Interval estimation

Idea: Determine an interval depending on the values of the sample variables \(X_1, \dots, X_n\) that contains the true value of the parameter \(\theta\) to be estimated with high probability

For given error probability \(\alpha \in [0,1]\), the functions \(G_u = g_u(X_1, \dots, X_n)\) and \(G_o = g_o(X_1, \dots, X_n)\) of the sample variables give a \((1 - \alpha)\) - confidence interval if: \[P(G_u \ge G_o) = 1\] \[P(G_u \ge \theta \ge G_o) = 1 - \alpha\] \(1 - \alpha\) is also called confidence level

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Confidence interval or the expected value

- Case 1: \(o^2\) known; normally distributed

\([\overline{X} - z_{1-\frac{\alpha}{2}}\frac{o}{\sqrt{n}}, \overline{X} + z_{1-\frac{\alpha}{2}}\frac{o}{\sqrt{n}}]\)

- Case 2: \(o^2\) unknown; normally distributed

\([\overline{X} - t_{1-\frac{\alpha}{2}}(n-1)\frac{S}{\sqrt{n}}, \overline{X} + t_{1-\frac{\alpha}{2}}(n-1)\frac{S}{\sqrt{n}}]\) with \(S = \sqrt{\frac{1}{n-1}\sum_{i}(X_i-\overline{X})^2}\)

- Case 3: \(o^2\) known; any distribution (n > 30)

\([\overline{X} - z_{1-\frac{\alpha}{2}}\frac{o}{\sqrt{n}}, \overline{X} + z_{1-\frac{\alpha}{2}}\frac{o}{\sqrt{n}}]\)

- Case 4: \(o^2\) unknown; any distribution (n > 30)

\([\overline{X} - z_{1-\frac{\alpha}{2}}\frac{S}{\sqrt{n}}, \overline{X} + z_{1-\frac{\alpha}{2}}\frac{S}{\sqrt{n}}]\)

Note that the values for \(z_p\) and \(t_p\) are given in a table (see book page 547 ff.)

For the variance \(o^2\); normally distributed

\([\frac{(n-1)S^2}{q_{1-\frac{1-\alpha}{2}}}, \frac{(n-1)S^2}{q_{\frac{\alpha}{2}}}]\)

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Width of an Interval

\(B = 2*t_{1-\frac{\alpha}{2}}\frac{o}{\sqrt{n}}\)