Asked by Isabel
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?
Answers
Answered by
PsyDAG
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
95% = mean ± 1.96 SEm
SEm = SD/√n
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