Question

Suppose that\({X_1},...,{X_n}\)form a random sample from the normal distribution with unknown mean μ and unknown variance\({\sigma ^2}\). Describe a method for constructing a confidence interval for\({\sigma ^2}\)with a specified confidence coefficient\(\gamma \left( {0 < \gamma < 1} \right)\) .

Short answer

\(P\left( {\frac{1}{{{c_2}}}\sum\limits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^{2}} \le {\sigma ^{2}} \le } \frac{1}{{{c_1}}}\sum\limits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^2}} } \right) = \gamma \)

Step-by-step solution

Given information

\({X_1},...,{X_n}\) form a random sample from the normal distribution with unknown mean μ and unknown variance \({\sigma ^2}\). We need to describe a method for constructing a confidence interval for \({\sigma ^{2}}\) with a specified confidence coefficient \(\gamma \left( {0 < \gamma < 1} \right)\) .

A method for constructing a confidence interval for \({\sigma ^2}\) with a specified confidence coefficient \(\gamma \left( {0 < \gamma  < 1} \right)\) .

Let\(U = \frac{1}{{{\sigma ^2}}}\sum\limits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^2}} \) has chi-square distribution with n-1 degrees of freedom, the constants \({c_1}\,\,\,\,{\rm{and}}\,\,\,\,{c_2}\) in such that \(P\left( {{c_1} < U < {c_2}} \right) = \gamma \) could be one of an infinite number of constants which satisfies the probability. However an interval with limits taken from

\(\begin{align}P\left( {{c_1} < U < {c_2}} \right)\\ &= P\left( {\frac{1}{{{c_2}}}\sum\limits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^2} \le {\sigma ^2} \le } \frac{1}{{{c_1}}}\sum\limits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^2}} } \right)\end{align}\)

Where the constants can be any of the infinite pairs satisfying the probability.

Hence a method for constructing a confidence interval for \({\sigma ^2}\) with a specified confidence coefficient \(\gamma \left( {0 < \gamma < 1} \right)\):

\(P\left( {\frac{1}{{{c_2}}}\sum\limits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^2} \le {\sigma ^2} \le } \frac{1}{{{c_1}}}\sum\limits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^2}} } \right) = \gamma \)