Question

1. Shown below is a partial Excel output from a regression analysis.

ANOVA
	df	SS	MS	F
Regression		60
Residual
Total	19	140
	Coefficients	Standard Error
Intercept	10.00	2.00
x1	-2.00	1.50
x2	6.00	2.00
x3	-4.00	1.00

a.	Use the above results and write the regression equation. [4 Marks]
b.	Compute the coefficient of determination and fully interpret its meaning. [4 Marks]
c.	Is the regression model significant? Perform the test at let α = 0.05. [4 Marks]
d.	At  = 0.05, test to see if there is a relation between x1 and y. [4 Marks]
e.	At  = 0.05, test to see if there is a relation between x3 and y. [4 Marks]

Answer 1

using Given output

a) Use the above results and write the regression equation

y = 10.00-2.00x1+6.00x2-4.00x3

b)

we define the coefficient of determination by

so here we need to complete the Anova table

	df	SS	MS	F
Regression		60
Residual
Total	19	140

here we use following information to fill above table

df for Regression = k-1 where k number of is independent variable

df for Regression = 3-1 = 2

df for Residual = df for Total - df for Regression

df for Residual =19-2=17

Now SS for Residual = SS for Total - SS for Regression

SS for Residual = 140 - 60 = 80

MS Regression = SS Regression/df for Regression

MS Regression = 60/2 = 30

MS Residual = SS Residual/df for Regression

MS Residual = 80/17 =4.7059

F = (SS Regression/df for Regression) /( SS for Residual/df for Residual)

F =30/4.7059

F =6.3750

So Completed anova table is

	df	SS	MS	F
Regression	2	60	30	6.3750
Residual	17	80	4.7059
Total	19	140

we define the coefficient of determination by

R2= 30/ 140

R2=0.2143

It can be interpreted as 21.43% variation explained by independent variable to the dependent variable.

Is the regression model significant? Perform the test at let α = 0.05.

F-Critical value = 4.619.(using statistical table)

Since F-statistic=6.3750 > F-Critical value =4.619 so we reject the null hypothesis and conclude that the regression model significant.

d)

test to see if there is a relation between x1 and y

t-cal =

t-cal = 10/2 =5

degree of freedom = n-k-1

degree of freedom = 20-3-1= 16

t-critical at = 0.025 and 16 degree of freedom = 2.12

so t-cal =5 > t-critical= 2.12 so reject the null hypothesis and conclude that there is significant relationship between x1 and y.