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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.

Respuesta :

MrRoyal

Answer:

(a) There is a significant relationship between y and [tex]x_1, x_2, x_3, x_4[/tex]

(b) [tex]SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45[/tex]

(c) [tex]SSE_{(x_2,x_3)} = 100[/tex]

(d) [tex]x_1[/tex] and [tex]x_4[/tex] are significant

Step-by-step explanation:

Given

[tex]y = 17.6+3.8x_1 - 2.3x_2 +7.6x_3 +2.7x_4[/tex] --- estimated regression equation

[tex]n = 30[/tex]

[tex]p = 4[/tex] --- independent variables i.e. x1 to x4

[tex]SSR = 1760[/tex]

[tex]SST = 1805[/tex]

[tex]\alpha = 0.05[/tex]

Solving (a): Test of significance

We have:

[tex]H_o :[/tex] There is no significant relationship between y and [tex]x_1, x_2, x_3, x_4[/tex]

[tex]H_a :[/tex] There is a significant relationship between y and [tex]x_1, x_2, x_3, x_4[/tex]

First, we calculate the t-score using:

[tex]t = \frac{SSR}{p} \div \frac{SST - SSR}{n - p - 1}[/tex]

[tex]t = \frac{1760}{4} \div \frac{1805- 1760}{30 - 4 - 1}[/tex]

[tex]t = 440 \div \frac{45}{25}[/tex]

[tex]t = 440 \div 1.8[/tex]

[tex]t = 244.44[/tex]

Next, we calculate the p value from the t score

Where:

[tex]df = n - p - 1[/tex]

[tex]df = 30 -4 - 1=25[/tex]

The p value when [tex]t = 244.44[/tex] and [tex]df = 25[/tex] is:

[tex]p =0[/tex]

So:

[tex]p < \alpha[/tex] i.e. [tex]0 < 0.05[/tex]

Solving (b): [tex]SSE(x_1 ,x_2 ,x_3 ,x_4)[/tex]

To calculate SSE, we use:

[tex]SSE = SST - SSR[/tex]

Given that:

[tex]SSR = 1760[/tex] ----------- [tex](x_1 ,x_2 ,x_3 ,x_4)[/tex]

[tex]SST = 1805[/tex]

So:

[tex]SSE_{(x_1 ,x_2 ,x_3 ,x_4)} = 1805 - 1760[/tex]

[tex]SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45[/tex]

Solving (c): [tex]SSE(x_2 ,x_3)[/tex]

To calculate SSE, we use:

[tex]SSE = SST - SSR[/tex]

Given that:

[tex]SSR = 1705[/tex] ----------- [tex](x_2 ,x_3)[/tex]

[tex]SST = 1805[/tex]

So:

[tex]SSE_{(x_2,x_3)} = 1805 - 1705[/tex]

[tex]SSE_{(x_2,x_3)} = 100[/tex]

Solving (d): F test of significance

The null and alternate hypothesis are:

We have:

[tex]H_o :[/tex] [tex]x_1[/tex] and [tex]x_4[/tex] are not significant

[tex]H_a :[/tex] [tex]x_1[/tex] and [tex]x_4[/tex] are significant

For this model:

[tex]y =11.1 -3.6x_2+8.1x_3[/tex]

[tex]SSE_{(x_2,x_3)} = 100[/tex]

[tex]SST = 1805[/tex]

[tex]SSR_{(x_2 ,x_3)} = 1705[/tex]

[tex]SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45[/tex]

[tex]p_{(x_2,x_3)} = 2[/tex]

[tex]\alpha = 0.05[/tex]

Calculate the t-score

[tex]t = \frac{SSE_{(x_2,x_3)}-SSE_{(x_1,x_2,x_3,x_4)}}{p_{(x_2,x_3)}} \div \frac{SSE_{(x_1,x_2,x_3,x_4)}}{n - p - 1}[/tex]

[tex]t = \frac{100-45}{2} \div \frac{45}{30 - 4 - 1}[/tex]

[tex]t = \frac{55}{2} \div \frac{45}{25}[/tex]

[tex]t = 27.5 \div 1.8[/tex]

[tex]t = 15.28[/tex]

Next, we calculate the p value from the t score

Where:

[tex]df = n - p - 1[/tex]

[tex]df = 30 -4 - 1=25[/tex]

The p value when [tex]t = 15.28[/tex] and [tex]df = 25[/tex] is:

[tex]p =0[/tex]

So:

[tex]p < \alpha[/tex] i.e. [tex]0 < 0.05[/tex]

Hence, we reject the null hypothesis

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