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 ____

Answer 1

Solution :

Given that,

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 =1.55 ( Using z table    )

sample size = n = [Z_/2* / E] ²

n = ( 1.55* 300 /25 )²

n =346

Sample size = n =346