Question

5. A group of 5 students scores on Exam 1 and Exam 2 in Math 1153 is recorded bellow: Вс Student 1 3 3 5 N 14 15 Exam 1 86 84 90 85 Exam 2 78 68 57 80 (a) Sketch the scatter plot for Exam 1 vs. Exam 2 scores and identify the form and direction of the association between them. FO CH Te (b) Calculate the correlation coefficient r and interpret its value in the context of the problem. (c) Calculate the equation of the regression line and use it to predict the score on Exam 2 for a student who scored 80 on Exam 1. (d) Calculate and interpret the value of r?, the coefficient of determination. 1 (e) Use the Regression line in (c) to find the residual of a score 90 on Exam 1.

Answer 1

	x	y	X^2	y^2	xy
	86	78	7396	6084	6708
	84	68	7056	4624	5712
	90	57	8100	3249	5130
	74	75	5476	5625	5550
	85	80	7225	6400	6800
average	83.8	71.6	7050.6	5196.4	5980

a. Scatter plot :

On X axis is Scores 1 and on Y axis is the scores of 2

90 80 70 60 50 40 Series1 30 20 10 0 0 20 40 60 80 100

The plot clearly shows a negative association among the scores, i.e in a downward direction.

b Correlation coefficient is calculated using the formula

$r = \frac{average(xy) - average(x)*average(y)}{{\sqrt{[average(x^2)-(average(x))^{2})]*[average(y^2)-(average(y))^{2})]}}}$Substituting the values, we get

r = -0.453,

ie. as shown in the scatter plot the correlation coefficient also agrees with the negative relation.

c. Using the method of least squares, the estimated regression line is given by

Y=a+bX, where a is the intercept and b is the slope estimate.

$b = \frac{average(xy) - average(x)*average(y)}{{\sqrt{[average(x^2)-(average(x))^{2})]}}}$

and intercept =a = average(y) - b*average(x)

Substituting the values, we get , b= -0.713 and a=131.355

Thus, the regression line is Y= 131.355 -0.713X

When X=80, Y=131.355-0.713*80 = 74.315.

d. The value of r^2, ie the coefficient of determination shows the percentage of variation explained by the model.

here, r^2 = (-0.453)^2 = 0.205

i.e 0.205*100 = 20.5% of the total variation is only explained by the model.

e. Residual of a score is given by the difference between the given y and the estimated Y,,

When X=90, y =57

Estimated Y =131.355-0.713*90 = 67.185

Residual = 57-67.185 = -10.185