Posted by victoria on Tuesday, June 19, 2012 at 8:13am.
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.

Statistics  PsyDAG, Tuesday, June 19, 2012 at 10:29am
95% = mean ± 1.96 SEm
SEm = SD/√n
Do you know the mean?