Question

In: Statistics and Probability

Q. A random sample of n = 10 had a mean of 1283 and a standard...

Q. A random sample of n = 10 had a mean of 1283 and a standard deviation of 18.31. Find a 95% confidence interval for the population standard deviation.

Critical values for the confidence interval :

Lower limit (provide 2 decimals) :

Upper limit (provide 2 decimals) :

Expert Solution

Solution :

Given that,

s = 18.31

²_R = ²_/2,df = 19.023

²_L = ²_{1 -} _/2,df = 2.700

The 95% confidence interval for is,

(n - 1)s² / ²_/2 < < (n - 1)s² / ²_{1 -} _/2

9 * 18.31 ² / 19.023 < < 9 * 18.31² / 2.700

12.59 < < 33.43

Lower limit = 12.59

Upper limit = 33.43

orchestra answered 2 years ago

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