Your question is difficult to read but I assume that the sample variance s2 is given as 12.
For this problem of finding the confidence interval for σ2 we need the ꭓ2 distribution table. The following table extract applies for a 90% CI:
| degrees of freedom (n-1) |
α=0.05 |
α=0.95 |
| 19 |
30.144 |
10.117 |
| 29 |
42.557 |
17.708 |
Note that the significance level for 90% is two-tailed so we need to halve the 10% to 5% for each tail to give us a 90% width for the interval.
We also need to calculate (n-1)s2 for the two sample sizes, n=20 and n=30.
Respectively this is 19×12=228 and 29×12=348.
CI's are [228/30.144, 228/10.117] and [348/42.557,348/17.708]=
n=20: [7.564,22.536]
n=30: [8.177,19.652]