A sample of 250 workers aged 16 and
older produced an average length of time with the current employer ('job tenure') of 4.4 years with a standard deviation of 3.8 years. Construct a 99% confidence interval for the mean job tenure of all workers aged 16 and older.
pls respect
Answers & Comments
The answer is
°EXPLANATION;
{ A sample of 250 workers an average of 4.4 years with a standard Deviation of 3.8 Years }
SE= 2.58 ×38/
[tex] \sqrt[]{250} = 0.619[/tex]
CI = ( 4.4 - 0.619 )
=( 3.781. 5.019