Answer
n = 6, s = 4.1 deaths, df = 6-1=5, α = 1-0.99 = 0.01
To find χ2 right ,
α/2=0.005
From the table,
χ2 right = 16.750
To find χ2 left,
1-0.05=0.995
From the table,
χ2 left =0.412
The Confidence Interval for a Variance:
$ \frac{(n-1)s^2}{χ2 right}$ < σ^2 < $ \frac{(n-1)s^2}{ χ2 left}$
$ \frac{5*4.1^2}{16.750}$ < σ^2 < $ \frac{5*4.1^2}{0.412}$
= 5.0179 < σ^2 < 204.0049
The Confidence Interval for a Standard Deviation:
$\sqrt 5.0179 $ < σ< $\sqrt 204.0049$
2.24 < σ < 14.283
Hence, we can be 99% confident that the true population variance for carbon monoxide deaths is between 5.0179 to 204.0049, while the true population standard deviation is between 2.24 to 14.283 based on a sample of 6 number of years.
