Question

Comparison of Regression Equations

Friendly Bank is attempting to determine the cost behavior of its small business lending operations. One of the major activities is the application activity. Two possible activity drivers have been mentioned: application hours (number of hours to complete the application) and number of applications. The bank controller has accumulated the following data for the setup activity:

Month		Application Costs		Application Hours		Number of Applications
February		$7,700		2,000		70
March		7,650		2,100		50
April		10,052		3,000		50
May		9,400		2,700		60
June		9,584		3,000		20
July		8,480		2,500		40
August		8,550		2,400		60
September		9,735		2,900		50
October		10,500		3,000		90

Required:

In the computations below, use intercepts rounded to the nearest whole number and use X variable coefficients rounded to 2 decimal places.

1. Estimate a regression equation with application hours as the activity driver and the only independent variable. If the bank forecasts 2,600 application hours for the next month, what will be the budgeted application cost? Round your answer to the nearest dollar.

$_____________________

2. Estimate a regression equation with number of applications as the activity driver and the only independent variable. If the bank forecasts 80 applications for the next month, what will be the budgeted application cost? Round your answer to the nearest dollar.

$_____________________

3. Which of the two regression equations do you think does a better job of predicting application costs? Choose the correct answer: 1. Applications hours or 2. Number of applications.

4. Run a multiple regression to determine the cost equation using both activity drivers. What are the budgeted application costs for 2,600 application hours and 80 applications? Round your answer to the nearest cent.

$_______________

Answer 1

X	Application hours
Y	Application Cost

	x	y	XY	x*x	Y*Y
February	2000	7700	15400000	4000000	59290000
March	2100	7650	16065000	4410000	58522500
April	3000	10052	30156000	9000000	101042704
May	2700	9400	25380000	7290000	88360000
June	3000	9584	28752000	9000000	91853056
July	2500	8480	21200000	6250000	71910400
August	2400	8550	20520000	5760000	73102500
September	2900	9735	28231500	8410000	94770225
October	3000	10500	31500000	9000000	110250000
Total	23600	81651	217204500	63120000	749101385

a = (?y)(?x*x)-(?x)(?xy)/n(?x*x)-(?x)*(?x)

b = n(?xy)-(?x)(?y)/n(?x*x)-(?x)*(?x)

a =

27784920000/11120000 = 2498.643885

b = 2.5

Regression Equation = y = 2498.64+2.5x

If x = 2600 hours then Application Cost(y) = 2498.64+2.5(2600)=$8999

2)

X = Number of Applications

Y = Application Cost

	x	y	XY	x*x	Y*Y
February	70	7700	539000	4900	59290000
March	50	7650	382500	2500	58522500
April	50	10052	502600	2500	101042704
May	60	9400	564000	3600	88360000
June	20	9584	191680	400	91853056
July	40	8480	339200	1600	71910400
August	60	8550	513000	3600	73102500
September	50	9735	486750	2500	94770225
October	90	10500	945000	8100	110250000
Total	490	81651	4463730	29700	749101385

a = (?y)(?x*x)-(?x)(?xy)/n(?x*x)-(?x)*(?x)

b = n(?xy)-(?x)(?y)/n(?x*x)-(?x)*(?x) a = 237807000/27200 = 8742.90 b = 6.05 Regression Equation = y = 8742.90+6.05x If x = 80 Applications then Application Cost(y) = 8742.90+6.05(80)=$9227 3) Application hours shall be taken as better job of predicting Application cost since Cost is incurred in proportionate to time consumed for the respective number of applications