Question

find the 90% confidence interval for the variance and standard deviation of a sample of 16 and a standard deviation of 2.1 Assume the variable is normally distributed.

Answer 1

There are different ways you can do this kind of problem, but the formula below might be one of the easier ways:

s/[1 + (1.645/√2n)]
..to..
s/[1 - (1.645/√2n)]

...where s = standard deviation, 1.645 represents the 90% confidence interval using a z-table, and n = sample size.

Substituting your values:
(2.1)/[1 + (1.645/√2*16)]
..to..
(2.1)/[1 - (1.645/√2*16)]


Note: After you finish the above calculations, square the standard deviation values to find the variance values.

I hope this will help.