College admissions administrators at a university are reevaluating their entrance requirements.They believe that the mean GPA for graduating high school seniors has increased dramatically over the last decade.They randomly sample 85 records of graduating seniors from 10 years ago (sample 1) and find an average GPA of 2.68.An independent random sample of 80 current graduating seniors yields an average GPA of 2.94.If the population variance of high school GPAs is 0.2601, find a 95% confidence interval for the difference in population mean GPA over the last decade.
A) (-0.42, -0.10)
B) (-0.29, -0.13)
C) (-0.36, -0.16)
D) (-0.46, -0.05)

Question

Quizplus · Accepted Answer

The correct answer is A because the 95% confidence interval for the difference in population mean GPA over the last decade is (-0.42, -0.10). This interval is calculated using the formula:CI = (x1 - x2) +/- z * sqrt((s1^2/n1) + (s2^2/n2))where x1 and x2 are the sample means, s1 and s2 are the sample standard deviations, n1 and n2 are the sample sizes, and z is the z-score corresponding to the desired confidence level.In this case, we have:x1 = 2.68x2 = 2.94s1 = sqrt(0.2601) = 0.5101s2 = sqrt(0.2601) = 0.5101n1 = 85n2 = 80z = 1.96 (for a 95% confidence level)Plugging these values into the formula, we get:CI = (2.68 - 2.94) +/- 1.96 * sqrt((0.5101^2/85) + (0.5101^2/80))CI = -0.26 +/- 1.96 * 0.123CI = -0.26 +/- 0.241CI = (-0.42, -0.10)

College Admissions Administrators at a University Are Reevaluating Their Entrance