Answer
Lower Bound = 6.61, Upper Bound = 31.36
We note that width of the interval increases by an increase in the confidence level.
Work Step by Step
Here n = 20, df = n-1 = 19, $s^{2} = 12$, Confidence Interval = 98%
α = 1 - 0.98 = 0.02, α/2 = 0.01, 1 - α/2 = 0.99
$χ2_{α/2} = 36.191$, $χ2_{1-α/2} = 7.633$
$\frac{(n-1)s^{2}}{χ2_{α/2}} < σ^{2} < \frac{(n-1)s^{2}}{χ2_{1-α/2}}$
$\frac{19 \times 12.6}{36.191} < σ^{2} < \frac{19 \times 12.6}{7.633}$
$ 6.61 < σ^{2} < 31.36$
