Question

In: Statistics and Probability

Production volume 400 450 550 600 700 750 Total cost 4000 5000 5400 5900 6400 7000...

Production volume

400

450

550

600

700

750

Total cost

4000

5000

5400

5900

6400

7000

An important application of regression analysis in accounting is in the estimation of cost. By collecting data on volume and cost and using the least squares method to develop an estimated regression equation relating volume and cost, an accountant can estimate the cost associated with a particular manufacturing volume. Consider the following sample of production volumes and total cost data for a manufacturing operation.

Excel File: data12-21.xls
  1. Compute b1 and b0 (to 2 decimals if necessary).
    b1
    b0

    Complete the estimated regression equation (to 2 decimals if necessary).
    ŷ =  +  x
  2. What is the variable cost per unit produced (to 1 decimal)?
  3. Compute the coefficient of determination (to 4 decimals). Note: report r2 between 0 and 1.
    r2 =

    What percentage of the variation in total cost can be explained by the production volume (to 2 decimals)?
    %
  4. The company's production schedule shows 500 units must be produced next month. What is the estimated total cost for this operation (to 2 decimals)?
    $

Solutions

Expert Solution

Answer:

Here let y is total cost and x is production volume.

First enter the given data in excel:

Production Volume(x) Total cost(y)
400 4000
450 5000
550 5400
600 5900
700 6400
750 7000

To use the regression analysis in excel use the following path:

Data> Data Analysis ( upper right corner) > select Regression > select range both x and y variables> Give output range> OK

The output as below:

SUMMARY OUTPUT
Regression Statistics
Multiple R 0.979127101
R Square 0.958689879
Adjusted R Square 0.948362349
Standard Error 241.5229458
Observations 6
ANOVA
df SS MS F Significance F
Regression 1 5415000 5415000 92.82857 0.00064897
Residual 4 233333.3333 58333.33
Total 5 5648333.333
Coefficients Standard Error t Stat P-value Lower 95% Upper 95%
Intercept 1246.666667 464.1599341 2.685856 0.054894 -42.04791038 2535.381244
Production Volume(x) 7.6 0.788810638 9.634759 0.000649 5.409910566 9.790089434

From output:

a)

The coefficients are

b0 = 1246.66667 = 1246.67 ... rounded to 2 decimals

b1 = 7.6

Hence regression is y^ = 1246.67 + 7.6*x

b)

Here slope coefficient is 7.6, hence variable cost is 7.6 per one unit is produced.

c)

From summary output in bold case

coefficient of determination = r2 = 0.95868 = 0.9587 ...( rounded to 4 decimals)

This interpreted as 95.87 % of variation of estimation of cost(y) is explained by production volume(x).

d)

The company's production schedule shows 500 units must be produced next month (i.e x = 500 )

Hence total cost = y = 1246.67 + 7.6 * 500

= 5046.67 $

R code and Output for reference:

> x=c(400,450,550,600,700,750)
> y=c(4000,5000,5400,5900,6400,7000)
> regression=lm(y~x)
> regression

Call:
lm(formula = y ~ x)

Coefficients:
(Intercept) x
1246.7 7.6

> summary(regression)

Call:
lm(formula = y ~ x)

Residuals:
1 2 3 4 5
-286.67 333.33 -26.67 93.33 -166.67
6
53.33

Coefficients:
Estimate Std. Error t value
(Intercept) 1246.6667 464.1599 2.686
x 7.6000 0.7888 9.635
Pr(>|t|)
(Intercept) 0.054894 .
x 0.000649 ***
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1
‘ ’ 1

Residual standard error: 241.5 on 4 degrees of freedom
Multiple R-squared: 0.9587,   Adjusted R-squared: 0.9484
F-statistic: 92.83 on 1 and 4 DF, p-value: 0.000649

Coefficients and R squared are in bold case , which gives same output.


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