Question

200 high school freshmen were randomly selected for a national survey. Among the survey participants, the mean grade-point average (GPA) was 2.7. If the data follows a normal distribution with a standard deviation of 0.4, what is the margin of error, assuming a 95% confidence level?

Answer 1

Solution :

Given that,

Point estimate = sample mean =     = 2.7

Population standard deviation =     = 0.4

Sample size n =200

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z / 2   = Z0.025 = 1.96   ( Using z table )

Margin of error = E =   Z/2    * ( /n)
= 1.96 * (0.4 / 200 )

=0.055