91Ó°ÊÓ

It is given that two variables \(\mu \,\,and\,\,\tau \)have the joint normal-gamma distribution such that \({\mu _0} = 183.95,{\lambda _0} = 20,{\alpha _0} = 11,{\beta _0} = 50925.37\)

Let \(\mu \,\,{\rm{and}}\,\,\tau \) be random variables. Suppose that the conditional distribution of \(\mu \,\,{\rm{given}}\,\,\tau \) is the normal distribution with mean \({\mu _0}\) and precision \({\lambda _0}\tau \).

Suppose also that the marginal distribution of \(\,\tau \) is the gamma distribution with parameters \({\alpha _0}\,\,{\rm{and}}\,\,{\beta _0}\).

Then we say that the joint distribution of\(\mu \,\,{\rm{and}}\,\,\tau \) is the normal-gamma distribution with hyperparameters \({\mu _0},{\lambda _0},{\alpha _0},{\beta _0}\).

Let,

\(\begin{align}U &= {\left( {\frac{{{\lambda _0}{\alpha _0}}}{{{\beta _0}}}} \right)^{\frac{1}{2}}}\left( {\mu - {\mu _0}} \right)\\ &= 0.0657\left( {\mu - 183.95} \right)\end{align}\)

Here U follows t distribution with \(2{\alpha _0}\) degrees of freedom, that is 22.

\(\begin{align}P\left( {a < \mu < b} \right) &= 0.90\\ \Rightarrow P\left( {0.0657\left( {a - 183.95} \right) < 0.0657\left( {\mu - 183.95} \right) < 0.0657\left( {b - 183.95} \right)} \right) &= 0.90\\ \Rightarrow P\left( {0.0657\left( {a - 183.95} \right) < U < 0.0657\left( {b - 183.95} \right)} \right) &= 0.90\end{align}\)

The confidence intervals derived for U are -1.717 and +1.717

Therefore, the value of a and b is,

\(\begin{align}0.0657\left( {a - 183.95} \right) &= - 1.717\\ \Rightarrow a &= 157.83\end{align}\)

And,

\(\begin{align}0.0657\left( {b - 183.95} \right) &= 1.717\\ \Rightarrow b &= 210.07\end{align}\)

Therefore, the interval is (157.83, 210.07)

To find this, we need to know the expectation and variance of \(\mu \) .

\(\begin{align}E\left( \mu \right) &= {\mu _0}\\Var\left( \mu \right) &= \frac{{{\beta _0}}}{{{\lambda _0}\left( {{\alpha _0} - 1} \right)}}\end{align}\)

Therefore,

\(\begin{align}E\left( \mu \right) &= 183.95\\Var\left( \mu \right) &= \frac{{50925.37}}{{20\left( {11 - 1} \right)}} &= 254.62\end{align}\)

Hence,

\(\begin{align}P\left( {a < \mu < b} \right) &= 0.90\\ \Rightarrow P\left( {\frac{{a - 183.95}}{{\sqrt {254.62} }} < Z < \frac{{b - 183.95}}{{\sqrt {254.62} }}} \right) &= 0.90\end{align}\)

The value for 0.90 confidence is \({Z_{\frac{\alpha }{2}}} = 1.645\)

Solving the inequality, we get a=152.55, b=211.79

Therefore, the interval is (152.55, 211.79)