Question

The following regression output was obtained from a study of architectural firms. The dependent variable is the total amount of fees in millions of dollars.

Predictor	Coefficient	SE	T	P-vaule
Constant	7.096	3.245	2.187	0.010
x1	0.222	0.117	1.897	0.000
x2	-1.024	0.562	-1.822	0.028
x3	-0.337	0.192	-1.755	0.114
x4	0.623	0.263	2.369	0.001
x5	-0.056	0.029	-2.000	0.112

Analysis of Variance

Source	DF	SS	MS	F	P-Value
Regression	5	2009.28	401.9	7.33	0.000
Residual Error	50	2741.54	54.83
Total	55	4750.81

x1 is the number of architects employed by the company.

x2 is the number of engineers employed by the company.

x3 is the number of years involved with health care projects.

x4 is the number of states in which the firm operates.

x5 is the percent of the firm’s work that is health care−related.

A) Write out the regression equation. (Negative answers should be indicated by a minus sign. Round your answers to 3 decimal places.)

B) How large is the sample? How many independent variables are there?

c-1.) At the 0.05 significance level, state the decision rule to test: H0: β1 = β2 = β3 =β4 = β5 = 0; H1: At least one β is 0. (Round your answer to 2 decimal places.)

c-2.) Compute the value of the F statistic. (Round your answer to 2 decimal places.)

c-3.) What is the decision regarding H0: β1 = β2 = β3 = β4 = β5 = 0?

d-1.) State the decision rule for each independent variable. Use the 0.05 significance level. (Round your answers to 3 decimal places.) For x1 For x2 For x3 For x4 For x5 H0: β1 = 0 H0: β2 = 0 H0: β3 = 0 H0: β4 = 0 H0: β5 = 0 H1: β1 ≠ 0 H1: β2 ≠ 0 H1: β3 ≠ 0 H1: β4 ≠ 0 H1: β5 ≠ 0

d-2.) Compute the value of the test statistic. (Negative answers should be indicated by a minus sign. Round your answers to 3 decimal places.)

d-3.) For each variable, make a decision about the hypothesis that the coefficient is equal to zero.

Answer 1

A) Write out the regression equation.

y = 7.096 + 0.222 x1 -1.024 x2 -0.337 x3 + 0.623 x4 - 0.056 x5

B) How large is the sample?

sample size = total degrees of freedom + 1 = 55 + 1 = 56

Number of independent variables = 5

c-1.) At the 0.05 significance level, state the decision rule to test

H0: β1 = β2 = β3 =β4 = β5 = 0;

H1: At least one β is not equal 0.

if p-value is less than 0.05 reject null hypothesis otherwise fail to reject.

c-2.) Compute the value of the F statistic.

From the ANOVA table F = 7.33

c-3.) What is the decision regarding H0: β1 = β2 = β3 = β4 = β5 = 0?

since p-value (0.000) is less than 0.05 we reject null hypothesis and there is a significant evidence to conclude that At least one β is not equal to 0.

d-1.) State the decision rule for each independent variable. Use the 0.05 significance level. (Round your answers to 3 decimal places.)

For x1

H0: β1 = 0 , H1: β1 ≠ 0

For x2

H0: β2 = 0

H1: β2 ≠ 0

For x3

H0: β3 = 0

H1: β3 ≠ 0

For x4

H0: β4 = 0 , H1: β4 ≠ 0

For x5

H0: β5 = 0 H1: β5 ≠ 0

decision rule for each of then are "if p-value is less than 0.05 reject null hypothesis otherwise fail to reject."

d-2.) Compute the value of the test statistic.

For x1

T = 1.897

For x2

T = -1.822

For x3

T = -1.755

For x4

T = 2.369

For x5

T = -2.000

d-3.) For each variable, make a decision about the hypothesis that the coefficient is equal to zero.

For x1

since p-value (0.000) is less than 0.05 we reject null hypothesis and there is a significant evidence to conclude that β₁ is not equal to 0.

For x2

since p-value (0.028) is less than 0.05 we reject null hypothesis and there is a significant evidence to conclude that β₂ is not equal to 0.

For x3

since p-value (0.114) is more than 0.05 we reject null hypothesis and there is a significant evidence to conclude that β₃ is equal to 0.

For x4

since p-value (0.001) is less than 0.05 we reject null hypothesis and there is a significant evidence to conclude that At least one β₄ is not equal to 0.

For x5

since p-value (0.112) is more than 0.05 we reject null hypothesis and there is a significant evidence to conclude that β₅ is equal to 0.