The annual earnings of 12 randomly selected computer software engineers have a sample standard deviation of
$ 3626 Assume the sample is from a normally distributed population. Construct a confidence interval for the population variance
sigma squared σ2 and the population standard deviation sigma σ. Use a 95%
level of confidence. Interpret the results.
What is the confidence interval for the population variance
sigma squared σ2?
What is the confidence interval for the population standard deviation
sigma σ?
Please show all steps
a)
Chi-square critical values at 0.05 level with 11 df = L = 3.816 , U = 21.920
95% confidence interval for is
(n-1) S2 / U < < (n-1) S2 / L
11 * 36262 / 21.920 < < 11 * 36262 / 3.816
6597930 < < 37900062
We are 95% confident that population variance is between 6597930 and 37900062
b)
95% confidence interval for is
Sqrt [ (n-1) S2 / U ] < < sqrt [ (n-1) S2 / L ]
sqrt [11 * 36262 / 21.920 ] < < sqrt [11 * 36262 / 3.816]
2569 < < 6156
We are 95% confident that population variance is between 2569 and 6156
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