In a survey, 20 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $30 and standard deviation of $2. Find the margin of error for a 99% confidence level.

Answer 1

Z = (score-mean)/SEm

SEm = SD/√n

The error would be .005 at each end, which gives you a Z score of ±1.645.

I'll let you do the calculations.