stadistics
posted by victoria .
If 200 of the adults surveyed were in the age category of 24 and under and they provided a standard deviation of $14.50, construct a 95% confidence interval for the weekly average expenditure on fast food for adults 24 years of age and under. Assume fast food weekly expenditures are normally distributed.

95% = mean ± 1.96 SEm
SEm = SD/√n
Do you know the mean?