Posted by Anon on Sunday, October 24, 2010 at 10:56pm.
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.

Statistics  MathGuru, Monday, October 25, 2010 at 7:30pm
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 ztable, 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.
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