Answer:
Given,
sample n = 6
Standard deviation = 1.3
alpha = 0.10
degree of freedom = n - 1 = 6 - 1 = 5
(1-alpha/2
, df) = 1.14547623
(alpha/2
, df) = 11.07049769
Consider,
(n-1)s^2/
(alpha/2
, df) <
< (n-1)s^2/
(1
- alpha/2 , df)
(6-1)*1.3^2 / 11.0705 <
< (6-1)*1.3^2 / 1.1455
0.7633 <
< 7.3767
0.8737 <
< 2.7160
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