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Comparison of Population Variances

There are situations in which we may wish to compare the variances of two populations with independent samples. In that case, the test statistic is the ratio of the sample variances, $s_1^2/s_2^2$. Statistical theory implies that if the populations are approximately normally distributed or the sample sizes are large, then under the assumption the population variances are equal, the sampling distribution of this ratio is an F-distribution. This distribution has two parameters, called numerator and denominator degrees of freedom, respectively, which are given by $n_1-1,n_2-1$. This implies that a test of the hypotheses,

\begin{eqnarray*}
H_0:&\ \sigma_1 \le \sigma_2\\
H_1:&\ \sigma_1 > \sigma_2,
\end{eqnarray*}

can be constructed based on the ratio of sample variances,

\begin{displaymath}
F = \frac{s_1^2}{s_2^2}.
\end{displaymath}

Strong evidence for this one-sided alternative would be an F-ratio that is much greater than 1. The p-value therefore would be

\begin{displaymath}
pvalue = 1 - pf(F,n_1-1,n_2-1).
\end{displaymath}

Note that we could have used

\begin{displaymath}
F_{21} = \frac{s_2^2}{s_1^2}
\end{displaymath}

in which case strong evidence for the alternative would be a value for this F-ratio that is much smaller than 1. The p-value then would be

\begin{displaymath}
pvalue = pf(F_{21},n_2-1,n_1-1).
\end{displaymath}

This follows from the fact that

\begin{displaymath}
pf(x,n_1-1,n_2-1) = 1 - pf(1/x,n_2-1,n_1-1).
\end{displaymath}

If the hypotheses had been two-sided,

\begin{eqnarray*}
H_0:&\ \sigma_1 = \sigma_2\\
H_1:&\ \sigma_1 \ne \sigma_2,
\end{eqnarray*}

then in practice we could divide the larger sample variance by the smaller sample variance. The corresponding p-value would be two times the area to the right of this ratio under the corresponding F-distribution.

For example, the data given above for the comparison of male and female financial analysts reported sample sd's $s_1=6000$, $s_2=9000$ based on sample sizes of 25,18. Suppose we wish to test the two-sided hypotheses,

\begin{eqnarray*}
H_0:&\ \sigma_1 = \sigma_2\\
H_1:&\ \sigma_1 \ne \sigma_2,
\end{eqnarray*}

Then the test statistic is

\begin{displaymath}
\left(\frac{9000}{6000}\right)^2 = 2.25.
\end{displaymath}

The p-value is taken from the F-distribution with 17,24 degrees of freedom. This can be obtained using R as follows:
pvalue = 2*(1 - pf(2.25,17,24))
which gives pvalue = 0.0673. Therefore, we would reject the null hypothesis at the 10% level of significance. This conclusion is based on the assumption that the populations are approximately normal, so that assumption should be checked.


next up previous
Next: Hypothesis tests to compare Up: Statistical Decisions Previous: Hypothesis tests to compare
ammann
2017-11-16