Question

Suppose that \({{\bf{X}}_{\bf{1}}}{\bf{, \ldots ,}}{{\bf{X}}_{\bf{n}}}\)form a random sample from the normal distribution with unknown mean μand unknown variance \({{\bf{\sigma }}^{\bf{2}}}\), and let the random variableLdenote the length of the shortest confidence interval forμthat can be constructed from the observed values in the sample. Find the value of \({\bf{E}}\left( {{{\bf{L}}^{\bf{2}}}} \right)\)for the following values of the sample sizenand the confidence coefficient\(\gamma \):

\(\begin{align}{\bf{a}}{\bf{.n = 5,}}\gamma {\bf{ = 0}}{\bf{.95}}\\{\bf{b}}{\bf{.n = 10,}}\gamma {\bf{ = 0}}{\bf{.95}}\\{\bf{c}}{\bf{.n = 30,}}\gamma {\bf{ = 0}}{\bf{.95}}\\{\bf{d}}{\bf{.n = 8,}}\gamma {\bf{ = 0}}{\bf{.90}}\\{\bf{e}}{\bf{.n = 8,}}\gamma {\bf{ = 0}}{\bf{.95}}\\{\bf{f}}{\bf{.n = 8,}}\gamma {\bf{ = 0}}{\bf{.99}}\end{align}\)

Short answer

1. For \(n = 5\;and\;\gamma = 0.95\), \(E\left( {{L^2}} \right) = 6.16{\sigma ^2}\)
2. For\(n = 10\;and\;\gamma = 0.95\),\(E\left( {{L^2}} \right) = 2.04{\sigma ^2}\)
3. For\(n = 30\;and\;\gamma = 0.95\),\(E\left( {{L^2}} \right) = 0.55{\sigma ^2}\)
4. For\(n = 8\;and\;\gamma = 0.90\),\(E\left( {{L^2}} \right) = 1.79{\sigma ^2}\)
5. For\(n = 8\;and\;\gamma = 0.95\),\(E\left( {{L^2}} \right) = 2.79{\sigma ^2}\)
6. For \(n = 8\;and\;\gamma = 0.99\), \(E\left( {{L^2}} \right) = 6.12{\sigma ^2}\)

Step-by-step solution

Given information

\({X_1}, \cdots {X_n}\) is a random sample from a normal distribution with an unknown mean and variance. The mean and variance are denoted by \(\mu \;and\;{\sigma ^2}\) respectively. L is the length of the shortest confidence interval for \(\mu \). \(\mu \)can be constructed from the observed values in the sample.

determine the expectation of the shortest confidence interval

find the value of \(E\left( {{L^2}} \right)\).

As there given that the variance is unknown, there can be concluded that the length of the interval can be expressed as, \(L = 2T_{n - 1}^{ - 1}\left( {\frac{{1 + \gamma }}{2}} \right)\frac{{\sigma '}}{{\sqrt n }}\)

So,\({L^2} = 4{\left( {T_{n - 1}^{ - 1}\left( {\frac{{1 + \gamma }}{2}} \right)} \right)^2}\frac{{{{\sigma '}^2}}}{n}\)

Therefore, there can be clearly seen that for the value of\(E\left( {{L^2}} \right)\), simply, need to calculate\(E\left( {{{\sigma '}^2}} \right)\).

Now, it is known that,\(E\left( {\frac{{\sum\nolimits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^2}} }}{{{\sigma ^2}}}} \right) = \left( {n - 1} \right)\)

So, the value of\(E\left( {{{\sigma '}^2}} \right)\),

\(\begin{align}E\left( {{{\sigma '}^2}} \right) &= E\left( {\frac{{\sum\nolimits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^2}} }}{{\left( {n - 1} \right)}}} \right)\\ &= \frac{{{\sigma ^2}}}{{\left( {n - 1} \right)}}E\left( {\frac{{\sum\nolimits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^2}} }}{{{\sigma ^2}}}} \right)\\ &= {\sigma ^2}\end{align}\)

Therefore, the required value of\(E\left( {{L^2}} \right)\)is,

\(\begin{align}E\left( {{L^2}} \right) &= 4{\left( {T_{n - 1}^{ - 1}\left( {\frac{{1 + \gamma }}{2}} \right)} \right)^2}\frac{{E\left( {{{\sigma '}^2}} \right)}}{n}\\ &= 4{\left( {T_{n - 1}^{ - 1}\left( {\frac{{1 + \gamma }}{2}} \right)} \right)^2}\frac{{{\sigma ^2}}}{n}\end{align}\)

Calculate the expectation values for fixed confidence coefficient

In the problem the variance is unknown, so, \(E\left( {{L^2}} \right)\) will be function of \({\sigma ^2}\). There can obtained the values of T by the t- table with the corresponding degrees of freedom.

a.

