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An institutional research administrator believes that there is a direct relation

ID: 3327763 • Letter: A

Question

An institutional research administrator believes that there is a direct relationship between a student’s GPA and their score on a senior Aptitude Test. The following data show the results of 10 student’s grade point averages (X) and their aptitude test score (Y).

GPA (X)

Aptitude Test

Score (Y)

1.8

26

2.3

31

2.6

28

2.4

30

2.8

34

3.0

38

3.4

41

3.2

44

3.6

40

3.8

43

Put the data above into an Excel spreadsheet. Use the output from a Regression Analysis to answer the questions below. Don’t forget to turn in your Excel spreadsheet with the output.

a. Develop an estimated regression equation relating GPA and Aptitude Test score (use excel printout above).

b.         If a student’s GPA is 3.5, predict their Aptitude Test score.

c.         Interpret the coefficient of determination. Make sure you provide the numeric value.

d.         Interpret the correlation coefficient. Make sure you provide the numeric value.

e.     Use a t test to determine whether there is a relationship between GPA and the Aptitude test. Thoroughly explain your findings.

GPA (X)

Aptitude Test

Score (Y)

1.8

26

2.3

31

2.6

28

2.4

30

2.8

34

3.0

38

3.4

41

3.2

44

3.6

40

3.8

43

Explanation / Answer

a)

b1= nE(xy)-ExEy/nE(x2)-(Ex)^2

=10*1059.7-(28.9*355)/10*87.09-(28.9*28.9)

=10597-10260/870.9-835.21

=337/35.69

=9.44

b0=Ey-b1Ex/n

355-9.44*28.9/10=8.21

y=b0+b1x

y=8.21+9.44x

b)

y=8.21+9.44(3.5)

y=41.25

c and d)

r (correlation)=n(Exy)-(Ex)(Ey)/ (nEx^2-(Ex)^2)(nEY^2-(Ey)^2)

=10(1059.7)-(28.9)(355)/(10*87.09-(28.9)^2)(10*12987-(355)^2)

=0.911 (There is a positive correlation amongst the two variables)

r^2=0.83 (83% of the variation in dependent variable is accounted for by the independent variable)

d)

H0: =0

H1: > 0

t = r * (n-2) / (1-r^2)

t = (0.911)*8) / 1- (0.911)^2)

t= 2.57669711/0.4124

t=6.24795688

The critical t with 8 degrees of freedom (.05 significance level) is 1.8595. Calculated t exceeds critical t, so reject H0 in favor of H1. It is reasonable that the coefficient is greater than 0

GPA (X) Aptitude Test Score (Y) x^2 y^2 xy 1.8 26 3.24 676 46.8 2.3 31 5.29 961 71.3 2.6 28 6.76 784 72.8 2.4 30 5.76 900 72 2.8 34 7.84 1156 95.2 3 38 9 1444 114 3.4 41 11.56 1681 139.4 3.2 44 10.24 1936 140.8 3.6 40 12.96 1600 144 3.8 43 14.44 1849 163.4 28.9 355 87.09 12987 1059.7 Total
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