\(\sigma^2\) is a nuisance parameter.

Two-side composite hypothesis test

\(X_1,\dotsc,X_n \sim N(\mu,\sigma^2)\)

\(H_0: \mu = \mu_0\)

\(H_1: \mu \neq \mu_0\)

\[\Lambda \left( x \right) = \frac{ sup \{ L(\theta | x) : \theta \in \Theta_0 \}}{sup \{ L(\theta | x) : \theta \in \Theta \}}=\frac{ sup \{ L(\mu,\sigma^2 | x) : \mu=\mu_0,0<\sigma^2<\infty \}}{sup \{ L(\mu,\sigma^2 | x) : 0<\mu<\infty,0<\sigma^2<\infty \}} = \frac{L_0}{L}\] For the denominator, we have

\[L= \left( \frac{1}{2\pi\widehat{\sigma}^2} \right)^{\frac{n}{2}} exp \left[ - \frac{\sum_{i=1}^{n} (x_i - \mu_0)^2 }{ 2 \widehat{\sigma}^2} \right]\] Given that \(\widehat{\mu}=\overline{x}\) and \(\widehat{\sigma}^2 = \frac{1}{n}\sum_{i=1}^{n} (x_i-\overline{x})^2\) (MLE estimates)

\[L=\left( \frac{1}{2\pi \frac{1}{n}\sum_{i=1}^{n} (x_i-\overline{x})^2} \right)^{\frac{n}{2}} exp \left( -\frac{n}{2}\right)=\left( \frac{n \, e^{-1}}{2\pi \sum_{i=1}^{n} (x_i-\overline{x})^2} \right)^{\frac{n}{2}}\] Similarly, for the numerator, we have \[L_0=\left( \frac{n \, e^{-1}}{2\pi \sum_{i=1}^{n} (x_i-\mu_0)^2} \right)^{\frac{n}{2}}\] Then \[\Lambda \left( x \right) = \left[ \frac{\sum_{i=1}^{n} (x_i-\overline{x})^2}{\sum_{i=1}^{n} (x_i-\mu_0)^2}\right]^{\frac{n}{2}}\] Given that

\[\sum_{i=1}^{n}(x_i-\mu_0)^2 = \sum_{i=1}^{n}(x_i-\overline{x})^2 + n(\overline{x}-\mu_0)^2\] The likelihood ratio is

\[\Lambda \left( x \right) = \left( \frac{1}{1 + \frac{n(\overline{x}-\mu_0)^2}{\sum_i(x_i-\overline{x})^2} }\right)^{\frac{n}{2}}\]

So

\[\left( \frac{1}{1 + \frac{n(\overline{x}-\mu_0)^2}{\sum_i(x_i-\overline{x})^2} }\right)\leq c^{\frac{2}{n}}\] which is equivalent to

\[\frac{n(\overline{x}-\mu_0)^2}{\frac{1}{n-1}\sum_i(x_i-\overline{x})^2} \geq (n-1)(c^{-2/n-1})=k^*\] Hence \[\frac{(\overline{x}-\mu_0)^2}{S^2/n} \geq k^*\] where \(S^2\) is the sample variance. That is

\[\frac{|\overline{x}-\mu_0|}{S/\sqrt{n}} \geq k^{**}\] \(\frac{\overline{x}-\mu_0}{S/\sqrt{n}}\) is known as a \(T\) statistic.

If \(X_1,\dotsc,X_n \sim N(\mu,\sigma^2)\) is normally distributed, then \(T\) follows a \(t\) distribution with \((n-1)\) degrees of freedom.

If we want to fix \(\alpha\) then

\[k^{**} = t_{\alpha/2,n-1}\]