The following regression analysis was designed to explain firm revenues from the number of employees, using a company database of 42 firms. Revenues are measured in millions of dollars, while employees represent the number of people.
Regression analysis to predict Revenue from Employees.
The prediction equation is:
Revenue = 23,431
+ 0.2573 Employees
0.366 R squared
79,413 Standard error of estimate
42 Number of observations
23.0723 F statistic
0.000022 p value
Coeff Lower CI Upper CI StdErr t p Signif?
Constant 23,431 -5,802 52,664 14,464 1.6199 0.1131 no(p>.05)
Employees 0.2573 .1490 .3655 .0536 4.8034 .000022 y(p<.001)
---Find the marginal value of one additional employee in terms of additional dollars of revenue, giving both the name of the usual statistical measure and its numerical value. Name: Value:
---What percentage of the variability of revenues is explained here by the number of employees? Name: Usual statistical measure and its numerical value:
-----Your co-worker raises the possibility that there is actually no real connection between employees and revenues in this data set, and that the apparent connection is entirely due to random chance )i.e., just noise in the data). How should you respond to this suggestion, based on the evidence from the regression analysis?
--What level of revenues would you expect for a firm with 50,000 employees?
Please see the attachment for solution. If you need any further clarification please ask me. Thank you.
1)Find the marginal value of one additional employee in terms of additional dollars of revenue, giving both the name of the usual statistical measure and its numerical value.
Name: Coefficient of Employees in the regression equation Value: 0.2573
The regression equation to predict Revenue from Employees is:
Revenue = 23,431 + 0.2573 Employees
In the regression equation, the coefficient of Employees (Slope of the regression equation) indicates the average change in the Revenue value for a unit increase in the number of Employees. Here the coefficient of Employees is 0.2573. Thus if ...
The solution contains an application of the regression analysis.