Question

A study was conducted to determine whether the final grade of a student in an introductory psychology course is linearly related to his or her performance on the verbal ability test administered before college entrance. The verbal scores and final grades for 10 students are shown in the table below.

Student / Verbal Score x / Final Grade y

1 / 42 / 88

2 / 68 / 65

3 / 31 / 83

4 / 68 / 64

5 / 42 / 82

6 / 65 / 65

7 / 72 / 68

8 / 35 / 89

9 / 76 / 76

10 / 42 / 68

Find the following:

(a) The correlation coefficient: r= -0.7231

(b) The least squares line: y^=

(c) Calculate the residual for the ninth student=

Answer 1

Solution:
Correlation coefficient can be calculated as
Correlation Coefficient = (n*Summation(XY) - Summation(X)*Summation(Y))/sqrt(((n*Summation(X^2)) - (Summation(X))^2))*((n*Summation(Y^2)) - (Summation(Y))^2)))

Student	Verbal Score(X)	Final Grade(Y)	X^2	Y^2	XY
1	42	88	1764	7744	3696
2	68	65	4624	4225	4420
3	31	83	961	6889	2573
4	68	64	4624	4096	4352
5	42	82	1764	6724	3444
6	65	65	4225	4225	4225
7	72	68	5184	4624	4896
8	35	89	1225	7921	3115
9	76	76	5776	5776	5776
10	42	68	1764	4624	2856
	541	748	31911	56848	39353

Correlation coefficient = ((10*39353) - (541*748))/sqrt(((10*31911 - 541*541)*(10*56848 - 748*748)) = -11138/sqrt(26429*8976) = -0.7231
Solution(b)
Regression equation can be calculated as
Y = a+bx
Here a is intercept of line and b is slope of line

Slope = (n*Summation(XY) - Summation(X)*Summation(Y))/((n*Summation(X^2)) - (Summation(X))^2)) = ((10*39353) - (541*748))/(10*31911 - 541*541) = -0.4214
Intercept can be calculated as
Intercept = (Summation(Y) - Slope*Summation(X))/n = (748 + 0.4214*541)/10 = 97.6
So regression equation is
Y = 97.6 - 0.42*X
Solution(c)
At X = 76, Y = 97.6 - 0.42*76 = 65.57
So residual can be calculated as
Residual = Predicted value - Observed value = 65.57 - 76 = -10.43