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)
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)
