Build a confidence interval for a variance

Question

You observe n = 20 i.i.d. \mathcal{N}(\mu, \sigma^2) values and compute a sample variance S^2 = 4. Build a 95% confidence interval for the true variance \sigma^2, and comment on its shape.

Accepted Answer

The pivotal quantity is \frac{(n-1)S^2}{\sigma^2} \sim \chi^2{n-1}. With n - 1 = 19 degrees of freedom, the central 95% of the \chi^2{19} distribution lies between \chi^2{19,\,0.025} \approx 8.91 \quad	ext{and}\quad \chi^2{19,\,0.975} \approx 32.85. Invert to bracket \sigma^2 (note the larger quantile goes in the lower bound): \left[\frac{(n-1)S^2}{\chi^2{19,\,0.975}},\ \frac{(n-1)S^2}{\chi^2{19,\,0.025}}ight] = \left[\frac{19 \cdot 4}{32.85},\ \frac{19 \cdot 4}{8.91}ight] = \left[\frac{76}{32.85},\ \frac{76}{8.91}ight] \approx [2.31,\ 8.53]. So a 95% CI for \sigma^2 is about [2.31, 8.…

Build a confidence interval for a variance

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