Question

SAT scores are normally distributed with a mean of 1,500 and a standard deviation of 300. An administrator at a college is interested in estimating the average SAT score of first-year students. If the administrator would like to limit the margin of error of the 88% confidence interval to 25 points, how many students should the administrator sample? Make sure to give a whole number answer.

Answer 1

Solution :

Given that,

Population standard deviation = = 300

Margin of error = E = 25

At 88% confidence level the z is ,

= 1 - 88% = 1 - 0.88 = 0.12

/ 2 = 0.12 / 2 = 0.06

Z_/2 = Z_0.06 = 1.555

sample size = n = (Z_/2* / E) ²

n = (1.555*300 / 25)²

n = 348.20

n = 349

Sample size = 349

349 students should the administrator sample.