Question

Armer Company is accumulating data to use 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 the use of linear regression to derive an equation for maintenance hours and costs. Data regarding the maintenance hours and costs for the last year and the results of the regression analysis follow: Maintenance Month Cost Machine Hours Jan. $ 5,150 860 Feb. 3,950 700 Mar. 4,550 780 Apr. 3,770 680 May 5,300 880 June 3,910 690 July 3,980 700 Aug. 5,420 1,000 Sept. 5,210 870 Oct. 5,000 850 Nov. 4,250 730 Dec. 4,110 720 Sum $54,600 9,460 Average $ 4,550 788 Average cost per hour $ 6.00 a (intercept) $125.3539 b (coefficient) 5.9307 Standard error of the estimate 171.8873 R-squared 0.9314 t-value for b 11.6480 If Armer Company uses the high-low method of analysis, the equation for the relationship between hours of activity and maintenance cost follows: (Round your answers to 1 decimal place.) Multiple Choice O y =4.550 - 788x 0 = 234 - 6x O y= 220 -5.2x

Answer 1

High low Method
Variable Cost per unit = (Cost at highest activity - Cost at lowest activity) / (Highest activity - Lowest Activity)
= ($5420 - $3910) / (1000 - 690)
= $4.87 per hour

Fixed Cost = $5420 - 1000 x $4.87 = $550

Equation
y = 550 + 4.87x