3. Students in a class were given two quizzes. The scores are presented in the following table.

 Student Quiz 1 Quiz 2 A 5 9 B 4 7 C 2 5 D 6 10 E 3 4

a) Find the least squares regression equation which will be used to predict the scores on the second test as a function of the scores on the first test. i.e., let the scores on the first test be the explanatory variable and let the scores on the second test be the response variable.

b) Find the correlation coefficient r.

c) Use the regression equation to predict the score that student E should have gotten on the second quiz.

d) What is the residual for student E?

a) The explanatory variable, the scores on the first quiz will be our x, and the response variable, the scores on the second quiz will be our y. Your calculator should be able to compute the coefficients in the least squares regression formula. For the calculation, first find the means of the x's and y's

 Student x y A 5 9 B 4 7 C 2 5 D 6 10 E 3 4 Totals 20 35

Since there are 5 scores, the means are

mx = 20/5 = 4

my = 35/5 = 7

If the regression equation is

y = ax + b

then

So we perform the following computations.

 x y x - mx y - my (x - mx)(y - my) (x - mx)2 5 9 1 2 2 1 4 7 0 0 0 0 2 5 -2 -2 4 4 6 10 2 3 6 4 3 4 -1 -3 3 1 20 35 0 0 15 10

So

a = 15/10 = 1.5

The formula for b is

b = my - amx

= 7 - 1.5(4)

= 7 - 6 = 1

So the equation is

y = 1.5x + 1.

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b) The formula for the correlation coefficient, r is

So we also need the sum of the squares of the deviations on the ys.

 x y x - mx y - my (x - mx)(y - my) (x - mx)2 (y - my) 2 5 9 1 2 2 1 4 4 7 0 0 0 0 0 2 5 -2 -2 4 4 4 6 10 2 3 6 4 9 3 4 -1 -3 3 1 9 20 35 0 0 15 10 26

So

= .93026

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c) Student E got a 3 on the first quiz, so the regression equation would predict that the score on the second quiz should have been

y = 1.5(3) + 1 = 4.5 + 1 = 5.5

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d) The student actually got a 4 on the second quiz, so the residual is

4 - 5.5 = -1.5

The student's score on the second quiz was 1.5 points lower than the equation predicted.

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