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		8.870			3.128			2.836			0.010
x1		0.212			0.050			4.240			0.000
x2	−	1.004			0.545		−	1.842			0.028
x3	−	0.246			0.387		−	0.636			0.114
x4		0.669			0.326			2.052			0.001
x5	−	0.062			0.030		−	2.067			0.112

Analysis of Variance
Source	DF		SS		MS		F		p-value
Regression	5		2,029.13		405.8		8.48		0.000
Residual Error	54		2,583.92		47.85
Total	59		4,613.05

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 values 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.

Answer 1

(a)

The regression model is

(b)

The sample size is:

n= 59+1 = 60

The number of independent variables is: k = 5

(c)

The degree of freedom of numerator: df1=5

The degree of freedom of denominator: df2 =54

The critical value of F using excel function "=FINV(0.05,5,54)" is 2.39.

Reject H0, if F > 2.39

c-2)

The test statistics is

F = 8.48

c-3)

Since F lies in the rejection region so we reject the null hypothesis and we can conclude that at least one coefficient is different from zero.

d-1)

The critical values of t for df = 54 are +/- 2.005.

Rejection region:

If t < -2.005 or t > 2.005, reject H0

Excel function used for critical value "=TINV(0.05,54)"

d-2)

	T
X1	4.240
X2	-1.842
X3	-0.636
X4	2.052
X5	-2.067

e)

Since t for X2 and X3 does not lie in the rejection region so these are not significant to the model and we can eliminate them.