For \(n = 5\;and\;\gamma = 0.95\),

\(\begin{align}E\left( {{L^2}} \right) &= 4{\left( {T_{n - 1}^{ - 1}\left( {\frac{{1 + \gamma }}{2}} \right)} \right)^2}\frac{{{\sigma ^2}}}{n}\\ &= 4{\left( {T_4^{ - 1}\left( {0.975} \right)} \right)^2}\frac{{{\sigma ^2}}}{5}\\ &= 4 \times {\left( {2.776} \right)^2} \times \frac{{{\sigma ^2}}}{5}\\ &= 6.16{\sigma ^2}\end{align}\)

b.

For \(n = 10\;and\;\gamma = 0.95\),

\(\begin{align}E\left( {{L^2}} \right) &= 4{\left( {T_{n - 1}^{ - 1}\left( {\frac{{1 + \gamma }}{2}} \right)} \right)^2}\frac{{{\sigma ^2}}}{n}\\ &= 4{\left( {T_9^{ - 1}\left( {0.975} \right)} \right)^2}\frac{{{\sigma ^2}}}{{10}}\\ &= 4 \times {\left( {2.262} \right)^2} \times \frac{{{\sigma ^2}}}{{10}}\\ &= 2.04{\sigma ^2}\end{align}\)

c.

For \(n = 30\;and\;\gamma = 0.95\),

\(\begin{align}E\left( {{L^2}} \right) &= 4{\left( {T_{n - 1}^{ - 1}\left( {\frac{{1 + \gamma }}{2}} \right)} \right)^2}\frac{{{\sigma ^2}}}{n}\\ &= 4{\left( {T_{29}^{ - 1}\left( {0.975} \right)} \right)^2}\frac{{{\sigma ^2}}}{{30}}\\ &= 4 \times {\left( {2.045} \right)^2} \times \frac{{{\sigma ^2}}}{{30}}\\ &= 0.55{\sigma ^2}\end{align}\)

So, there can be noticed that for a fixed confidence coefficient \(\gamma \), if the sample size increases the confidence interval became more precise.

 Step 4: Calculate the expectation values for fixed sample size

d.

For \(n = 8\;and\;\gamma = 0.90\),

\(\begin{align}E\left( {{L^2}} \right) &= 4{\left( {T_{n - 1}^{ - 1}\left( {\frac{{1 + \gamma }}{2}} \right)} \right)^2}\frac{{{\sigma ^2}}}{n}\\ &= 4{\left( {T_7^{ - 1}\left( {0.950} \right)} \right)^2}\frac{{{\sigma ^2}}}{8}\\ &= 4 \times {\left( {1.894} \right)^2} \times \frac{{{\sigma ^2}}}{8}\\ &= 1.79{\sigma ^2}\end{align}\)

e.

For \(n = 8\;and\;\gamma = 0.95\),

\(\begin{align}E\left( {{L^2}} \right) &= 4{\left( {T_{n - 1}^{ - 1}\left( {\frac{{1 + \gamma }}{2}} \right)} \right)^2}\frac{{{\sigma ^2}}}{n}\\ &= 4{\left( {T_7^{ - 1}\left( {0.975} \right)} \right)^2}\frac{{{\sigma ^2}}}{8}\\ &= 4 \times {\left( {2.364} \right)^2} \times \frac{{{\sigma ^2}}}{8}\\ &= 2.79{\sigma ^2}\end{align}\)

f.

For \(n = 8\;and\;\gamma = 0.99\),

\(\begin{align}E\left( {{L^2}} \right) &= 4{\left( {T_{n - 1}^{ - 1}\left( {\frac{{1 + \gamma }}{2}} \right)} \right)^2}\frac{{{\sigma ^2}}}{n}\\ &= 4{\left( {T_7^{ - 1}\left( {0.995} \right)} \right)^2}\frac{{{\sigma ^2}}}{8}\\ &= 4 \times {\left( {3.499} \right)^2} \times \frac{{{\sigma ^2}}}{8}\\ &= 6.12{\sigma ^2}\end{align}\)

So, there can be noticed that for a fixed sample size n, if the confidence interval increases the confidence interval became less precise.