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		9.387			3.069			3.059			0.010
x1		0.232			0.204			1.137			0.000
x2	−	1.214			0.584		−	2.079			0.028
x3	−	0.273			0.424		−	0.644			0.114
x4		0.642			0.362			1.773			0.001
x5	−	0.060			0.028		−	2.143			0.112

Analysis of Variance
Source	DF		SS		MS		F		p-value
Regression	5		2,364.50		472.9		10.29		0.000
Residual Error	53		2,436.07		45.96
Total	58		4,800.57

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.)

Answer 1

a)  
Y^= 9.387+0.232*X1 -1.214*X2 -0.273*X3+0.642*X4 -0.060*X3  
  
b)  
total sample=59  
independent variables=5  
  
c-1)  
F(0.05,5,53)=   1.00
reject Ho if test stat > 1.00  
  
c2)  
value of F=10.29  
  
c3)  
decision: reject Ho  
  
d1)  
t critical value,t(.05/2,53)=   2.006
reject Ho if t > 2.006  
  
d2)  

Predictor

t

x1

1.137

x2

−

2.079

x3

−

0.644

x4

1.773

x5

−

2.143