Question

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.

Production Volume (units)	Total Cost ($)
400	3800
500	4300
600	5300
650	5900
750	6300
800	7100

The data on the production volume  and total cost  for particular manufacturing operation were used to develop the estimated regression equation.

a. The company’s production schedule shows that  units must be produced next month. What is the point estimate of the total cost for next month?

(to 2 decimals)

b. Develop a  prediction interval for the total cost for next month.

	(to 2 decimals)
-value	(to 3 decimals)
	(to 2 decimals)

Prediction Interval for an Individual Value:

( ,   ) (to whole number)

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. Production Volume (units) Total Cost ($) 400 3800 500 4300 600 5300 650 5900 750 6300 800 7100 The data on the production volume x and total cost y for particular manufacturing operation were used to develop the estimated regression equation y = 416.91 +8.16x. a. The company's production schedule shows that 550 units must be produced next month. What is the point estimate of the total cost for next month? (to 2 decimals) b. Develop a 95% prediction interval for the total cost for next month. (to 2 decimals) S t-value (to 3 decimals) Spred (to 2 decimals) Prediction Interval for an Individual Value: (to whole number)

Answer 1

a)

Answer:

4905.88

Explanation:

The regression analysis is done in excel by following these steps,

Step 1: Write the data values in excel,

Step 2: DATA > Data Analysis > Regression > OK. The screenshot is shown below,

IE INSERT PAGE LAYOUT FORMULAS DATA REVIEW 000 - Data Analysis Х ? OK Cancel AZA Data Analysis Analysis Tools Fourier Analysis Histogram Moving Average Random Number Generation Rank and Percentile Regression Sampling t-Test: Paired Two Sample for Means t-Test: Two-Sample Assuming Equal Variances t-Test: Two-Sample Assuming Unequal Variances Help 36.6

Step 3: Select Input Y Range: 'Y' column, Input X Range: 'X' column then OK. The screenshot is shown below,

A1 X ✓ fx A B C E F H — 1 x iy 4001 2 ! 3 ? X 4 500! 600! 650 750 OK 5 $B$1:$B$7 S 3800 4300 Regression 5300 Input 5900 Input Y Range: 6300 Input X Range: 7100 Labels Confidence Level: 6 Cancel $A$1:$A$7| 7 8001 Help 8 Constant is Zero 95 % 9 10 11 SJS1 12 13 14 15 16 Output options Qutput Range: New Worksheet Ply: New Workbook Residuals Residuals Standardized Residuals Normal Probability Normal Probability Plots Residual Plots Line Fit Plots 17 18 19

The result is obtained. The screenshot is shown below,

J K L M N 0 Р 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.989225 5 R Square 0.978565 6 Adjusted R Square 0.973207 7 Standard Error 203.3271 8 Observations 6 9 df 10 ANOVA 11 12 Regression 13 Residual 14 Total SS MS F gnificance F 1 7549632 7549632 182.6145 0.000174 4 165367.6 41341.91 5 7715000 UT 16 17 Intercept 18 x Coefficientandard Err t Stat P-value Lower 95%Upper 95% 416.9118 381.587 1.092573 0.335972 -642.544 1476.367 8.161765 0.603972 13.51349 0.000174 6.484871 9.838658

The regression equation is,

Y = 416.9118 +8.1618X

For X = 550

Y = 416.9118 + 8.1618 x 550 4905.882

d)

Answer:

$s=203.33$

$t-\text{value}=2.776$

$s_{\text{Pred}} =223.28$

$\text{95\% PI}=(4285.96,5525.80)$

Explanation:

From the regression output summary,

The standard error of the regression is,

$s=203.3271\approx 203.33$

The t-critical value is obtained from t-distribution table for significance level = 0.05 and degree of freedom = n - 2 = 6 - 2 = 4

$t_{1-\alpha/2}=2.7764$

The standard error of the prediction is obtained using the following formula,

$s_{\text{Pred}} =s\sqrt{1+\frac{1}{n}+\frac{(X_{\text{predictor}}-\overline{X})^2}{\sum X^2-\frac{1}{n}(\sum X)^2}}$

From the data values,

x	x^2
400	160000
500	250000
600	360000
650	422500
750	562500
800	640000
Sum=3700	2395000

$\overline{X}=\frac{3700}{6}=616.6667$

$\sum X^2=2395000$

$s_{\text{Pred}} =s\sqrt{1+\frac{1}{6}+\frac{(550-616.6667)^2}{2395000-\frac{1}{6} (3700)^2}}$

$s_{\text{Pred}} =223.2789$

Now, the confidence interval for X_predictor = 550 is obtained using the formula,

$\text{PI}=\widehat{Y}\pm t_{1-\alpha/2}\times s_{\text{Pred}}$

$\text{PI}=4905.882\pm 2.7764\times 223.2789$

$\text{95\% PI}=(4285.96,5525.80)$