A random sample of 100 teachers in a large metropolitan area revealed a mean weekly salary of $487 with a standard deviation of $48. With what degree of confidence can we assert the average weekly salary of all teachers in the metropolitan area is between $472 and $502?
I can't figure out how to calculate the degree of confidence, i.e. Z alpha over 2...anyone good with statistics?
2006-12-20 15:57:23 · 3 answers · asked by Danderson P. Maxwell 2 in Science & Mathematics ➔ Mathematics
Consider how the 472 and 502 compare to the mean weekly salary of 487. Each endpoint of this confidence interval is 15 dollars from the mean.
That means that $15 is the margin of error (m) for the confidence interval. m is z*SD/sqrt(n), so z = (m*sqrt(n))/SD. Thus, z = (15*sqrt(100))/48 = 150/48 = 3.125.
The degree of confidence will be the area between z = -3.125 and z = +3.125, which is about 99.8%.
2006-12-21 04:43:42 · answer #1 · answered by jbm616 2 · 0⤊ 0⤋
Those teachers need a raise!
Normalize...z=(x-mu)/sigma=(502-487)/(48/sqrt(100)) <<<<---note use of central limit theorem so z=150/48=25/8 then the area outside of between +/- 25/8 on the standard normal about .001778050600 so about 99.8 % confident
2006-12-20 16:00:48 · answer #2 · answered by a_math_guy 5 · 1⤊ 0⤋
5 to 1
2006-12-20 15:59:09 · answer #3 · answered by Anonymous · 0⤊ 3⤋