Find the critical values $\chi^{2} L$ and $\chi^{2}{ }_{R}$ for the given confidence level $c$ and sample size $n$.
\[
c=0.95, n=26
\]
$\chi_{L}^{2}=13.12$ (Round to three decimal places as needed.)
$\chi_{R}^{2}=\square$ (Round to three decimal places as needed.)

Step 1 ：Given the confidence level \(c=0.95\) and sample size \(n=26\), we need to find the critical values \(\chi^{2}_{L}\) and \(\chi^{2}_{R}\).

Step 2 ：The lower critical value \(\chi_{L}^{2}\) corresponds to the value of the random variable such that the area to its left under the chi-square distribution curve is \((1-c)/2\).

Step 3 ：The upper critical value \(\chi_{R}^{2}\) corresponds to the value of the random variable such that the area to its left under the chi-square distribution curve is \((1+c)/2\).

Step 4 ：Here, \(c\) is the confidence level and \(n\) is the degrees of freedom (sample size - 1).

Step 5 ：We are given \(c=0.95\) and \(n=26\), so we need to find \(\chi_{R}^{2}\).

Step 6 ：The degrees of freedom is \(df = n - 1 = 25\).

Step 7 ：Using the chi-square percent point function (ppf), which is the inverse of the cumulative distribution function, we find that \(\chi_{R}^{2} = 40.6464691202752\).

Step 8 ：Rounding to three decimal places, we get \(\chi_{R}^{2} = 40.646\).

Step 9 ：Final Answer: The upper critical value \(\chi_{R}^{2}\) for the given confidence level \(c=0.95\) and sample size \(n=26\) is \(\boxed{40.646}\).