1200 high school freshman were randomly selected for a national survey. Among survey participants, the mean grade-point average was 2.8, and standard deviation was 0.6. What is the margin of error, assuming a 95% confidence level?
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (±.025) and its Z score.
95% = mean ± 1.96 SEm
SEm = SD/√n