Use this fact to derive an exact 95% confidence interval for the population variance o?.

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### Theorem: Variance and Chi-Square Distribution

Let \( X_1, X_2, \ldots, X_n \) be a random sample from a normal distribution with mean \( \mu \) and variance \( \sigma^2 \). The expression for the sample variance \( s^2 \) is given by:

\[
\frac{(n-1)s^2}{\sigma^2} = \frac{\sum_{i=1}^n (X_i - \overline{X})^2}{\sigma^2}
\]

This expression follows a chi-square (\(\chi^2\)) distribution with \( n-1 \) degrees of freedom. Here, \( \overline{X} \) is the sample mean, and \( s^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i - \overline{X})^2 \) is the unbiased estimator of the population variance. Importantly, \( \overline{X} \) and \( s^2 \) are independent random variables.

### Application

Use this theorem to derive an exact 95% confidence interval for the population variance \( \sigma^2 \). Consider the properties of the chi-square distribution and the relation of sample variance to the population variance in your calculations.
Transcribed Image Text:### Theorem: Variance and Chi-Square Distribution Let \( X_1, X_2, \ldots, X_n \) be a random sample from a normal distribution with mean \( \mu \) and variance \( \sigma^2 \). The expression for the sample variance \( s^2 \) is given by: \[ \frac{(n-1)s^2}{\sigma^2} = \frac{\sum_{i=1}^n (X_i - \overline{X})^2}{\sigma^2} \] This expression follows a chi-square (\(\chi^2\)) distribution with \( n-1 \) degrees of freedom. Here, \( \overline{X} \) is the sample mean, and \( s^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i - \overline{X})^2 \) is the unbiased estimator of the population variance. Importantly, \( \overline{X} \) and \( s^2 \) are independent random variables. ### Application Use this theorem to derive an exact 95% confidence interval for the population variance \( \sigma^2 \). Consider the properties of the chi-square distribution and the relation of sample variance to the population variance in your calculations.
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