Question

21. Complete a simple linear regression on the following data from a random survey of a random sample of free throws made out of 100:

Age (in years)          # of Free throws made (out of 100)

20                               30

22                               36

26                               28

28                               20

33                               25

33                               15

38                               10

42                               25

49                               8

54                               15

55                               22

55                               18

57                               35

60                               12

Provide the equation of the linear regression line (3 pts):
Y = a + bX

Sum of x is 572

Sum of y is 299

Mean of x is 40.8571

Mean of y is 21.3571

Sum of squares is 2575.7143

Sum of products is -669.2857
Provide the coefficient of correlation (2 pts):
How many would free throws would you expect a 42 year old to make (2 pts):
What is the residual for the 42 year old we sampled (2 pts):

Answer 1

Solution:
Regression equation can be calculated as
Y = a+ bX
Here Y is dependent Variable i.e. No. of free throws
X is an independent Variable i.e. Age
a is Y-intercept of Line
b is Slope of regression line
Slope of regression line can be calculated as
Slope = ((n*Xi*Yi) - (Xi * Yi))/((n*Xi^2) - (Xi)^2) =

Age(X)	No. of Free throws made(Y)	X^2	Y^2	XY
20	30	400	900	600
22	36	484	1296	792
26	28	676	784	728
28	20	784	400	560
33	25	1089	625	825
33	15	1089	225	495
38	10	1444	100	380
42	25	1764	625	1050
49	8	2401	64	392
54	15	2916	225	810
55	22	3025	484	1210
55	18	3025	324	990
57	35	3249	1225	1995
60	12	3600	144	720
572	299	25946	7421	11547

Slope = ((14*11547) - (572*299))/((14*25946) -(572*572)) = -9370/36060 = -0.26
Intercept of regression line a = ((Yi) - Slope*(Yi))/n = (299 - (-0.26*572))/14 = 31.97
So regression equation is
Y = 31.97 - 0.26*X
Solution(b)
Coefficient of correlation can be calculated as
Slope = ((n*Xi*Yi) - (Xi * Yi))/sqrt(((n*Xi^2) - (Xi)^2)*((n*Yi^2) - (Yi)^2))) = ((14*11547) - (572*299))/sqrt(((14*25946) -(572*572))*((14*7421)-(299*299))) = -9370/sqrt(36060*14493) = -0.41

Solution(c)
From Regression equation Y = 31.97 - 0.26*X
At X= 42, we need to calculate Free throws which can be calculated as
Y = 31.97 - 0.26*X = 31.97 - 0.26*42 = 21.05 after round off 21
Solution(d)
Residual for the 42 year = No. of free throws predicted from regression equation - No. of free throws from actual data = 21 - 25 = -4
that means according to the regression line we can say that the number of free throws made is less by 4 as compared to actual.