Solution :
Given that,
s2 = 0.012
n = 20
Degrees of freedom = df = n - 1 = 20 - 1 = 19
At 95% confidence level the 2 value is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
1 - / 2 = 1 - 0.025 = 0.975
2L = 2/2,df = 32.853
2R = 21 - /2,df = 8.907
The 95% confidence interval for is,
(n - 1)s2 / 2/2 < < (n - 1)s2 / 21 - /2
(19)(0.012) / 32.853 < < (19)(0.012) / 8.907
0.0833 < < 0.1600
(0.0833 , 0.1600)
option A. is correct
(21) A sample of size 20 is drawn from a normal distribution with unknown variance and...
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