91Ó°ÊÓ

A random sample \({X_1}, \ldots ,{X_n}\) follows a normal distribution with mean \(\mu \) and variance \({\sigma ^2}\). \({\hat \sigma ^2}\) Is the sample variance.

So, \({\hat \sigma ^2} = \frac{1}{n}\sum\nolimits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^2}} \)

Here, \({X_i} \sim N\left( {\mu ,{\sigma ^2}} \right)\), by the theory of joint distribution of the sample mean and sample variance, there can be concluded that \(Y = {\sum\nolimits_{i = 1}^n {\left( {\frac{{{X_i} - {{\bar X}_n}}}{\sigma }} \right)} ^2}\).

Hence, \(Y \sim \chi _{\left( {n - 1} \right)}^2\).

Now,

\(\begin{align}\Pr \left( {\frac{{{{\hat \sigma }^2}}}{{{\sigma ^2}}} \le 1.5} \right) &= \Pr \left( {\frac{Y}{n} \le 1.5} \right)\\ &= \Pr \left( {Y \le 1.5n} \right)\end{align}\)

Let us consider some values of n such that the 0.95 quantiles of \(\chi _{\left( {n - 1} \right)}^2\) is \(1.5n\).

Therefore, trying the different values n, there can be concluded that, if n=21 then,

\(\Pr \left( {Y \le 1.5 \times 21} \right) = 0.9510 \ge 0.95\), where \(Y \sim \chi _{\left( {20} \right)}^2\).

Here, \({X_i} \sim N\left( {\mu ,{\sigma ^2}} \right)\), , by the theory of joint distribution of the sample mean and sample variance, there can be concluded that \(Y = {\sum\nolimits_{i = 1}^n {\left( {\frac{{{X_i} - {{\bar X}_n}}}{\sigma }} \right)} ^2}\).

Hence,\(Y \sim \chi _{\left( {n - 1} \right)}^2\).

Now,

\(\begin{align}\Pr \left( {\left| {{{\hat \sigma }^2} - {\sigma ^2}} \right| \le \frac{1}{2}{\sigma ^2}} \right) &= \Pr \left( { - \frac{1}{2} \le \frac{{{{\hat \sigma }^2}}}{{{\sigma ^2}}} - 1 \le \frac{1}{2}} \right)\\ &= \Pr \left( {\frac{1}{2} \le \frac{{{{\hat \sigma }^2}}}{{{\sigma ^2}}} \le \frac{3}{2}} \right)\\ &= \Pr \left( {\frac{n}{2} \le Y \le \frac{{3n}}{2}} \right)\end{align}\)

To \(\Pr \left( {\left| {{{\hat \sigma }^2} - {\sigma ^2}} \right| \le \frac{1}{2}{\sigma ^2}} \right) \ge 0.8\), let us consider some values of n.

Therefore, trying the different values n, there can be concluded that, if n=13 then,

\(\Pr \left( {\frac{n}{2} \le Y \le \frac{{3n}}{2}} \right) = 0.8116 \ge 0.8\), where\(Y \sim \chi _{\left( {12} \right)}^2\).