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 Coefficient			t			p-value
Constant		7.166			3.075			2.330			0.010
x1		0.228			0.302			0.755			0.000
x2	−	1.184			0.586		−	2.020			0.028
x3	−	0.193			0.110		−	1.755			0.114
x4		0.572			0.293			1.952			0.001
x5	−	0.056			0.022		−	2.545			0.112

Analysis of Variance
Source	DF		SS		MS		F		p-value
Regression	5		1,847.24		369.4		5.60		0.000
Residual Error	41		2,704.20		65.96
Total	46		4,551.44

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.

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

2. How large is the sample? How many independent variables are there?

1. 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.)

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

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

1. 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

1. 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.)

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

Next Visit question map

Question 3 of 10 Total3 of 10

Prev

Answer 1

a. Regression equation y = 7.166 + 0.288x₁ - 1.184x₂ - 0.193x₃ +0.572x₄  - 0.056x₅

a. Sample size = df Total+1 = 46+1 = 47

Independent variables = 5

c.1. Reeject the null hypothesis if p-value < 0.05

c.2. F = 5.60

c.3. p-value (ANOVA) = 0.00

Since p-value is less than 0.05, we reject the null hypothesis and conclude that at least one β is 0.

c.d. Critical t (Using Excel function T.INV.2T(probability,n-2)) = T.INV.2T(0.05,45) = 2.014

	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-1. Decision rule	Reject null hypothesis if Coefficien t > 2.104	Reject null hypothesis if Coefficien t > 2.104	Reject null hypothesis if Coefficien t > 2.104	Reject null hypothesis if Coefficien t > 2.104	Reject null hypothesis if Coefficien t > 2.104
d-2. Test statistic	0.755	-2.02	-1.755	1.952	-2.545
d-3. Decision rule	Do not reject null hypothesis, β1 = 0	Do not reject null hypothesis, β2 = 0	Do not reject null hypothesis, β3 = 0	Do not reject null hypothesis, β4 = 0	Do not reject null hypothesis, β5 = 0