An important application of regression analysis in accounting is in the estimati
ID: 3351444 • Letter: A
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.
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 ($) 450 3900 550 4900 650 5600 700 6400 750 7300 850 7900 The data on the production volume and total cost y for particular manufacturing operation were used to develop the estimated regression equation y--835.92 + 10.38z. a. The company's production schedule shows that 800 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) (to 3 decimals) (to 2 decimals) t-value pred Prediction Interval for an Individual Value: (to whole number) C. If an accounting cost report at the end of next month shows that the actual production cost during the month was 86000, should managers be concerned about incurring such a high total cost for the month? Discuss. Based on one month, 86000 select your ans e out of the prediction interval A sequence of five to seven months it consistent high costs should cause concern.Explanation / Answer
y= -835.92 + 10.38x
y= -835.92+10.38*800
point estimate of y= 202.08
s=standard deviation=0.30
calculation
t=2.571 for 95% confidence interval anf 5 degrees of freedom
c)
9.11 for 1 unit
1->1/9.11
6000->1/9.11*600 =658.61 units
so they should not concern as this is more than the average units produced.
mean difference square 1) 450 3900 8.666667 0.447257 0.200039 2) 550 4900 8.909091 0.204833 0.041957 3) 650 5600 8.615385 0.498539 0.248542 4) 700 6400 9.142857 -0.02893 0.000837 5) 750 7300 9.733333 -0.18019 0.03247 6) 850 7900 9.294118 -0.18019 0.03247 3950 36000 0.556314=sum mean 9.113924 0.092719=variance 0.304498=standard deviationRelated Questions
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