Answer:
a) 0.2218 to 0.7057
b) 0.4710 to 0.8401
Step-by-step explanation:
Given:
sample = n = 18
sample variance = s² = 0.36
To find:
a) 90% confidence interval estimate of the population variance.
b) 90% confidence interval estimate of the population standard deviation.
a)
Compute degree of freedom
degree of freedom = df = n - 1 = 18 - 1 = 17
Compute value of α for a 90% confidence interval
The confidence level for 90% is:
c = 0.90
So
α = 1 - c = 1 - 0.90 = 0.1
α = 0.1
Now use the table of critical values of the chi-square distribution in order to find critical values. Search for the 17th row of table using df = 17 and find the column corresponding to 1 - α/2 i.e. 0.95 of table for upper tail critical values and column corresponding to α /2 i.e. 0.05 of table for lower tail critical values.
Using the table:
[tex]X^{2}_{0.95}[/tex] = 8.672
[tex]X^{2}_{0.05}[/tex] = 27.587
90% Confidence interval estimate of the population variance
The boundaries of CI are computed using formula:
(n−1) s² / [tex]X^{2}_{\alpha/2}[/tex] ≤ σ² ≤ (n−1) s² / [tex]X^{2}_{1-\alpha/2}[/tex]
(n−1) s² / [tex]X^{2}_{\alpha/2}[/tex] = (18-1) 0.36 / 27.587
= (17) 0.36 / 27.587
= 6.12 / 27.587
(n−1) s² / [tex]X^{2}_{\alpha/2}[/tex] = 0.2218
(n−1) s² / [tex]X^{2}_{1-\alpha/2}[/tex] = (18-1) 0.36 / 8.672
= (17) 0.36 / 8.672
= 6.12 / 8.672
= 0.7057
This results in inequalities 0.2218 ≤ σ² ≤ 0.7057 for the variance
b)
90% Confidence interval estimate of the population standard deviation
The boundaries of CI are computed using formula:
√(n−1) s² / [tex]X^{2}_{\alpha/2}[/tex] ≤ σ ≤ √(n−1) s² / [tex]X^{2}_{1-\alpha/2}[/tex]
√(n−1) s² / [tex]X^{2}_{\alpha/2}[/tex] = √((18-1) 0.36 / 27.587)
= √((17) 0.36 / 27.587)
= √(6.12 / 27.587)
= √0.2218
= 0.4709
= 0.4710
√(n−1) s² / [tex]X^{2}_{1-\alpha/2}[/tex] = √((18-1) 0.36 / 8.672)
= √((17) 0.36 / 8.672)
= √ (6.12 / 8.672)
= √0.7057
= 0.8401
0.4710 ≤ σ ≤ 0.8401 for the standard deviation.
