Answer
Lower Bound = 7.55, Upper Bound = 102.71
Hence, we see that the width of the interval increases with an increase in the confidence level.
Work Step by Step
Here n = 10, df = n-1 = 9, $s^{2} = 19.8$, Confidence Interval = 95%
α = 1 - 0.99 = 0.01, α/2 = 0.005, 1 - α/2 = 0.995
$χ2_{α/2} = 23.589$, $χ2_{1-α/2} = 1.735$
$\frac{(n-1)s^{2}}{χ2_{α/2}} < σ^{2} < \frac{(n-1)s^{2}}{χ2_{1-α/2}}$
$\frac{9 \times 19.8}{23.589} < σ^{2} < \frac{9 \times 19.8}{1.735}$
$ 7.55 < σ^{2} < 102.71$
