Question

part 3 of 5
(c) when ( alpha = 0.02 ) and ( n = 25 ),
( chi _ { \text { left } } ^ { 2 } = 10.856 )
( chi _ { \text { right } } ^ { 2 } = 42.980 )
part 4 of 5
(d) when ( alpha = 0.10 ) and ( n = 28 ),
( chi _ { \text { left } } ^ { 2 } = 16.151 )
( chi _ { \text { right } } ^ { 2 } = 40.113 )
part: ( 4 / 5 )
part 5 of 5
(e) when ( alpha = 0.02 ) and ( n = 11 ),
( chi _ { \text { left } } ^ { 2 } = )
( chi _ { \text { right } } ^ { 2 } = )

Explanation:

Step1: Determine the degrees of freedom

The degrees of freedom \(df=n - 1\). Given \(n = 11\), then \(df=11-1 = 10\).

Step2: Find \(\chi_{left}^{2}\)

For a two - tailed test with \(\alpha=0.02\), the left - tail area is \(\frac{\alpha}{2}=0.01\). Looking up the \(\chi^{2}\) distribution table with \(df = 10\), \(\chi_{left}^{2}=\chi_{0.01,10}^{2}=2.558\).

Step3: Find \(\chi_{right}^{2}\)

The right - tail area is \(1-\frac{\alpha}{2}=0.99\). Looking up the \(\chi^{2}\) distribution table with \(df = 10\), \(\chi_{right}^{2}=\chi_{0.99,10}^{2}=23.209\).

Answer:

\(\chi_{left}^{2}=2.558\), \(\chi_{right}^{2}=23.209\)