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The following regression output was obtained from a study of architectural firms. The dependent variable is...

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.)
  1. 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 = β34 = 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.

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Expert Solution

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

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