Use technology to construct the confidence intervals for the population variance sigmaσsquared2 and the population standard deviation sigmaσ. Assume the sample is taken from a normally distributed population.cequals=0.99, sequals=35, nequals=19The confidence interval for the population variance is ​( nothing​, nothing​).

Respuesta :

Answer: (593.44, 3519.67).

Step-by-step explanation:

As per given , we have

Confidence level : [tex]1-\alpha=0.99[/tex]

Significance level : [tex]\alpha=0.01[/tex]

Sample size : n= 19

Sample standard deviation : s= 35

Using chi-square distribution table , the critical values are:

[tex]\chi^2_{\alpha/2, n-1}=\chi^2_{0.005, 18}=37.1565\\\\\chi^2_{1-\alpha/2, n-1}=\chi^2_{0.995, 18}=6.2648[/tex]

Now, the Confidence interval for the population variance is given by :-

[tex]\dfrac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}<\sigma^2<\dfrac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}}\\\\ \dfrac{(18)(35)^2}{37.1565}<\sigma^2< \dfrac{(18)(35)^2}{6.2648}\\\\ 593.44<\sigma^2< 3519.67[/tex]

Hence, the confidence interval for the population variance is (593.44, 3519.67).