z* = ME/(σ/sqrt(n)) with z* = z(95% CI) = z(<97.5)
Rearranging:
n = (z*σ/ME)2
n = (1.96*300/25)2 = 554 (rounded up)
Please consider a tutor. Take care.
Shqipton H.
asked 03/08/23SAT scores are distributed with a mean of 1,500 and a standard deviation of 300. You are interested in estimating the average SAT score of first year students at your college. If you would like to limit the margin of error of your 95% confidence interval to 25 points, how many students should you sample?
z* = ME/(σ/sqrt(n)) with z* = z(95% CI) = z(<97.5)
Rearranging:
n = (z*σ/ME)2
n = (1.96*300/25)2 = 554 (rounded up)
Please consider a tutor. Take care.
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