Question

A sample of 250 workers aged 16 and older produced an average length of time with the current employer of 4.4 years with standard deviation of 3.8 years. Construct a 99.9% confidence interval for mean job tenure of all workers aged 16 or older​

Answer 1

$99.9\%CI=(4.4-3.33\frac{3.8}{\sqrt{250}}, 4.4+3.33\frac{3.8}{\sqrt{250}})=(3.60,5.20).$