In: Statistics and Probability
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] |
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.