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Name Economics 5 Ch 13 Practice The follo wing data are the monthly salaries y and the grade point averages x for students wh

Name Economics 5 Ch 13 Practice The follo wing data are the monthly salaries y and the grade point averages x for students who obtained a bachelor's degree in business administ ration Obser vation index xi 2.6 3300 3.4 3600 3.6 4000 3.2 3500 3.5 3900 2.9 3600 TSS Totals SSR SSE 1. Calculate and y 2. Use the least squares method to develop the estimated regression equation. Use two decimal points in your answers for bo and bi. 3. Calculate the values of for each x in the data. Show the answers in the table 4. Compute the Sum of Squares Due to Regression, known as SSR. 5. Compute the Sum of Squares Due to Error, known as SSE. 6. Compute the Total Sum of Squares, known as TSS (or SST) 7. Compute r 8. What is the value of the sample correlation coefficient? 9. Is the relationship between GPA and salary positive or negative? 10. Based on your regression analysis, how much does predicted salary change whern GPA increases by 1.0?
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Answer #1
x y x-xbar y-ybar (x-xbar)(y-ybar) (x-xbar)^2 y^ (y^ - ybar) (y^ - ybar)^2 ei (yi - y^)^2 (yi-ybar)^2
2.6 3300 -0.6 -350 210 0.36 3301.351 -348.649 121555.9 -1.35135 1.82615 122500
3.4 3600 0.2 -50 -10 0.04 3766.216 116.2162 13506.21 -166.216 27627.83 2500
3.6 4000 0.4 350 140 0.16 3882.432 232.4324 54024.84 117.5676 13822.13 122500
3.2 3500 0 -150 0 0 3650 0 0 -150 22500 22500
3.5 3900 0.3 250 75 0.09 3824.324 174.3243 30388.97 75.67568 5726.808 62500
2.9 3600 -0.3 -50 15 0.09 3475.676 -174.324 30388.97 124.3243 15456.54 2500
3.2 3650 430 0.74 249864.9 85135.14 335000

1)xbar = 3.2
ybar = 3650

2)
slope = sum ((x-xbar)(y-ybar)) / sum (x-xbar)^2
= 430/0.74
= 581.0811

b0 = ybar - slope*xbar
= 1790.541

3) See in the table
4)
SSR = 249864.9
5)
SSE = 85135.14
6)
SST = 335000

7)
r^2 = SSR/SST =
8)
r = 0.8636
9)
there is positive relationship

10)this is slope
= 581.0811

11)

MSE = sqrt(SSE/(n-2)) = 21283.78

12)s = sqrt(MSE) = 145.8896
13)sb1 = sqrt(MSE/SSx)
= 169.5932

14)
t = 3.4263
15)
df = 4
16)
p-value = 0.0266

0 < p-value < 0.05

17)

p-value < alpha
hence we reject the null hypothesis

we conclude that there is statistically significant relationship

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