Answer: [tex]1.97<\sigma<4.67[/tex]
Step-by-step explanation:
The confidence interval for population standard deviation is given by :-
[tex]s\sqrt{\dfrac{(n-1)}{\chi^2_{n-1,\alpha/2}}}<\sigma<s\sqrt{\dfrac{(n-1)}{\chi^2_{n-1,1-\alpha/2}}}[/tex]
Given : Sample size : [tex]n=20[/tex]
Standard deviation : [tex]s=2.8\text{ years}[/tex]
Significance level : [tex]\alpha: 1-0.99=0.01[/tex]
Critical value using chi-square distribution table :
[tex]\chi^2_{n-1,\alpha/2}=\chi^2_{19,0.005}=38.58[/tex]
[tex]\chi^2_{n-1,1-\alpha/2}=\chi^2_{19,0.995}=6.84[/tex]
Then , 99% confidence interval for the standard deviation of the replacement times of all washing machines of this type will be :
[tex]2.8\sqrt{\dfrac{(19)}{38.58}}<\sigma<2.8\sqrt{\dfrac{(19)}{6.84}}\\\\\\=1.9649600278<\sigma<4.66666666667\\\\\approx1.97<\sigma<4.67[/tex]
