Question

In a regression analysis involving 30 observations, the following estimated regressionequation was obtained.y^ =17.6+3.8x 1 −2.3x 2 +7.6x 3 +2.7x 4​For this estimated regression equation SST = 1805 and SSR = 1760. a. At \alpha =α= .05, test the significance of the relationship among the variables.Suppose variables x 1 and x 4 are dropped from the model and the following estimatedregression equation is obtained.y^ =11.1−3.6x 2​ +8.1x 3For this model SST = 1805 and SSR = 1705.b. Compute SSE(x 1 ,x 2 ,x 3 ,x 4 )c. Compute SSE (x2 ,x3 ) d. Use an F test and a .05 level of significance to determine whether x1 and x4 contribute significantly to the model.

Answer 1

(a) There is a significant relationship between y and
`x_1, x_2, x_3, x_4`

(b)
`SSE_((x_1 ,x_2 ,x_3 ,x_4) )= 45`

(c)
`SSE_((x_2,x_3)) = 100`

(d)
`x_1` and
`x_4` are significant

Explanation:

Given

`y = 17.6+3.8x_1 - 2.3x_2 +7.6x_3 +2.7x_4` --- estimated regression equation

`n = 30`

`p = 4` --- independent variables i.e. x1 to x4

`SSR = 1760`

`SST = 1805`

`\alpha = 0.05`

Solving (a): Test of significance

We have:

`H_o :` There is no significant relationship between y and
`x_1, x_2, x_3, x_4`

`H_a :` There is a significant relationship between y and
`x_1, x_2, x_3, x_4`

First, we calculate the t-score using:

`t = (SSR)/(p) / (SST - SSR)/(n - p - 1)`

`t = (1760)/(4) / (1805- 1760)/(30 - 4 - 1)`

`t = 440 / (45)/(25)`

`t = 440 / 1.8`

`t = 244.44`

Next, we calculate the p value from the t score

Where:

`df = n - p - 1`

`df = 30 -4 - 1=25`

The p value when
`t = 244.44` and
`df = 25` is:

`p =0`

So:

`p < \alpha` i.e.
`0 < 0.05`

Solving (b):
`SSE(x_1 ,x_2 ,x_3 ,x_4)`

To calculate SSE, we use:

`SSE = SST - SSR`

Given that:

`SSR = 1760` -----------
`(x_1 ,x_2 ,x_3 ,x_4)`

`SST = 1805`

So:

`SSE_((x_1 ,x_2 ,x_3 ,x_4)) = 1805 - 1760`

`SSE_((x_1 ,x_2 ,x_3 ,x_4) )= 45`

Solving (c):
`SSE(x_2 ,x_3)`

To calculate SSE, we use:

`SSE = SST - SSR`

Given that:

`SSR = 1705` -----------
`(x_2 ,x_3)`

`SST = 1805`

So:

`SSE_((x_2,x_3)) = 1805 - 1705`

`SSE_((x_2,x_3)) = 100`

Solving (d): F test of significance

The null and alternate hypothesis are:

We have:

`H_o :`
`x_1` and
`x_4` are not significant

`H_a :`
`x_1` and
`x_4` are significant

For this model:

`y =11.1 -3.6x_2+8.1x_3`

`SSE_((x_2,x_3)) = 100`

`SST = 1805`

`SSR_((x_2 ,x_3)) = 1705`

`SSE_((x_1 ,x_2 ,x_3 ,x_4) )= 45`

`p_((x_2,x_3)) = 2`

`\alpha = 0.05`

Calculate the t-score

`t = (SSE_((x_2,x_3))-SSE_((x_1,x_2,x_3,x_4)))/(p_((x_2,x_3))) / (SSE_((x_1,x_2,x_3,x_4)))/(n - p - 1)`

`t = (100-45)/(2) / (45)/(30 - 4 - 1)`

`t = (55)/(2) / (45)/(25)`

`t = 27.5 / 1.8`

`t = 15.28`

Next, we calculate the p value from the t score

Where:

`df = n - p - 1`

`df = 30 -4 - 1=25`

The p value when
`t = 15.28` and
`df = 25` is:

`p =0`

So:

`p < \alpha` i.e.
`0 < 0.05`

Hence, we reject the null hypothesis