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 98% confidence interval to 5 points, how many students should the administrator sample? Make sure to give a whole number answer.

Answer 1

Solution :

Given that,

standard deviation = = 300

margin of error = E = 5

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z/2 = Z0.01 = 2.326

Sample size = n = ((Z/2 * ) / E)2

= ((2.326* 300) / 5)2

=  19476.9

Sample size = 19477