posted by Anon .
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.
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)]
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)]
(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.