Practice Problem 500 respondent report spending an average of $243 per month eating in restaurants. The variance of the mean is $47. What is the standard error of the mean? You will need one of the following formulae: 68% (If N did not change, but variance was only 27, what would the SEm do?) '95%--------- SEm = SD/VN Var = [(0-M) 99% N-1 Mean = [ O / N -3 SD -2 SD 1 SD Statistic +1 SD +2 SD +3 SD Confidence Interval = Mean + SEm x 1.96 Std Dev = 1.96 SEs 2.56 SEs SD = VVar Variance = SEm = SD / VN