Answer: . a. 36.08 to 78.80
Step-by-step explanation:
The confidence interval for population standard deviation is given by :-
[tex]\sqrt{\dfrac{(n-1)s^2}{\chi_{\alpha/2}}}<\sigma\sqrt{\dfrac{(n-1)s^2}{\chi_{1-\alpha/2}}}[/tex]
Given : Sample size = [tex]n=24[/tex]
Significance level : [tex]\alpha:1-0.99=0.01[/tex]
Using the chi-square distribution table , the critical values would be :-
[tex]\chi_{n-1,\alpha/2}=\chi_{23, 0.005}=44.18127525[/tex]
[tex]\chi_{n-1,1-\alpha/2}=\chi_{23, 0.995}=9.26042478[/tex]
Then , a 99% confidence interval for the population standard deviation will be :-
[tex]\sqrt{\dfrac{(23)(50)^2}{44.18127525}}<\sigma\sqrt{\dfrac{(23)(50)^2}{9.26042478}}\\\\=36.075702658<\sigma<78.7985939401\\\\\approx36.08<\sigma<78.80[/tex]