Answer: A 95 % confidence interval for [tex]\sigma^2[/tex] :-
[tex]2.90<\sigma^2<67.21[/tex]
A 95 % confidence interval for [tex]\sigma[/tex] :-
[tex]1.70<\sigma<8.20[/tex]
Step-by-step explanation:
Given : Sample size : [tex]n=5[/tex]
Standard deviation : [tex]\sigma : 2.84[/tex]
Significance level : [tex]\alpha=1-0.95=0.05[/tex]
We assume this is normal distribution
By using chi-square distribution, the critical value will be :-
[tex]\chi^2_{n-1,\alpha/2}=\chi^2_{4,0.025}=11.14[/tex]
[tex]\chi^2_{n-1,1-\alpha/2}=\chi^2_{4,0.975}=0.48[/tex]
The confidence interval for population variance:-
[tex]\dfrac{(n-1)s^2}{\chi^2_{n-1\alpha/2}}<\sigma^2<\dfrac{(n-1)s^2}{\chi^2_{n-1, 1-\alpha/2}}[/tex]
[tex]=\dfrac{(2.84)^2(4)}{11.14}<\sigma^2<\dfrac{(2.84)^2(4)}{0.48}\\\\\approx2.90<\sigma^2<67.21[/tex]
For standard deviation just take square root in the above inequality,
[tex]\sqrt{2.90}<\sigma^2<\sqrt{67.21}\\\\\approx1.70<\sigma<8.20[/tex]
