Question

A random sample of size *17* is taken from a normally
distributed population, and a sample variance of *23* is
calculated.

If we are interested in creating a 95% confidence interval for
*σ*2, the population variance, then:

a) What is the appropriate degrees of freedom for the
*χ*2distribution?

b) What are the appropriate *χ*2*R*and
*χ*2*L* values, the right and left Chi-square
values?

Round your responses to at least 3 decimal places.

*χ*2*R*=

*χ*2*L*=

Answer #1

Solution :

Given that,

Point estimate = s^{2} = 23

n = 17

(a)

Degrees of freedom = df = n - 1 = 17 - 1 = 16

(b)

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

^{2}_{L}
=
^{2}_{/2,df}
= 28.845

^{2}_{R}
=
^{2}_{1 -} _{
/2,df} = 6.908

Suppose a random sample of size 11 was taken from a normally
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at...

A random sample of size 15 taken from a normally
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aa
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