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 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.

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