Question

In Professor Krugman’s economics course, the correlation between the students’ total scores prior to the final examination and their final‑examination scores is ?=0.5.r=0.5. The pre‑exam totals for all students in the course have mean 280280 and standard deviation 40.40. The final‑exam scores have mean 7575 and standard deviation 8.8. Professor Krugman has lost Julie’s final exam but knows that her total before the exam was 300.300. He decides to predict her final‑exam score from her pre‑exam total.

(a) Which of the choices is the least‑squares regression line of final‑exam scores on pre‑exam total scores in this course?

a. ?̂ =47+0.1?y^=47+0.1x

b. ?̂ =−0.1−47?y^=−0.1−47x

c. ?̂ =−47+0.1?y^=−47+0.1x

d. ?̂ =47−0.1?y^=47−0.1x

Which of the choices is correct interpretation of the slope in the context of the problem?

a. For each point of a student's pre‑exam score means an drop of 4747 points on the final exam, on average.

b. For each point of a student's pre‑exam score means an additional 4747 points on the final exam, on average.

c. For each point of a student's pre‑exam score means an additional 0.10.1 point on the final exam, on average.

d. For each point of a student's pre‑exam score means an drop of 0.10.1 point on the final exam, on average.

(b) Use a regression line to predict Julie’s final‑exam score. Give your answer as a whole number.

Predicted final‑score =

(c) Julie does not think this method accurately predicts how well she did on the final exam. Select the statement that correctly uses ?2r2 to argue that her actual score could have been much higher (or much lower) than the predicted value.

a. Since ?2=0.25r2=0.25 is close to 0,0, then the regression line does not predict final scores very well. Julie could have a much higher or lower score.

b. Since ?2=0.25r2=0.25 is not exactly 0,0, then the regression line does predict final scores very well. Julie could have a much higher or lower score.

c. Since ?2=0.25r2=0.25 is positive, then the regression line does not predict final scores very well. Julie could have a much higher score.

d. Since ?2=0.5r2=0.5 is close to 0,0, then the regression line does not predict final scores very well. Julie could have a much higher or lower score.

Answer 1

Given, r= 0.5 , x-bar=280, Sxx= 40^2= 1600

y-bar= 75, Syy= 8^2= 64

Now, r= Sxy/√Sxx*Syy, So, Sxy= 0.5*√1600*64= 160

Now, Regression equation is given as y = A + Bx

Where B= Sxy/Sxx= 160/1600= 0.1, A= y-bar- B*x-bar= 75-0.1*280 = 47

So the required regression equation is: y= 47 + 0.1x

So option a is correct.

a-2) The slope above is 0.1 which means that for every increase in pre exam score by 1, the final exam score increases by 0.1. So option c is the correct interpretation of the slope.

b) Putting x= 300 in the regression equation we get,

Y= 47 + 0.1*300 = 77

So Julie's predicted final exam score is 77

c) r^2 = 0.05*0.5= 0.25 (This is a low r^2, all values of r^2 >0.6 are considered strong)

Since this r^2 is close to 0, we can say that the regression line does not predict the final scores very well. Julie's true score could be higher or lower than that predicted. So option d is correct.