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?