Question

Find the critical values X2L and X2R that correspond to the given confidence level and sample size: 99%; n = 28 This involves using chi-square, critical vaules and area. I will need the full answer and the answer must be correct for my quiz for full points to be awared please and thank you in advance. I am happy to provide additional information if this isn't clear. :)

My numbers just need to be changed to n=18 and 98%

Answer 1

Concepts and reason

The critical value is derived from the significance level, $\left( \alpha \right)$ and the probability distribution of a test statistic. That is, $z,t,F,{\rm{ or chi - square}}{\rm{.}}$

The critical value helps to make a decision about the statistical test. The critical value decides whether to accept the null hypothesis or to reject the null hypothesis at the specified significant level $\alpha .$

Fundamentals

The Excel formula for finding the left and right tailed critical values is,

$\begin{array}{c}\\\chi _L^2 = \left( { = CHIINV\left( {\frac{\alpha }{2},df} \right)} \right)\\\\\chi _R^2 = \left( { = CHIINV\left( {1 - \frac{\alpha }{2},df} \right)} \right)\\\end{array}$

The formula for level of significance is,

${\rm{Level of significance}}\,{\rm{ = 1 - confidence level}}$

The formula for degree of freedom is,

${\rm{Degree of freedom }} = n - 1$

The provided information is,

$\begin{array}{l}\\n = 28,\\\\{\rm{confidence level = 0}}{\rm{.99}}\\\end{array}$

The level of significance $\left( \alpha \right)$ is,

$\begin{array}{c}\\{\rm{Level of significance}}\,{\rm{ = 1 - confidence level}}\\\\ = 1 - 0.99\\\\ = 0.01\\\end{array}$

The degrees of freedom is,

$\begin{array}{c}\\df = n - 1\\\\ = 28 - 1\\\\ = 27\\\end{array}$

The left tailed critical value is,

$\begin{array}{c}\\\chi _L^2 = \left( { = CHIINV\left( {\frac{\alpha }{2},df} \right)} \right)\\\\ = \left( { = CHIINV\left( {\frac{{0.01}}{2},27} \right)} \right)\\\\ = 49.645\\\end{array}$

The right tailed critical value is,

$\begin{array}{c}\\\chi _R^2 = \left( { = CHIINV\left( {1 - \frac{\alpha }{2},df} \right)} \right)\\\\ = \left( { = CHIINV\left( {1 - \left( {\frac{{0.01}}{2}} \right),27} \right)} \right)\\\\ = 11.808\\\end{array}$

The provided information is,

$\begin{array}{l}\\n = 18,\\\\{\rm{confidence level = 0}}{\rm{.98}}\\\end{array}$

The level of significance $\left( \alpha \right)$ is,

$\begin{array}{c}\\{\rm{Level of significance}}\,{\rm{ = 1 - confidence level}}\\\\ = 1 - 0.98\\\\ = 0.02\\\end{array}$

The degrees of freedom is,

$\begin{array}{c}\\df = n - 1\\\\ = 18 - 1\\\\ = 17\\\end{array}$

The left tailed critical value is,

$\begin{array}{c}\\\chi _L^2 = \left( { = CHIINV\left( {\frac{\alpha }{2},df} \right)} \right)\\\\ = \left( { = CHIINV\left( {\frac{{0.02}}{2},17} \right)} \right)\\\\ = 33.409\\\end{array}$

The right tailed critical value is,

$\begin{array}{c}\\\chi _R^2 = \left( { = CHIINV\left( {1 - \frac{\alpha }{2},df} \right)} \right)\\\\ = \left( { = CHIINV\left( {1 - \left( {\frac{{0.02}}{2}} \right),17} \right)} \right)\\\\ = 6.408\\\end{array}$

Ans:

The left and right tailed critical​ values are ${\bf{\chi }}_{\bf{L}}^{\bf{2}}{\bf{ = 49}}{\bf{.645 and \chi }}_{\bf{R}}^{\bf{2}}{\bf{ = 11}}{\bf{.808}}{\bf{.}}$

The left and right tailed critical values are ${\bf{\chi }}_{\bf{L}}^{\bf{2}}{\bf{ = 33}}{\bf{.409 and \chi }}_{\bf{R}}^{\bf{2}}{\bf{ = 6}}{\bf{.408}}{\bf{.}}$