To validate whether a candidate cost driver has a strong cause-and-effect relationship with a cost pool, plant managers use Simple Linear Regression. This statistical method tests how well changes in the driver (independent variable, $X$) explain changes in total activity cost (dependent variable, $Y$).
Linear Regression Equation
The relationship is expressed as:
$$Y = a + bX$$
Where:
- $Y$ = Total Activity Cost (Dependent Variable)
- $X$ = Cost Driver Volume (Independent Variable)
- $a$ = Fixed Cost component (Y-intercept)
- $b$ = Variable Cost per unit of driver (Slope)
Step-by-Step Example: Machine Maintenance Cost vs. Machine Hours
A plant manager wants to evaluate if Machine Hours ($X$) is a valid cost driver for Monthly Maintenance Cost ($Y$).
1. Historical Plant Data (6 Months)
| Month | Machine Hours (X) | Maintenance Cost (Y) | X2 | XY |
| Jan | 100 | $2,200 | 10,000 | $220,000 |
| Feb | 120 | $2,500 | 14,400 | $300,000 |
| Mar | 150 | $3,100 | 22,500 | $465,000 |
| Apr | 180 | $3,700 | 32,400 | $666,000 |
| May | 200 | $4,000 | 40,000 | $800,000 |
| Jun | 220 | $4,500 | 48,400 | $990,000 |
| Sum ($\sum$) | 970 | $20,000 | 167,700 | $3,441,000 |
- $n = 6$ months
- Mean of $X$ ($\bar{X}$) = $\frac{970}{6} = 161.67 \text{ hours}$
- Mean of $Y$ ($\bar{Y}$) = $\frac{20,000}{6} = \$3,333.33$
2. Calculate Slope ($b$) and Fixed Intercept ($a$)
$$\text{Slope } (b) = \frac{n \sum (XY) - (\sum X)(\sum Y)}{n \sum (X^2) - (\sum X)^2}$$
$$b = \frac{6(3,441,000) - (970)(20,000)}{6(167,700) - (970)^2} = \frac{20,646,000 - 19,400,000}{1,006,200 - 940,900} = \frac{1,246,000}{65,300} \approx \mathbf{19.08}$$
$$\text{Intercept } (a) = \bar{Y} - b\bar{X} = 3,333.33 - (19.08 \times 161.67) = 3,333.33 - 3,084.66 = \mathbf{248.67}$$
Cost Function:
$$\text{Maintenance Cost } (Y) = \$248.67 + \$19.08(X)$$
3. Evaluate Statistical Validity
Plant managers evaluate two metrics to confirm whether Machine Hours is a valid driver:
A. Coefficient of Determination ($R^2$)
$R^2$ measures the percentage of cost variance explained by the driver.
$$R^2 = \frac{[\sum (X - \bar{X})(Y - \bar{Y})]^2}{\sum (X - \bar{X})^2 \sum (Y - \bar{Y})^2} \approx \mathbf{0.995}$$
- Rule of Thumb: An $R^2 \ge 0.80$ indicates a strong cost driver.
- Result: $R^2 = 0.995$ ($99.5\%$), meaning machine hours almost completely explain maintenance cost variations.
B. Visual / Residual Check
Plotted data points must closely align along the linear regression line without wild scatter or non-linear curves.
Cost ($)
^
| * (Jun: 220 hrs, $4,500)
| * (May: 200 hrs, $4,000)
| * (Apr: 180 hrs, $3,700)
| * (Mar: 150 hrs, $3,100)
| * (Feb: 120 hrs, $2,500)
|* (Jan: 100 hrs, $2,200)
+--------------------------------------------> Machine Hours (X)
Decision Criteria Summary
| Metric | Calculated Value | Acceptance Threshold | Verdict |
| $R^2$ (Correlation Strength) | $0.995$ | $\ge 0.80$ | Valid Driver |
| Slope Sign ($b$) | $+19.08$ | Positive | Logical Relationship |
| Fixed Overhead ($a$) | $\$248.67$ | $\ge 0$ | Economically Plausible |
Because all criteria are met, Machine Hours is validated as a statistically reliable cost driver for Activity-Based Costing allocation.