99% = mean ± 2.575 SEm
SEm = SD/√n
2.575 SEm = 2.575 SD/√n = 15
Solve for n.
A statistician wants to estimate the mean weekly family expenditure on clothes. He believes that the standard deviation of the weekly expenditure is $125. Determine with 99% confidence the number of families that must be sampled to estimate the mean weekly family expenditure on clothes to within $15.
2 answers
463