Question

A company is interested on average wages of employee and variance. A sample of 20 workers is taken and it is found that mean wage was $9.5 and sample standard deviation was. $1.30. Construct a 95% confidence interval for the population variance. (pls show work if possible)
A. (1.79, 3.119)

B. (0.977, 3.605)

C. (0.752, 2.773)

D. (0.79, 2.919)

E. (1.029, 3.795)

Answer 1

Given that the mean is $9.5 and the standard deviation is $1.30, the standard error will be given by:
σ/√n
where
σ-standard deviation
n=sample size

thus, we shal have:
1.30/√20
=0.2906
Next we find the margin error
0.2906*2=0.581
thus the confidence interval will be:
(9.5+0.581, 9.5-0.581)
=(10.081,8.919)