Question

Exhibit 15-6
Below you are given a partial computer output based on a sample of 16 observations.

	Coefficient		Standard Error
Constant	12.924		4.425
X1	-3.682		2.630
X2	45.216		12.560
Analysis of Variance
Source of Variation	Degrees of Freedom		Sum of Squares	Mean Square	F
Regression			4,853	2,426.5
Error				485.3

Refer to Exhibit 15-6. The estimated regression equation is

Refer to Exhibit 15-6. The interpretation of the coefficient of X1 is that

Refer to Exhibit 15-6. We want to test whether the parameter b1 is significant. The test statistic equals Refer to Exhibit 15-6. The t value obtained from the table which is used to test an individual parameter at the 1% level is Refer to Exhibit 15-6. Carry out the test of significance for the parameter b1 at the 1% level. The null hypothesis should be Refer to Exhibit 15-6. The degrees of freedom for the sum of squares explained by the regression (SSR) are Refer to Exhibit 15-6. The sum of squares due to error (SSE) equals Refer to Exhibit 15-6. The test statistic used to determine if there is a relationship among the variables equals Refer to Exhibit 15-6. The test statistic used to determine if there is a relationship among the variables equals Refer to Exhibit 15-6. Carry out the test to determine if there is a relationship among the variables at the 5% level. The null hypothesis should Please show work with equations, I have answers to this problem but I want to make sure I have the correct steps.

Answer 1

From the partial computer output, the estimates are: bo = 12.924, b = -3.682, b = 45.216 The form of the estimated regression equation with two independent variables is: Y= b +0X +b,x, =12.924+(-3.682) X1 +45.216X2 Hence, the estimated regression equation is: Y =12.924–3.682X1 + 45.216X2 The interpretation of the coefficient of X, is that a one unit change in X, will lead to a 3.682 unit decrease in Y when all other variables are held constant. Suppose we want to test whether the parameter B, is significant. Null hypothesis, Ho:B2 = 0 Alternative hypothesis, H.:B. 70 From the partial computer output, the estimates for the coefficient and standard error of Xare: b =-3.682, sel 6) = 2.630 Test statistic, t=4-8 se(b) -3.682–0 -=-1.4 2.630 Thus, test statistic equals to -1.4. Level of significance, a =1% =0.01 Sample size, n=16 Number of independent variables, k = 2 Degrees of freedom for error, df (error )= n-k-1=16-2-1 =13 At a=0.01 with 13 degrees of freedom, the two-tailed critical value from the t-table, t* =ta/2.d4 = 10.01/2,13 = 3.012

Thus, the t value obtained from the table which is used to test an individual parameter at the 1% level is 3.012. Carry out the test of significance for the parameter B. at the 1% level. The null hypothesis should be Ho: B = 0 Null hypothesis, H : B=0 Alternative hypothesis, H :B, 70 Level of significance, a =1% =0.01 Degrees of freedom for error , df (error)=n-k-1=16-2-1 =13 At a=0.01 with 13 degrees of freedom, the two-tailed critical value from the t-table, ** =ta/2.04 = 10.01/2,13 = 3.012 Decision rule: Reject H, if ||23.012 Test statistic, t=4-8.-3.682-0 =-1.4 seb, 2.630 Clearly, \ 3.012, thus fail to reject H, at 1% significance level. There is no sufficient evidence to conclude that the parameter B. is significant. Number of independent variables, k = 2 Implies, degrees of freedom for regression, df (reg) = k = 2 Thus, the degrees of freedom for the sum of squares explained by the regression (SSR) are: 2. From the partial computer output: Mean square error is, MS (error) = 485.3 And, degrees of freedom for error, df (error)=n-k-1=16-2-1 =13 Then, SSE = df (error )MS (error ) = 13x 485.3 = 6308.9

Thus, the sum of squares due to error (SSE) equals to 6308.9. From the partial computer output: Mean square due to regression is, MS (reg) = 2426.5 Mean square due to error is, MS (error) = 485.3 MS (reg)_2426.5 - 5 Test statistic, F =- MS (error) 485.3 The test statistic used to determine if there is a relationship among the variables equals 5. Carry out the test to determine if there is a relationship among the variables at the 5% level. The null hypothesis should there is no relationship among the variables. Null hypothesis, H.: There is no relationship among the variables. Alternative hypothesis, H,: There is a relationship among the variables. Level of significance, a = 5% = 0.05 Degrees of freedom, df = (k, n-k-1)=(2, 16-2-1)=(2, 13) From F-table, at a=0.05 with (2, 13) degrees of freedom, the one-tailed (right) critical value is F*= Fadd = F0.05.2.13 = 3.81 Decision rule: Reject Hif F>3.81 From the earlier calculation the value of test statistic F is 5, which is greater than the critical value 3.81. Thus, reject H. at 5% significance level. There is sufficient evidence to conclude that there is a relationship among the variables.