Discussion
Suppose \(\{X_{1}, X_{2}, \ldots, X_{m}\}\) is a random sample from the distribution of X with mean \(\mu_{1}\) and variance \(\sigma_{1}^{2}\), \(\{y_{1}, \ldots, y_{n}\}\) is a random sample from the distribution of Y with mean \(\mu_{2}\) and variance \(\sigma_{2}^{2}\). The test statistic to test \(H_{0} : \sigma_{1}^{2} = \sigma_{2}^{2}\) against \(H_{1} : \sigma_{1}^{2} = \sigma_{2}^{2}\) is \(T_{n} = \dfrac{S_{X}^{2}}{S_{Y}^{2}}\), where \(S_{X}^{2}\) and \(S_{Y}^{2}\) are sample variances. Then the null distribution of \(T_{n}\) is :
Answer: A
The null distribution is F if X and Y are independent and have normal distributions.
For normal samples, \((m - 1)S_{X}^{2}/\sigma_{1}^{2}\) and \((n - 1)S_{Y}^{2}/\sigma_{2}^{2}\) are chi-square variables. If the samples are independent, their ratio, each divided by its degrees of freedom, has an F distribution; under \(H_{0} : \sigma_{1}^{2} = \sigma_{2}^{2}\) the variances cancel and \(S_{X}^{2}/S_{Y}^{2} \sim F_{m-1, n-1}\).
Both conditions are needed: normality gives the chi-square laws, and independence makes the ratio F. Equal means are not needed. (The printed alternative repeats '=' where '\(\ne\)' is meant; it does not affect the null distribution.)
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