Question

Determine the following:

a. The 95th percentile of the chi-squared distribution with \({\rm{v = 10}}\)

b. The 5th percentile of the chi-squared distribution with\({\rm{v = 10}}\)

\(\begin{array}{l}{\rm{c}}{\rm{.P}}\left( {{\rm{10}}{\rm{.98£ }}{{\rm{\chi }}^{\rm{2}}}{\rm{£ 36}}{\rm{.78}}} \right){\rm{,where }}{{\rm{\chi }}^{\rm{2}}}{\rm{ is achi - squared rv with \nu = 22}}\\{\rm{d}}{\rm{.P}}\left( {{{\rm{\chi }}^{\rm{2}}}{\rm{ < 14}}{\rm{.611}}} \right.{\rm{ or }}\left. {{{\rm{\chi }}^{\rm{2}}}{\rm{ > 37}}{\rm{.652}}} \right){\rm{,where }}{{\rm{\chi }}^{\rm{2}}}{\rm{ is achi squared rv with \nu = 25}}\end{array}\)

Short answer

The values are

\(\begin{array}{l}{\rm{a}}{\rm{. 18}}{\rm{.307; }}\\{\rm{b}}{\rm{. 3}}{\rm{.94; }}\\{\rm{c}}{\rm{. P}}\left( {{\rm{\chi }}_{{\rm{0}}{\rm{.975,22}}}^{\rm{2}}{\rm{£}}{{\rm{\chi }}^{\rm{2}}}{\rm{£\chi }}_{{\rm{0}}{\rm{.025,22}}}^{\rm{2}}} \right){\rm{ = 0}}{\rm{.95; }}\\{\rm{d}}{\rm{. P}}\left( {{\rm{\chi }}_{{\rm{0}}{\rm{.95,25}}}^{\rm{2}}{\rm{ > }}{{\rm{\chi }}^{\rm{2}}}{\rm{ or \chi }}_{{\rm{0}}{\rm{.5,25}}}^{\rm{2}}{\rm{ < }}{{\rm{\chi }}^{\rm{2}}}} \right){\rm{ = 0}}{\rm{.1}}\end{array}\)

Step-by-step solution

To determine

(a):

The \({\rm{9}}{{\rm{5}}^{{\rm{th}}}}\)percentile of the chi-squared distribution with \({\rm{\nu = 10}}\)degrees of freedom is the value for which the area to the left of it is \(0.95\). This indicates that \({\rm{\alpha = 1 - 0}}{\rm{.95 = 0}}{\rm{.05}}\)and the critical value is

\({\rm{\chi }}_{{\rm{0}}{\rm{.05,10}}}^{\rm{2}}{\rm{ = 18}}{\rm{.307,}}\)

which is the \({\rm{9}}{{\rm{5}}^{{\rm{th}}}}\)percentile. See the following picture where the area to the left of the value is \(0.95\).

(b):

The \({{\rm{5}}^{{\rm{th}}}}\)percentile of the chi-squared distribution with \({\rm{\nu = 10}}\)degrees of freedom is the value for which the area to the left of it is \(0.05\)This indicates that \({\rm{\alpha = 0}}{\rm{.05}}\)and the critical value is

\({\rm{\chi }}_{{\rm{1 - 0}}{\rm{.05,10}}}^{\rm{2}}{\rm{ = \chi }}_{{\rm{95,10}}}^{\rm{2}}{\rm{ = 3}}{\rm{.94}}\)

which is the \({{\rm{5}}^{{\rm{th}}}}\)percentile. See the following picture where the area to the left of the value is \(0.05\)

\({\rm{\chi }}_{{\rm{1 - \alpha ,22}}}^{\rm{2}}{\rm{ = 10}}{\rm{.987}}\)

To determine

(c):

one should at the table in the appendix (this can also be computed with a software). When the value \(10.987\)is found in the table, see that for \({\rm{\alpha = 0}}{\rm{.975}}\)the value is produced. Similarly, from

\({\rm{\chi }}_{{\rm{\alpha ,22}}}^{\rm{2}}{\rm{ = 36}}{\rm{.78}}\)

you get \({\rm{\alpha = 0}}{\rm{.025}}\).This means that the area to the left of \({\rm{10}}{\rm{.987 is 0}}{\rm{.025}}\), and the area to the right of \({\rm{36}}{\rm{.78 is 0}}{\rm{.025}}\). This means that the area between those values, and the probability are

\({\rm{P}}\left( {{\rm{\chi }}_{{\rm{0}}{\rm{.975,22}}}^{\rm{2}}{\rm{£}}{{\rm{\chi }}^{\rm{2}}}{\rm{£\chi }}_{{\rm{0}}{\rm{.025,22}}}^{\rm{2}}} \right){\rm{ = 1 - 0}}{\rm{.025 - 0}}{\rm{.025 = 0}}{\rm{.95}}\)

(d):

The mentioned probability can be computed as the sum of the area which is to the left of value \(14.611\)and the area to the right of \(37.652\)of chi squared distribution with \({\rm{\nu = 25}}\) degrees of freedom. In order to find \({\rm{\alpha }}\)for which

\({\rm{\chi }}_{{\rm{1 - \alpha ,25}}}^{\rm{2}}{\rm{ = 14}}{\rm{.611}}\)

one should at the table in the appendix (this can also be computed with a software). When the value $10.987$ is found in the table, see that for \({\rm{\alpha = 0}}{\rm{.95}}\) the value is produced. Similarly, from

\({\rm{\chi }}_{{\rm{\alpha ,25}}}^{\rm{2}}{\rm{ = 37}}{\rm{.652}}\)

you get \({\rm{\alpha = 0}}{\rm{.05}}\). This means that the area to the left of \({\rm{14}}{\rm{.611 is 0}}{\rm{.05}}\), and the area to the right of \({\rm{37}}{\rm{.652 is 0}}{\rm{.05}}\). This means that the mentioned sum (the probability) is

\({\rm{P}}\left( {{\rm{\chi }}_{{\rm{0}}{\rm{.95,25}}}^{\rm{2}}{\rm{ > }}{{\rm{\chi }}^{\rm{2}}}{\rm{ or \chi }}_{{\rm{0}}{\rm{.5,25}}}^{\rm{2}}{\rm{ < }}{{\rm{\chi }}^{\rm{2}}}} \right){\rm{ = 0}}{\rm{.05 + 0}}{\rm{.05 = 0}}{\rm{.1}}{\rm{.}}\)

The shaded area on the picture is \(0.9\).

Hence

\(\begin{array}{l}{\rm{ a}}{\rm{. 18}}{\rm{.307; }}\\{\rm{b}}{\rm{. 3}}{\rm{.94; }}\\{\rm{c}}{\rm{. P}}\left( {{\rm{\chi }}_{{\rm{0}}{\rm{.975,22}}}^{\rm{2}}{\rm{£}}{{\rm{\chi }}^{\rm{2}}}{\rm{£\chi }}_{{\rm{0}}{\rm{.025,22}}}^{\rm{2}}} \right){\rm{ = 0}}{\rm{.95; }}\\{\rm{d}}{\rm{. P}}\left( {{\rm{\chi }}_{{\rm{0}}{\rm{.95,25}}}^{\rm{2}}{\rm{ > }}{{\rm{\chi }}^{\rm{2}}}{\rm{ or \chi }}_{{\rm{0}}{\rm{.5,25}}}^{\rm{2}}{\rm{ < }}{{\rm{\chi }}^{\rm{2}}}} \right){\rm{ = 0}}{\rm{.1}}\end{array}\)