Question

A nationwide, standardized test taken by high-school juniors and seniors may or may not measure academic potential, but we can nonetheless attempt to predict academic performance in college from scores on this test. We have chosen a random sample of 47 students just finishing their first year of college, and for each student we've recorded her score on this standardized test (from 400 to 1600) and her grade point average (on a four point scale) for her first year in college. Letting x denote score on the standardized test and y denote first-year college grade point average, the least squares regression equation computed from the data is 0.9027 +0,0013.x. We're now interested in predicting the first year grade point average of a student who scored 1050 on the standardized test. We're also interested in a prediction Interval for this grade point average and a confidence interval for the mean first-year grade point average of students who scored 1050 on the standardized test. We have already computed the following for our data: • mean square error (MSE) - 0.352; 1 (1050-x) - = 0.0284, 60,-) where x, denote the standardized test scores in the sample, and x denotes their mean. Explanation Check 20 Merw ca Alghes Reserved on the

Answer 1

1) lower limit =estimate - t*std error =(0.9027+0.0013*1050)-2.690*sqrt(0.352*(1+0.0284))=0.65

upper limit =estimate + t*std error =(0.9027+0.0013*1050)+2.690*sqrt(0.352*(1+0.0284))=3.89

2)

prediction interval would have same center and will be wider

3)

would be wider and have a different center