Question

find the critical values, $x_{r}^{2}$ and $x_{l}^{2}$, for $c = 0.95$ and $n = 12$.

a. 4.575 and 26.757

b. 2.603 and 19.675

c. 3.816 and 21.920

d. 3.053 and 24.725

Explanation:

Step1: Calculate the degrees of freedom

The degrees of freedom \( df=n - 1\). Given \(n = 12\), then \(df=12-1 = 11\).

Step2: Calculate the right - tail and left - tail probabilities

The confidence level \(c = 0.95\), so the significance level \(\alpha=1 - c=1 - 0.95 = 0.05\).
The right - tail probability \(P(\chi^{2}>\chi_{R}^{2})=\frac{\alpha}{2}=0.025\), and the left - tail probability \(P(\chi^{2}<\chi_{L}^{2})=\frac{\alpha}{2}=0.025\), \(P(\chi^{2}>\chi_{L}^{2})=1-\frac{\alpha}{2}=0.975\).

Step3: Look up the critical values in the \(\chi^{2}\) - distribution table

Looking up the \(\chi^{2}\) - distribution table with \(df = 11\):
For the right - tail critical value \(\chi_{R}^{2}\) (where the area to the right is \(0.025\)), \(\chi_{R}^{2}=21.920\).
For the left - tail critical value \(\chi_{L}^{2}\) (where the area to the right is \(0.975\)), \(\chi_{L}^{2}=3.816\).

Answer:

C. \(3.816\) and \(21.920\)