Answer: [tex]\approx1.49<\sigma<3.97[/tex]
Step-by-step explanation:
We know that the confidence interval for population standard deviation 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]= Chi-square critical value for degree of freedom (n-1) and significance level ([tex]\alpha[/tex]).
Given : [tex]\alpha=1-0.99=0.01[/tex]
n= 16
Critical ch-square values for degree of freedom 15 and [tex]\alpha=0.01[/tex]will be :
[tex]\chi_{\alpha/2}=\chi_{0.005}=32.80132[/tex]
[tex]\chi_{1-\alpha/2}=\chi_{0.995}=4.6009[/tex]
Then , the required 99% confidence interval for the population standard deviation will be :
[tex]\sqrt{\dfrac{(15)(2.2)^2}{32.80132}}<\sigma<\sqrt{\dfrac{(15)(2.2)^2}{4.6009}}[/tex]
[tex]\sqrt{2.21332556129}<\sigma<\sqrt{15.779521398}[/tex]
[tex]1.48772496157<\sigma<3.97234457191[/tex]
[tex]\approx1.49<\sigma<3.97[/tex]
Hence, the a 99% confidence interval for the population standard deviation :
