26.What is the standard error of the estimate? A) 1.31. B) 0.81. C) 2.52. D) 1.71. The correct answer was A) The standard error of the estimate is equal to [SSE/(n – k – 1)]1/2 = [267.00/156]1/2 = approximately 1.31. 27.Consider the following regression equation: Salesi = 10.0 + 1.25 R&Di + 1.0 ADVi – 2.0 COMPi + 8.0 CAPi
where Sales is dollar sales in millions, R&D is research and development expenditures in millions, ADV is dollar amount spent on advertising in millions, COMP is the number of competitors in the industry, and CAP is the capital expenditures for the period in millions of dollars. Which of the following is NOT a correct interpretation of this regression information? A) If a company spends $1 million more on capital expenditures (holding everything else constant), Sales are expected to increase by $8.0 million. B) One more competitor will mean $2 million less in Sales (holding everything else constant). C) Increasing advertising dollars by $1 million (holding everything else constant), will result in $1 million additional Sales. D) If R&D and advertising expenditures are $1 million each, there are 5 competitors, and capital expenditures are $2 million, expected Sales are $8.25 million. The correct answer was D) Predicted sales = $10 + 1.25 + 1 – 10 + 16 = $18.25 million. 28.Consider the following estimated regression equation, with the standard errors of the slope coefficients as noted: Salesi = 10.0 + 1.25 R&Di + 1.0 ADVi – 2.0 COMPi + 8.0 CAPi where the standard error for the estimated coefficient on R&D is 0.45, the standard error for the estimated coefficient on ADV is 2.2 , the standard error for the estimated coefficient on COMP is 0.63, and the standard error for the estimated coefficient on CAP is 2.5. The equation was estimated over 40 companies. Using a 5 percent level of significance, which of the estimated coefficients are significantly different from zero? A) R&D, ADV, COMP, and CAP. B) COMP and CAP only. C) R&D, COMP, and CAP only. D) ADV and CAP only. The correct answer was C) The critical t-values for 40-4-1 = 35 degrees of freedom and a 5 percent level of significance are ± 2.03. The calculated t-values are:
t for R&D = 1.25 / 0.45 = 2.777
t for ADV = 1.0/ 2.2 = 0.455
t for COMP = -2.0 / 0.63 = -3.175
t for CAP = 8.0 / 2.5 = 3.2
Therefore, R&D, COMP, and CAP are statistically significant. 29.Toni Williams, CFA, has determined that commercial electric generator sales in the Midwest U.S. for Self-Start Company is a function of several factors in each area: the cost of heating oil, the temperature, snowfall, and housing starts. Using data for the most currently available year, she runs a cross-sectional regression where she regresses the deviation of sales from the historical average in each area on the deviation of each explanatory variable from the historical average of that variable for that location. She feels this is the most appropriate method since each geographic area will have different average values for the inputs, and the model can explain how current conditions explain how generator sales are higher or lower from the historical average in each area. In summary, she regresses current sales for each area minus its respective historical average on the following variables for each area. §
The difference between the retail price of heating oil and its historical average. §
The mean number of degrees the temperature is below normal in Chicago. §
The amount of snowfall above the average. §
The percentage of housing starts above the average. Williams used a sample of 26 observations obtained from 26 metropolitan areas in the Midwest U.S. The results are in the tables below. The dependent variable is in sales of generators in millions of dollars. Coefficient Estimates Table
| Variable
| Estimated Coefficient
| Standard Error of the Coefficient
| Intercept | 5.00 | 1.850 | $ Heating Oil | 2.00 | 0.827 | Low Temperature | 3.00 | 1.200 | Snowfall | 10.00 | 4.833 | Housing Starts | 5.00 | 2.333 | Analysis of Variance Table (ANOVA)
| | Source
| Degrees of Freedom
| Sum of Squares
| Mean Square
| | Regression | 4 | 335.20 | 83.80 | | Error | 21 | 606.40 | 28.88 | | Total | 25 | 941.60 |
| | | | | | | | |
One of her goals is to forecast the sales of the Chicago metropolitan area next year. For that area and for the upcoming year, Williams obtains the following projections: heating oil prices will be $0.10 above average, the temperature in Chicago will be 5 degrees below normal, snowfall will be 3 inches above average, and housing starts will be 3 percent below average. In addition to making forecasts and testing the significance of the estimated coefficients, she plans to perform diagnostic tests to verify the validity of the model’s results. According to the model and the data for the Chicago metropolitan area, the forecast of generator sales is: A) $65 million above the average. B) $35.2 million above the average. C) The average. D) $55 million above average. The correct answer was B) The model uses a multiple regression equation to predict sales by multiplying the estimated coefficient by the observed value to get: [5 + (2 * 0.10) + (3 * 5) + (10 * 3) + (5 * -3)] * $1,000,000 = $35.2 million. 30.Williams proceeds to test the hypothesis that none of the independent variables has significant explanatory power. He concludes that, at a 5 percent level of significance: A) none of the independent variables has explanatory power, because the calculated F-statistic does not exceed its critical value. B) at least one of the independent variables has explanatory power, because the calculated F-statistic exceeds its critical value. C) all of the independent variables have explanatory power, because the calculated F-statistic exceeds its critical value. D) at least one of the independent variables has explanatory power, because the calculated F-statistic does not exceed its critical value. The correct answer was B) From the ANOVA table, the calculated F-statistic is (mean square regression / mean square error) = (83.80 / 28.88) = 2.9017. From the F distribution table (4 df numerator, 21 df denominator) the critical F value is 2.84. Because 2.9017 is greater than 2.84, Williams rejects the null hypothesis and concludes that at least one of the independent variables has explanatory power. | 
发表于 2008-4-8 16:58
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