The Armer Company is accumulating data to be used in preparing its annual profit plan for the coming year. The cost behavior pattern of the maintenance costs must be determined. The accounting staff has suggested that linear regression be employed to derive an equation in the form of y = a + bx for maintenance costs. Data regarding the maintenance hours and costs for last year and the results of the regression analysis are as follows: Hours of Activity Maintenance Costs January 440 $ 4,400 February 290 2,800 March 450 3,400 April 340 3,050 May 480 4,190 June 340 2,600 July 340 3,100 August 460 4,780 September 430 4,970 October 550 3,980 November 320 3,600 December 340 3,500 Sum 4,780 44,370 Average 398 3,698 A coefficient 1,110.22 B coefficient 6.4953 Standard error of the a cofficient 912.565 Standard error of the a cofficient 2.24976 Standard error of the estimate 596.788 R2 0.45460 T-value a 1.217 T-value b 2.887 What would be the cost equation if regression analysis is used?
Cost Equation = $1110.22 + $6.50 x
SUMMARY OUTPUT | ||||||||
Regression Statistics | ||||||||
Multiple R | 0.674243 | |||||||
R Square | 0.454603 | |||||||
Adjusted R Square | 0.400063 | |||||||
Standard Error | 596.7885 | |||||||
Observations | 12 | |||||||
ANOVA | ||||||||
df | SS | MS | F | Significance F | ||||
Regression | 1 | 2968660 | 2968660 | 8.335268 | 0.016188 | |||
Residual | 10 | 3561565 | 356156.5 | |||||
Total | 11 | 6530225 | ||||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | |
Intercept | 1110.22 | 912.5652 | 1.216593 | 0.251694 | -923.102 | 3143.542 | -923.102 | 3143.542 |
X Variable 1 | 6.495263 | 2.249764 | 2.887086 | 0.016188 | 1.482476 | 11.50805 | 1.482476 | 11.50805 |
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