Discussion
Suppose \(\{X_{1}, X_{2}, ......., X_{n}\}\) is a random sample from the distribution of X with mean \(\mu\) and variance \(\sigma^{2}\). The test statistic to test \(H_{0} : \sigma^{2} = \sigma_{0}^{2}\) against \(H_{1} : \sigma^{2} \gt \sigma_{0}^{2}\) is given by \(T_{n} = \dfrac{\Sigma(X_{i} - \bar{X})^{2}}{\sigma_{0}^{2}}\). Then the null distribution of \(T_{n}\) is :
Answer: A
The null distribution is \(\chi_{n-1}^{2}\) if X has a normal distribution.
For a normal sample, \(\sum(X_{i} - \bar{X})^{2}/\sigma^{2}\) \(= (n - 1)S^{2}/\sigma^{2} \sim \chi_{n-1}^{2}\), with one degree of freedom lost to estimating the mean. Under \(H_{0}\), \(\sigma^{2} = \sigma_{0}^{2}\), so \(T_{n} \sim \chi_{n-1}^{2}\).
The result depends on normality, so 'irrespective of distribution' is wrong. The \(t\) distribution is for a standardised mean, and F is for a ratio of two independent variance estimates.
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