Question

Need help with all three please!

QUESTION 15 Suppose the regression model below was fit to n 32 data points with SSE 1,400. The value of the F test statistic requited to conduct the test, to two decimal places, is -A. 12.67 O B. 8.00 C. 5.33 D. 10.71

Answer 1

13) solution:

Given data :

Y = $\beta _{0}$ + $\beta _{1}X1$ + $\beta _{2}X2$ + $\beta _{3}X3$ + $\beta _{4}$$\beta _{4}X1X2$+ ?₅$\beta _{5}X1X3$ + $\beta _{6}X2X3$

SSE = 1400

n = 32

Now, we find out the:

The Mean Square Error (MSE) for the full model :

We already know the formula for the Mean Square Error (MSE) is ,

MSE = SSE / ( n-(k+1 ) )

Here, k = no. of constants

k = 6

MSE = 1400/ ( 32- ( 6+1))

= 1400 / ( 32- 7 )

= 1400 / 25

  = 56

$\therefore$ Mean Square Error (MSE) = 56

14) solution:

Given data :

Y = $\beta _{0}$ + $\beta _{1}X1$ + $\beta _{2}X2$ + $\beta _{3}X3$ + $\beta _{4}$$\beta _{4}X1X2$+ ?₅$\beta _{5}X1X3$ + $\beta _{6}X2X3$

SSE = 1400

n = 32

Sum of Square Error (SSE) for the reduced model =3,200

Now, we find out the:

The value of the numerator for the F statistic required to conduct the test:

We already know the formula for the numerator for the F statistic is ,

The numerator for the F statistic is = ( (SSE For reduced - SSE ) /q) / ( SSE /( n-k-1 ))

but now we are using FINV table

Here, df1 = no of items in the equation

df1 = 3

df2 = n-k-1

= 32-6-1

= 25

df2 = 25

now, F statistic = FINV ( significance , df1, df2 )

Here there is no significance value , so we can assume as 0.05

= FINV ( 0.05,3,25 )

From the  FINV table , we can get the value i.e., " 2.76 "

The numerator for the F statistic is = 2.76

$\therefore$ The value of the numerator for the F statistic required to conduct the test = 2.76

15) solution:

Given data :

Y = $\beta _{0}$ + $\beta _{1}X1$ + $\beta _{2}X2$ + $\beta _{3}X3$ + $\beta _{4}$$\beta _{4}X1X2$+ ?₅$\beta _{5}X1X3$ + $\beta _{6}X2X3$

SSE = 1400

n = 32

Now, we find out the:

The value of the F test statistic required to conduct the test, to two decimal places:

The value of the F test statistic =  $\frac{SSR/ (p-1)}{ SSE / ( n-p))}$

Here SSR and SSE are same

df1 = p-1

= 6-1

df1 = 5

df2 = n-p

   = 32-6

   df2 = 26

Now,   F test statistic =  $\frac{1400/ 5)}{1400 / (26))}$

=  $\frac{280}{53.84}$

= 5.2005

$\therefore$ F test statistic = 5.20

Note: From the question our calculating F- value is not there in options, but the process is correct .