How to do a hypothesis test for the difference between means when both population variances are known
- How to do a hypothesis test for the difference between two proportions
- How to do a hypothesis test for the mean with known standard deviation
- How to do a hypothesis test for the ratio of two population variances
- How to do a hypothesis test of a coefficient’s significance
- How to do a one-sided hypothesis test for two sample means
- How to do a two-sided hypothesis test for a sample mean
- How to do a two-sided hypothesis test for two sample means

## 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 look at the variability of Germany’s DAX closing prices.

Let’s load the dataset. (See how to quickly load some sample data.)
If using your own data, place it into the `values`

variable instead of using
the code below.

1
2
3
4

# install.packages("datasets") # If you have not already done this
library(datasets)
EuStockMarkets <- data.frame(EuStockMarkets)
values <- EuStockMarkets$DAX

### Two-tailed test

We may ask whether the population variance is significantly different from a hypothesized value. Let’s test against a variance of 1,000,000.

Our null hypothesis states that the population variance is equal to 1,000,000, $H_0: \sigma^2 = 1,000,000$. We calculate the test statistic and $p$-value as follows, using a $\chi^2$ distribution. We can use any $\alpha$ between 0.0 and 1.0 as our Type I Error Rate; we will use $\alpha=0.05$ here.

1
2
3
4

hyp.var <- 1000000 # hypothesized variance
df <- length(values) - 1 # degrees of freedom
test.statistic <- df*var(values)/hyp.var # test statistic
2*pchisq(test.statistic, df=df, lower.tail=FALSE) # two-tailed p-value

1

[1] 3.189769e-07

Our $p$-value, $3.189769\times10^{-7}$, is smaller than $\alpha$, so we have sufficient evidence to reject the null hypothesis. The variance of closing prices on Germany’s DAX is signficantly different from 1,000,000.

### Left-tailed test

What if we wanted to determine if the population variance were significantly less than 1,000,000? Our null hypothesis would therefore be $H_0: \sigma^2 \ge 1,000,000$.

The computations are very similar to the previous case, but with a different formula for the $p$-value. We repeat the code that’s in common, for ease of use when copying and pasting.

1
2
3
4

hyp.var <- 1000000 # hypothesized variance
df <- length(values) - 1 # degrees of freedom
test.statistic <- df*var(values)/hyp.var # test statistic
pchisq(test.statistic, df=df, lower.tail=TRUE) # left-tailed p-value

1

[1] 0.9999998

Our p-value, 0.9999998, is greater than $\alpha$, so we do not have sufficient evidence to reject the null hypothesis. We should continue to assume that the variance of closing prices on Germany’s DAX is greater than or equal to 1,000,000.

### Right-tailed test

What if we wanted to determine if the population variance were significantly less than 1,000,000? Our null hypothesis would therefore be $H_0: \sigma^2 \ge 1,000,000$.

The computations are very similar to the previous case, but with a different formula for the $p$-value. We repeat the code that’s in common, for ease of use when copying and pasting.

1
2
3
4

hyp.var <- 1000000 # hypothesized variance
df <- length(values) - 1 # degrees of freedom
test.statistic <- df*var(values)/hyp.var # test statistic
pchisq(test.statistic, df=df, lower.tail=FALSE) # right-tailed p-value

1

[1] 1.594884e-07

Our p-value, $1.594884\times10^{-7}$, is smaller than $\alpha$, so have sufficient evidence to reject the null hypothesis. We conclude that the variance of closing prices on Germany’s DAX is significantly greater than 1,000,000.

Content last modified on 24 July 2023.

See a problem? Tell us or edit the source.

Contributed by Elizabeth Czarniak (CZARNIA_ELIZ@bentley.edu)