Answer:
The 98% CI for the variance is
[tex]2.59\leq\sigma^2\leq 5.86[/tex]
Step-by-step explanation:
We have a sample of size n=67.
The sample variance is 3.75.
We want to construct a 98% confidence interval.
The degrees of freedom are N-1=66.
We can write the 98% confidence interval as:
[tex]\frac{(n-1)s^2}{\chi^2_{0.01}} \leq\sigma^2\leq \frac{(n-1)s^2}{\chi^2_{0.99}}[/tex]
We calculate (n-1)s^2 as:
[tex](n-1)s^2=66*3.75=247.5[/tex]
The values for the chi-square for both extremes of the 98% CI are:
[tex]\chi^2_{0.01}=95.63\\\\\chi^2_{0.99}=42.24[/tex]
Then, we replace the values in the CI
[tex]\frac{(n-1)s^2}{\chi^2_{0.01}} \leq\sigma^2\leq \frac{(n-1)s^2}{\chi^2_{0.99}}\\\\\frac{247.5}{95.63} \leq\sigma^2\leq \frac{247.5}{42.24}\\\\2.59\leq\sigma^2\leq 5.86[/tex]
