Question

1. Suppose we have the following information on the examination scores and weekly employment hours of eight students:

Exam Score      Employment Hours

      80                           10

      80                           5

      68                           15

      95                           5

      75                           21

      60                           40

      90                           0

    100                           0

Assuming Exam Score is the dependent variable (Y) and Employment Hours is the independent variable (X), calculate the simple linear regression equation by hand. Using this, test to see whether the slope on X is significant at the 0.05 level (do this by hand, as well). Make sure to show your work.

Answer 1

	X	Y	X * Y	X2	Ŷ	Sxx =Σ (Xi - X̅ )	Syy = Σ( Yi - Y̅ )	Sxy = Σ (Xi - X̅ ) * (Yi - Y̅)
	10	80	800	100	82.75	4	1	2
	5	80	400	25	87.125	49	1	7
	15	68	1020	225	78.375	9	169	-39
	5	95	475	25	87.125	49	196	-98
	21	75	1575	441	73.125	81	36	-54
	40	60	2400	1600	56.5	784	441	-588
	0	90	0	0	91.5	144	81	-108
	0	100	0	0	91.5	144	361	-228
Total	96	648	6670	2416	136	1264	1286	-1106

X̅ = Σ (Xi / n ) = 96/8 = 12
Y̅ = Σ (Yi / n ) = 648/8 = 81

Equation of regression line is Ŷ = a + bX
b = ( n Σ(XY) - (ΣX* ΣY) ) / ( n Σ X² - (ΣX)² )
b = ( 8 * 6670 - 96 * 648 ) / ( 8 * 2416 - ( 96 )²)
b = -0.875

a =( ΣY - ( b * ΣX ) ) / n
a =( 648 - ( -0.875 * 96 ) ) / 8
a = 91.5
Equation of regression line becomes Ŷ = 91.5 + -0.875 X

To Test :-

H0 :-  

H1 :-  

Test Statistic :-


t = -4.2714

Test Criteria :-
Reject null hypothesis if

= 4.2714 > 2.4469
Result :- Reject null hypothesis

Decision based on P value
P - value = P ( t > 4.2714 ) = 0.0053
Reject null hypothesis if P value < level of significance
P - value = 0.0053 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

There is statistically significant relationship between variables