Question: Burmer Co. has accumulated data to use in preparing its annual profit plan for the upcoming year. The cost behavior pattern of the maintenance costs
Burmer Co. has accumulated data to use in preparing its annual profit plan for the upcoming year. The cost behavior pattern of the maintenance costs must be determined. Data regarding the machine hours and maintenance costs for the last year and the results of the regression analysis are as follows:
| Month | Maintenance Cost | Machine Hours | ||||
| Jan. | $ | 5,040 | 620 | |||
| Feb. | 3,600 | 420 | ||||
| Mar. | 4,320 | 520 | ||||
| Apr. | 3,380 | 390 | ||||
| May | 5,220 | 650 | ||||
| June | 3,550 | 400 | ||||
| July | 3,640 | 430 | ||||
| Aug. | 5,360 | 680 | ||||
| Sept. | 5,110 | 640 | ||||
| Oct. | 4,860 | 610 | ||||
| Nov. | 3,960 | 460 | ||||
| Dec. | 3,790 | 440 | ||||
| Sum | $ | 51,830 | 6,260 | |||
| Average | $ | 4,319 | 522 | |||
A staff assistant has run regression analyses on the data and obtained the following output using Excel: REGRESSION ANALYSIS Y (Dependent) Variable: Maintenance Cost X (Independent) Variable: Maintenance Hours
| Regression Statistics | |||||||||||
| Multiple R | 0.998210294 | ||||||||||
| R Square | 0.996423791 | ||||||||||
| Adjusted R Square | 0.99606617 | ||||||||||
| Standard Error | 47.0629563 | ||||||||||
| Observations | 12 | ||||||||||
| ANOVA | |||||||||||
| df | SS | MS | F | Significance F | |||||||
| Regression | 1 | 6171342.448 | 6171342 | 2786.257 | 1.44166E-13 | ||||||
| Residual | 10 | 22149.21856 | 2214.922 | ||||||||
| Total | 11 | 6193491.667 | |||||||||
| Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | ||||||
| Intercept | 783.7782188 | 68.34114772 | 11.46861 | 4.47E-07 | 631.504653 | 936.051785 | |||||
| Hours | 6.777102456 | 0.12839066 | 52.78501 | 1.44E-13 | 6.491030239 | 7.06317467 | |||||
A key statistic that indicates reliability of the regression is:
Multiple Choice
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12.
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47.0630.
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0.9964.
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783.338.
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6.777.
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