Question

(21) A sample of size 20 is drawn from a normal distribution with unknown variance and mean. The sample variance s2 = 0.012.

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Answer #1

Solution :

Given that,

s2 = 0.012

n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

At 95% confidence level the \chi 2 value is ,

\alpha = 1 - 95% = 1 - 0.95 = 0.05

\alpha / 2 = 0.05 / 2 = 0.025

1 - \alpha / 2 = 1 - 0.025 = 0.975

\chi2L = \chi 2\alpha/2,df = 32.853

\chi2R = \chi 21 - \alpha /2,df = 8.907

The 95% confidence interval for \sigma is,

\sqrt(n - 1)s2 / \chi 2\alpha/2 < \sigma < \sqrt (n - 1)s2 / \chi 21 - \alpha /2

\sqrt(19)(0.012) / 32.853 < \sigma < \sqrt (19)(0.012) / 8.907

0.0833 < \sigma < 0.1600

(0.0833 , 0.1600)

option A. is correct

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