Answer: [tex]0.20<\sigma^2<0.48[/tex]
Step-by-step explanation:
Given : A A sample of 73 packages of blueberries has a variance of 0.30.
i.e. n= 73 and [tex]s^2=0.30[/tex]
Confidence level: [tex]\alpha=0.99[/tex]
⇒Significance level : [tex]\alpha=1-0.99=0.01[/tex]
Critical values using chi-square distribution:-
[tex]\chi^2_{\alpha/2, n-1}=\chi^2_{0.005, 72}=106.65[/tex]
[tex]\chi^2_{1-\alpha/2, n-1}=\chi^2_{0.995, 72}=44.84[/tex]
99% confidence interval for population variance will be :-
[tex]\dfrac{s^2(n-1)}{\chi^2_{\alpha/2, n-1}}<\sigma^2<\dfrac{s^2(n-1)}{\chi^2_{1-\alpha/2, n-1}}\\\\\\\dfrac{(0.30)(72)}{106.65}<\sigma^2<\dfrac{(0.30)(72)}{44.84}\\\\ \approx0.20<\sigma^2<0.48[/tex]
Hence, the 99% confidence interval to estimate the variance of the weights of the packages prepared by the machine: [tex]0.20<\sigma^2<0.48[/tex]
