The McGraw 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: (CMA adapted) $$\begin{array} { l r c }
& \begin{array} { c }
\text { Hours of } \
\text { Activity }
\end{array} & \begin{array} { c }
\text { Maintenance } \
\text { Costs }
\end{array} \
\text { January } & 480 & \$ 4,200 \
\text { February } & 320 & 3,000 \
\text { March } & 400 & 3,600 \
\text { April }& 300 & 2,820 \
\text { May } & 500 & 4,350 \
\text { June }& 310 & 2,960 \
\text { July } & 320 & 3,030 \
\text { August } & 520 & 4,470 \
\text { September } & 490 & 4,260 \
\text { October } & 470 & 4,050 \
\text { November } & 350 & 3,300 \
\text { December }& 340 & 3,160 \
\text { Sum } & 4,800 & \$ 3,200 \
\text { Average }& 400 & 3,600 \
\text { A coefficient } & & 684.65 \
\text { B coefficient } & & 7.2884 \
\text { Standard error of the a coefficient } & 49.515 & \
\text { Standard error of the b coefficient } & 0.12126 & \
\text { Standard error of the estimate } & & 34.469 \
\mathrm { R } ^ { 2 } & & 0.99724 \
\text { T-value a } & & 13.827 \
\text { T-value b } & & 60.105
\end{array}$$
- What would be the cost equation if the high-low method is used?
A) Maintenance Costs = $9.00 × Hours of Activity.
B) Maintenance Costs = 3,600 + (400 × Hours of Activity).
C) Maintenance Costs = $570 + ($7.50 × Hours of Activity).
D) Maintenance Costs = $34.469 + ($0.99724 × Hours of Activity).

Question

The McGraw 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: (CMA adapted) $$\begin{array} { l r c } 
& \begin{array} { c } 
\text { Hours of } \
\text { Activity }
\end{array} & \begin{array} { c } 
\text { Maintenance } \
\text { Costs }
\end{array} \
\text { January } & 480 & \$ 4,200 \
\text { February } & 320 & 3,000 \
 \text { March } & 400 & 3,600 \
\text { April }& 300 & 2,820 \
\text { May }  & 500 & 4,350 \
\text { June }& 310 & 2,960 \
\text { July }  & 320 & 3,030 \
\text { August } & 520 & 4,470 \
\text { September } & 490 & 4,260 \
\text { October } & 470 & 4,050 \
\text { November } & 350 & 3,300 \
\text { December }& 340 & 3,160 \
 \text { Sum } & 4,800 & \$ 3,200 \
 \text { Average }& 400 & 3,600 \
\text { A coefficient } & & 684.65 \
\text { B coefficient } & & 7.2884 \
\text { Standard error of the a coefficient } & 49.515 & \
\text { Standard error of the b coefficient } & 0.12126 & \
\text { Standard error of the estimate } & & 34.469 \
\mathrm { R } ^ { 2 } & & 0.99724 \
\text { T-value a } & & 13.827 \
\text { T-value b } & & 60.105
\end{array}$$ 
- What would be the cost equation if the high-low method is used?
A) Maintenance Costs = $9.00 × Hours of Activity.
B) Maintenance Costs = 3,600 + (400 × Hours of Activity).
C) Maintenance Costs = $570 + ($7.50 × Hours of Activity).
D) Maintenance Costs = $34.469 + ($0.99724 × Hours of Activity).

Quizplus · Accepted Answer

The high-low method uses the highest and lowest activity levels to estimate the variable cost per unit and the fixed cost. The highest activity level is 520 hours with a cost of $4,470, and the lowest is 300 hours with a cost of $2,820. The variable cost per hour is ($4,470 - $2,820) / (520 - 300) = $7.50. The fixed cost is calculated as $4,470 - (520 × $7.50) = $570. Thus, the cost equation is Maintenance Costs = $570 + ($7.50 × Hours of Activity).

The McGraw Company Is Accumulating Data to Be Used in Preparing