Question

Suppose that\({{\bf{X}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}\)form a random sample from the normal distribution with unknown mean μ and unknown standard deviation σ, and let\({\bf{\hat \mu }}\,\,{\bf{and}}\,\,{\bf{\hat \sigma }}\)denote the M.L.E.’s of μ and σ. For the sample size n = 17, find a value of k such that

\({\bf{Pr}}\left( {{\bf{\hat \mu > \mu + k\hat \sigma }}} \right){\bf{ = 0}}{\bf{.95}}\)

Short answer

The value of k is -0.4365.

Step-by-step solution

Given information

Suppose that \({X_1},...,{X_n}\) from a random sample from the normal distribution with unknown mean is \(\mu \) and unknown standard deviation is \(\sigma \). And the sample size is\(n = 17\).

Finding the value of k

Since,

\(\hat \mu = {\bar X_n}\,\,and\,\,{\hat \sigma ^2} = \frac{{S_n^2}}{n}\)

It follows the t-distribution of the U has the t-distribution with n-1 degree of freedom.

It can be rewritten as,

\(U = \frac{{{n^{\frac{1}{2}}}\left( {{{\bar X}_n} - \mu } \right)}}{{{{\left( {\frac{{S_n^2}}{{n - 1}}} \right)}^{\frac{1}{2}}}}}\)

Then,

\(\begin{align}\Pr \left( {\hat \mu > \mu + k\hat \sigma } \right) &= \Pr \left( {\frac{{{{\bar X}_n} - \mu }}{{\hat \sigma }} > k} \right)\\ &= \Pr \left( {U > k{{\left( {n - 1} \right)}^{\frac{1}{2}}}} \right)\end{align}\)

Since U has the t-distribution with n-1 degrees of freedom and \(n = 17\)

Then,

\begin{align}\Pr \left[ {U > \,k{{\left( {n - 1} \right)}^{\frac{1}{2}}}} \right] = \,0.95\\\Pr \left[ {U\, > \,k{{\left( {17 - 1} \right)}^{\frac{1}{2}}}} \right] = \,0.95\\\Pr \left[ {U > k{{\left( {16} \right)}^{\frac{1}{2}}}} \right] = 0.95\\ \Pr \left[ {U > 4k} \right] = 0.95\end{align}

It is found from a table t-distribution with 16 degrees of freedom that,

\(\Pr \left( {U < 1.746} \right) = 0.95\)

Hence, by symmetry,

\(\Pr \left( {U > - 1.746} \right) = 0.95\)

So,

\(\begin{align}4k &= - 1.746\\k &= - \frac{{1.746}}{4}\\k &= - 0.4365\end{align}\)

Therefore, the value of k is -0.4365.