Question

The annual earnings of 14 randomly selected computer software engineers have a sample standard deviation of $3630. Assume the sample is from a normally distributed population. Construct a confidence interval for the population variance σ²   the population standard deviation σ. Use a 99% level of confidence. Interpret the results.

The waiting times​ (in minutes) of a random sample of 22 people at a bank have a sample standard deviation of 3.3 minutes. Construct of confidence interval for the population variance σ² and the population standard deviation σ. Use a 90% level of confidence. Assume the sample is from a normally distribution population.

What is the confidence interval for the population variance σ^2?

Answer 1

1)

Here s = 3630 and n = 14
df = 14 - 1 = 13

α = 1 - 0.99 = 0.01
The critical values for α = 0.01 and df = 13 are Χ^2(1-α/2,n-1) = 3.565 and Χ^2(α/2,n-1) = 29.819

CI = (13*3630^2/29.819 , 13*3630^2/3.565)
CI = (5744649.38 , 48050406.73)

Here s = 3630 and n = 14
df = 14 - 1 = 13

α = 1 - 0.99 = 0.01
The critical values for α = 0.01 and df = 13 are Χ^2(1-α/2,n-1) = 3.565 and Χ^2(α/2,n-1) = 29.819

CI = (sqrt(13*3630^2/29.819) , sqrt(13*3630^2/3.565))
CI = (2396.8 , 6931.84)

2)

Here s = 3.3 and n = 22
df = 22 - 1 = 21

α = 1 - 0.9 = 0.1
The critical values for α = 0.1 and df = 21 are Χ^2(1-α/2,n-1) = 11.591 and Χ^2(α/2,n-1) = 32.671

CI = (21*3.3^2/32.671 , 21*3.3^2/11.591)
CI = (7 , 19.73)

Here s = 3.3 and n = 22
df = 22 - 1 = 21

α = 1 - 0.9 = 0.1
The critical values for α = 0.1 and df = 21 are Χ^2(1-α/2,n-1) = 11.591 and Χ^2(α/2,n-1) = 32.671

CI = (sqrt(21*3.3^2/32.671) , sqrt(21*3.3^2/11.591))
CI = (2.65 , 4.44)