[Solution Library] Juan Martinez wants to develop a good regression model that can predict the market value of a firm. He has data that is given in the following
Question:
Juan Martinez wants to develop a good regression model that can predict the market value of a firm. He has data that is given in the following table.
| Firm |
Market Value
(in millions) |
Assets
(in millions) |
Sales
(in millions) |
Employees |
| Goodyear | $3,023 | $8,610 | $10,328 | 121,586 |
| Firestone | 1,386 | 2,593 | 3,712 | 55,000 |
| GenCorp | 1,949 | 2,119 | 3,099 | 26,700 |
| Goodrich | 1,319 | 1,820 | 2,553 | 11,914 |
| Premark Intl | 867 | 1,407 | 1,959 | 21,900 |
| Dayco | 254 | 511 | 911 | 4,000 |
| Armstrong | 191 | 521 | 800 | 7,505 |
| Rubbermaid | 1,969 | 579 | 795 | 6,103 |
| Cooper | 375 | 368 | 578 | 5,398 |
| Dorsey | 159 | 190 | 527 | 4,400 |
| Danaher | 252 | 741 | 524 | 6,950 |
| Carlisle | 314 | 322 | 466 | 4,949 |
| Millipore | 1,778 | 398 | 443 | 4,868 |
| Standard Products | 468 | 199 | 443 | 5,700 |
| Lancaster Colony | 267 | 250 | 423 | 5,100 |
The following are the results of a single variable regression run in Excel in which he modeled market value in terms of assets only.
| Regression Statistics | ||||||||
| Multiple R | 0.7551 | |||||||
| R Square | 0.5702 | |||||||
| Adjusted R Square | 0.5371 | |||||||
| Standard Error | 601.7672 | |||||||
| Observations | 15 | |||||||
| Analysis of Variance | ||||||||
| df | Sum of Squares | Mean Square | F | Significance F | ||||
| Regression | 1 | 6245258.03 | 6245258.03 | 17.2462 | 0.0011 | |||
| Residual | 13 | 4707609.57 | 362123.813 | |||||
| Total | 14 | 10952867.6 | ||||||
| Coefficients | Standard Error | t Statistic | P-value | Lower 95% | Upper 95% | |||
| Intercept | 542.5593 | 186.5611 | 2.9082 | 0.0115 | 139.5185 | 945.6001 | ||
| Assets | 0.3118 | 0.0751 | 4.1529 | 0.0010 | 0.1496 | 0.4741 | ||
- [10 points] Based on the computer output, evaluate the regression model.
-
[10 points] Suppose that a firm had assets of $1,800 million. Calculate a 95.4% confidence interval for this firm’s market value.
Subsequently, Juan decided to see if he could get an improved regression model by adding sales and
employees as independent variables. The results of the regression are presented in the following Excel
spreadsheet.Regression Statistics Multiple R 0.7697 R Square 0.5925 Adjusted R Square 0.4814 Standard Error 636.9900 Observations 15 Analysis of Variance df Sum of Squares Mean Square F Significance F Regression 3 6489548.83 2163182.94 5.3312 0.0164 Residual 11 4463318.77 405756.251 Total 14 10952867.6 Coefficients Standard Error t Statistic P-value Lower 95% Upper 95% Intercept 451.8043 242.1464 1.8658 0.0832 -81.1566 984.7652 Assets -0.0100 0.8770 -0.0115 0.9910 -1.9402 1.9201 Sales 0.4698 0.7479 0.6281 0.5400 -1.1764 2.1159 employees -0.0169 0.0275 -0.6148 0.5486 -0.0776 0.0437 - [10 points] Based on the computer outputs from both the first and second regression models, which of the two models would you prefer and why?
- [10 points] What is multicollinearity? How would you check for multicollinearity?
Price: $2.99
Solution: The downloadable solution consists of 5 pages
Deliverable: Word Document 