Testing Populations for Equal Variances

The choice among various difference-of-means tests depends partially on the assumption that two populations have a common variance. In the chapter, we did not discuss how to test whether this assumption holds or not. Here we offer an F test for whether the variances of growth are the same between 1960 and 1990.

Null hypothesis $H_0$: $\sigma^2_{1960}= \sigma^2_{1990}$,i.e., $\frac{\sigma^2_{1960}}{\sigma^2_{1990}}=1$

Alternative hypothesis $H_a$: $\sigma^2_{1960} \not= \sigma^2_{1990}$

The test statistic for the null hypothesis is: $F = \frac{\sigma^2_{1960}}{\sigma^2_{1990}}$

We reject the null hypothesis if the $p$ value for the $F$ test statistic is smaller than the acceptable Type I error, 0.05.

The R code and output below indicate that the $p$ value is much smaller than 0.05. Hence, we reject the null hypothesis that growth in 1960 and growth in 1990 are of equal variance.