91Ó°ÊÓ

(a):

As mentioned in the example, random variable

\({\rm{2\lambda }}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{X}}_{\rm{i}}}} \)

has chi square distribution with degrees of freedom \({\rm{2n}}{\rm{. Since n = 15,}}\)

\({\rm{2\lambda }}\sum\limits_{{\rm{i = 1}}}^{{\rm{15}}} {{{\rm{X}}_{{\rm{15}}}}} \)has chi square distribution with degrees of freedom \({\rm{v = 2 \times 15 = 30}}\)

From the table in the appendix for \(30\) degrees of freedom of chi square distribution, for the area of \(0.025\)lower and upper tails, where the area between is in total \(1\)\(0.05 = 0.95\)are

\({\rm{16}}{\rm{.791 and 46}}{\rm{.976}}\)

The blue area on the graph is \(0.95\). Upper and lower tail areas are \(0.025\)as mentioned.

From the mentioned example, the confidence interval for parameter \({\rm{\mu = 1/\lambda }}\)

\(\left( {\frac{{{\rm{2}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} }}{{{\rm{UB}}}}{\rm{,}}\frac{{{\rm{2}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} }}{{{\rm{LT}}}}} \right)\)

where \({\rm{U B = 46}}{\rm{.976 and L T = 16}}{\rm{.791}}\). Therefore, the \(95\% \)confidence interval for expected values \({\rm{\mu = 1/\lambda }}\)is

\(\begin{array}{l}\left( {\frac{{{\rm{2}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} }}{{{\rm{UB}}}}{\rm{,}}\frac{{{\rm{2}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} }}{{{\rm{LT}}}}} \right){\rm{ = }}\left( {\frac{{{\rm{2 \times 63}}{\rm{.2}}}}{{{\rm{46}}{\rm{.979}}}}{\rm{,}}\frac{{{\rm{2 \times 63}}{\rm{.2}}}}{{{\rm{16}}{\rm{.791}}}}} \right)\\{\rm{ = (2}}{\rm{.69,7}}{\rm{.53)}}\end{array}\)

where the sum is calculated by summing all observations on lifetime of heat pumps.

Hence the CI for expected (true average) lifetime is \(\left( {2.69,7.53} \right).\)

(b):

The only difference would be in the area of the tails, the critical values that capture the area \({\rm{100(1 - \alpha )/2 = 0}}{\rm{.05}}\). Those values can be found in the appendix of the bo or calculated by a software. The lower bound is \(13.787\)and the upper bound is \(53.672\)Just by substituting those two values in

\(\left( {\frac{{{\rm{2}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} }}{{{\rm{UB}}}}{\rm{,}}\frac{{{\rm{2}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} }}{{{\rm{LT}}}}} \right)\)

for UB and LB, the \(99\% \)confidence interval would be obtained. See the following graph.

Hence altered to achieve a confidence level of \(99\% \)by Change critical values

(c):

The variance of an exponential distributed random variable X is

\({\rm{V(X) = }}\frac{{\rm{1}}}{{{{\rm{\lambda }}^{\rm{2}}}}}\)

which means that the standard deviation is

\({{\rm{\sigma }}_{\rm{X}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{\lambda }}}{\rm{ = \mu }}\)

Therefore, the confidence interval for the standard deviation would be identical to the confidence interval of the mean. Which means that \(95\% \)confidence interval for \({{\rm{\sigma }}_{\rm{X}}}\)is

\(\begin{array}{l}\left( {\frac{{{\rm{2}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} }}{{{\rm{UB}}}}{\rm{,}}\frac{{{\rm{2}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} }}{{{\rm{LT}}}}} \right){\rm{ = }}\left( {\frac{{{\rm{2 \times 63}}{\rm{.2}}}}{{{\rm{46}}{\rm{.979}}}}{\rm{,}}\frac{{{\rm{2 \times 63}}{\rm{.2}}}}{{{\rm{16}}{\rm{.791}}}}} \right)\\{\rm{ = (2}}{\rm{.69,7}}{\rm{.53)}}\end{array}\)

Hence \(95\% \)CI for the standard deviation is\(\left( {2.69,7.53} \right).\)