the complete DATA is hese: https://web.njit.edu/~wguo/Math644_2012/GPA.txt pleas
ID: 2921620 • Letter: T
Question
the complete DATA is hese: https://web.njit.edu/~wguo/Math644_2012/GPA.txt
please give the detailed answer.
1.19. Grade point average. The director of admissions of a small college selected 120 students at random from the new freshman class in a study to determine whether a student's grade point average (GPA) at the end of the freshman year (Y) can be predicted from the ACT test score (X). The results of the study follow. Assume that first-order regression model (1.1) is appropriate. i: 2 3 118 119 120 X,: 21 28 28 16 28 3.897 3.885 3.778 3.914 1.860 2.948 2.13. Refer to Grade point average Problem 1.19. a. Obtain a 95 percent interval estimate of the mean freshman GPA for students whose ACT b. Mary Jones obtained a score of 28 on the entrance test. Predict her freshman GPA using a c. Is the prediction interval in part (b) wider than the confidence interval in part (a)? Should it d. Determine the boundary values of the 95 percent confidence band for the regression line test score is 28. Interpret your confidence interval. 95 percent prediction interval. Interpret your prediction interval be? when Xh 28. Is your confidence band wider at this point than the confidence interval in part (a)? Should it be?Explanation / Answer
The regression line and standard errors are as shown below:
a. When ACT score is 28, mean GPA = 2.12 + 0.038*28 = 3.184
95% CI for mean GPA = mean - t0.975*sqrt(MSRes*(1/n + (x-x_bar)2/Sxx)), mean + t0.975*sqrt(MSRes*(1/n + (x-x_bar)2/Sxx))
= 3.184 - 1.96*sqrt(3.54(1/121 + 0.3862 / 48.89)), 3.184 + 1.96*sqrt(3.54(1/121 + 0.3862 / 48.89))
= 2.79, 3.58
b. For 95% Prediction interval we have :
=mean - t0.975*sqrt(MSRes*(1+1/n + (x-x_bar)2/Sxx)), mean + t0.975*sqrt(MSRes*(1+1/n + (x-x_bar)2/Sxx))
= 3.184 - 1.96*sqrt(3.54(1+1/121 + 0.3862 / 48.89)), 3.184 + 1.96*sqrt(3.54(1+1/121 + 0.3862 / 48.89))
= -0.52, 6.89
c. The prediction interval is more than the confidence interval and that is expected as the variance for the prediction interval is more and hence the interval should be wider
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.