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Stat 102 Solutions for Homework3-2012-2

Stat 102 Solutions for Homework3-2012-2 - Solutions for...

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Solutions for Homework 3, STAT 102, Spring 2012 0. Problem 6 on Page 90. (We already did this problem for HW, so you don’t need to do it again. It has the same data set as problem 14, below, so you may want to look at it again.) 1. Problem 12 on Page 111. Preparatory analyses: The regression analysis of Overhead costs(in \$1000) on Production Units (in 10000) gives: Bivariate Fit of Overhead costs(in \$1000) By Production Units (in 10000) 11 12 13 14 15 16 17 18 Overhead costs(in \$1000) 4 5 6 7 8 9 10 11 12 Production Units (in 10000) Linear Fit Linear Fit Overhead costs(in \$1000) = 7.1 + 0.9107143*Production Units (in 10000) Summary of Fit RSquare 0.889486 RSquare Adj 0.867383 Root Mean Square Error 0.759652 Mean of Response 14.38571 Observations (or Sum Wgts) 7 Analysis of Variance Source DF Sum of Squares Mean Square F Ratio Model 1 23.223214 23.2232 40.2432 Error 5 2.885357 0.5771 Prob > F C. Total 6 26.108571 0.0014 Parameter Estimates Term Estimate Std Error t Ratio Prob>|t| Intercept 7.1 1.183832 6.00 0.0018 Production Units (in 10000) 0.9107143 0.143561 6.34 0.0014 1

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Moments Mean 8 Std Dev 2.1602469 Std Err Mean 0.8164966 upper 95% Mean 9.9978952 lower 95% Mean 6.0021048 N 7 (a) The point estimate of the overhead cost on average, for production runs of 80,000 units is: Overhead costs(in \$1000) = 7.1 + 0.9107143*Production Units (in 10000) =7.1 + 0.9107143*8=14.3857 (\$1000). (b) The standard error of the estimate is : and , hence the 95% confidential interval estimate of the overhead cost on average, for production runs of 80,000 units is: in \$ 1000. (c) The point prediction of overhead costs for a single production run of 80,000 is the same as the point estimate of the average overhead cost for all production runs of 80,000 units, which is around \$14,386. (d)* The prediction ( individual ) standard error is and . So the 95% prediction interval for the overhead costs of this particular production run is: in \$ 1000. (e) That is because that s p must allow for the variability of the individual overhead cost, as well as the variability of point estimate of the overhead cost on average about the true 2
mean. In fact, s p is bigger than s m , due to the extra 1 in the formula and . 2. Problem 14 on Page 111. (Notes: Problem 14 should read, “An analyst wants an estimate of the average dividend yield for all firms with EPS = \$3. For these problems you may either use the data tables in the text, or you may produce your own using JMP. The data is available in webcafe. If you use the book’s tables be sure to explain where your answer comes from. Preparatory analyses: A. The regression analysis of Dividends on EPS gives. Bivariate Fit of DIVYIELD By EPS Bivariate Fit of DIVYIELD By EPS 0 1 2 3 4 5 6 7 DIVYIELD 0 1 2 3 4 5 EPS Linear Fit DIVYIELD = 2.0336398 + 0.3739551*EPS Summary of Fit RSquare 0.057468 RSquare Adj 0.033905 Root Mean Square Error 1.849754 Mean of Response 2.750476 Observations (or Sum Wgts) 42 Analysis of Variance Source DF Sum of Squares Mean Square F Ratio Model 1 8.34483 8.34483 2.4389 Error 40 136.86356 3.42159 Prob > F 3

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Source DF Sum of Squares Mean Square F Ratio C. Total 41 145.20839 0.1262 Parameter Estimates Term Estimate Std Error t Ratio Prob>|t| Intercept 2.0336398 0.540518 3.76 0.0005 EPS 0.3739551 0.239455 1.56 0.1262 From this graph, we can get that for EPS=3, the estimate of Dividends is: Dividends=2.034+0.3740*3=3.156 he value of S e is 3.4216 1.850 e S MSE = = = . (Or see Summary of Fit table.) Later, we’ll also need the sample mean and standard deviation of EPS. We can get
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