Answer:
0.16 <\sigma <0.13
Step-by-step explanation:
given
S=0.28
n=22
d_f=n-1 =22-1 =21
[tex]\chi_{L}^{2}=\chi _{0.995}^{2}[/tex]=64.278
[tex]\chi_{U}^{2}=\chi _{0.005}^{2}[/tex]=96.578
now
80% confidence interval for Standard deviation is given by
[tex]\sqrt{\frac{(n-1)S^{2}}{\chi _{U}^{2}}}<\sigma<\sqrt{\frac{(n-1)S^{2}}{\chi _{L}^{2}}}[/tex]
=[tex]\sqrt{\frac{(22-1)(0.28)^{2}}{64.27}}<\sigma<\sqrt{\frac{(22-1)(0.28)^{2}}{96.578}}[/tex]
=0.16 <\sigma <0.13
Answer: Degree of freedom = [tex]df=21[/tex]
Critical values :
[tex]\chi^2 _{\alpha/2 , df}=29.6151[/tex]
[tex]\chi^2 _{1-\alpha/2 , df}=13.2396[/tex]
Confidence interval : [tex]2358<\sigma<0.3526[/tex]
Step-by-step explanation:
We know that the confidence interval for population standard deviation [tex](\sigma)[/tex] is given by :-
[tex]\sqrt{\dfrac{(n-1)s^2}{\chi^2_{\alpha/2}}}<\sigma<\sqrt{\dfrac{(n-1)s^2}{\chi^2_{1-\alpha/2}}}[/tex]
, where n = Sample size.
s= sample standard deviation.
[tex]\chi^2_{\alpha/2}[/tex] and [tex]\chi^2_{1-\alpha/2}[/tex] = Critical values.
Given : Confidence level : c= 80%=0.80
Then, significance level = [tex]\alpha=1-0.80=0.20[/tex]
Sample size : n= 22
Degree of freedom = [tex]df=n-1=22-1=21[/tex]
Then, by using Chi-square distribution table ,
[tex]\chi^2 _{\alpha/2 , df}=\chi^2_{{0.10,\ 21}=29.6151[/tex]
[tex]\chi^2 _{1-\alpha/2 , df}=\chi^2_{{0.90,\ 21}=13.2396[/tex]
Sample standard deviation is given to be s= 0.28 mg.
Then , the 80% confidence interval estimate of [tex]\sigma[/tex] will be :-
[tex]\sqrt{\dfrac{(21)(0.28)^2}{29.6151}}<\sigma<\sqrt{\dfrac{(21)(0.28)^2}{13.2396}}\\\\=\sqrt{0.05559326}<\sigma<\sqrt{0.12435421}\\\\=0.23578223<\sigma<0.3526389\\\\\approx0.2358<\sigma<0.3526[/tex]
Hence, the required confidence interval : [tex]2358<\sigma<0.3526[/tex]
