Question

625 high school freshmen were randomly selected for a national survey. Among survey participants, the mean grade-point average (GPA) was 2.9, and the standard deviation was 0.5. What is the approximate margin of error, assuming a 95% confidence level?

Answer 1

Solution :

sample size = n = 625

Degrees of freedom = df = n - 1 = 624

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,624 = 1.964

Margin of error = E = t/2,df * (s /n)

= 1.964 * (0.5 / 624)

= 0.039