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### Homework4Solution

Course: STAT 102, Spring 2012
School: UPenn
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Problem 1. 4.8 (pg 169) Summary of Fit 0.088012 RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) 0.074194 798.2601 2821.296 135 Analysis of Variance Source Model Error C. Total Parameter Estimates Term Intercept EXCHRATE PRICE DF Sum of Squares 2 132 8117338 4058669 84112922 637219 92230260 134 Mean Square F Ratio 6.3693 Prob &gt; F 0.0023 Estimate Std...

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Problem 1. 4.8 (pg 169) Summary of Fit 0.088012 RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) 0.074194 798.2601 2821.296 135 Analysis of Variance Source Model Error C. Total Parameter Estimates Term Intercept EXCHRATE PRICE DF Sum of Squares 2 132 8117338 4058669 84112922 637219 92230260 134 Mean Square F Ratio 6.3693 Prob > F 0.0023 Estimate Std Error t Ratio Prob>|t| 3361.9317 1.8691544 -2413.837 633.1943 4.223047 846.4798 5.31 0.44 -2.85 <.0001 0.6588 0.0051 (a) From the result of JMP above, the estimated regression equation is: SHIPMENT=3361.9317+1.8691544* EXCHRATE -2413.837* PRICE (b) Hypothesis H0: EXCHRATE = PRICE = 0 Ha: At least one of the slopes is not zero Decision Rule F(0.05, 2, 135-2-1) = 3.06 Reject H0 if the sample F-statistic f > 3.06 Test-Statistic Sample F-ratio = 6.3693 Decision Since 6.3693> 3.06, we reject H0 and conclude that some of the predictors coefficients are significantly different from 0. (c) Hypothesis H0: EXCHANGE =0 Ha: EXCHANGE is not equal to 0 in the full model. Decision Rule t(0.025, 135-2-1) = 1.98 Reject H0 if the sample t-ratio t > 1.98 or t<-1.98 Test-Statistic Sample t-ratio = 0.44 Decision We retain H0 and conclude that the coefficient of EDUC is not significantly different from 0 when taking account of PRICE. (d) The R-square value is 0.088012, so 8.8% of variation has been explained. (e) From the result of jump, PRICE=-2413.837 and SE( )= 846.4798. Also t(0.025, 135-2-1) = 1.98 and then the 95% CI for the regression coefficient of PRICE is (-2413.837-846.4798*1.98, -2413.837+846.4798*1.98) = (-4089.867, -737.807). PRICE 2. Problem 4.8 (pg 169) (a) From the result of JMP, the estimated regression equation is: Longevity = 3.2438212+0.4508583* Mother + 0.4111835* Father +0.016553* Gmothers+0.0868583* Gfathers (b and c) Using JMP it is easy to compute the 95% prediction interval for the longevity of the individual man; it is (67.76, 79.72); and the 95% confidence interval for the average longevity of the men in this group is (70.95, 76.53). (d) Hypothesis H0: gmother=gfather=0 Ha: At least one of them is zero For the JMP result in the beginning of the solution, in the full model =674.2430. When excluding the factors of gmothers and/or gfathers, using JMP we can get: Analysis of Variance DF Source Model Error C. Total Sum of Squares 2 97 99 Mean Square 1917.0052 958.503 686.7548 7.080 2603.7600 F Ratio 135.3828 Prob > F <.0001 So the F-statistic is =()/2/95=(686.7548674.2430)/2674.2430/95=0.88 Testing at level .05, F(.05, 2, 100-4-1) = 3.09. We reject H0 if the sample F-statistic f > 3.09 Decision Since 0.88< 3.09, we retain H0 and conclude that under level .05, gmothers and gfathers are not useful for predicting longevity once mother and father have been taken into account. (e) i.e. (h) We fit a model with only gmother and father into account Summary of Fit 0.356257 RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) 0.342984 4.156913 72.32 100 Analysis of Variance Source Model Error C. Total DF Sum of Squares Mean Square 2 97 927.6075 1676.1525 463.804 17.280 99 2603.7600 H0: gmother=gfather=0 Ha: At least one coefficient is not equal to 0. Decision Rule F(.05, 2, 100-2-1) = 3.09 F Ratio 26.8406 Prob > F <.0001 Reject H0 if the sample F-statistic f > 3.09 Test-Statistic Sample F-ratio = 26.8406 Decision Since 26.8406> 3.09, we reject H0 and conclude that some of the predictors coefficients are significantly different from 0 and this model is useful for predicting mens longevity, which means that if the values of mother and father arent known, but the values of gmothers and gfathers are known, then these values of gmothers and gfathers are useful for predicting longevity. Explanation Why are gmothers and gfathers insignificant when mother and father are known, but become significant when mother and father are unknown? From the former result, we can conclude that gmothers and gfathers are useful to describe the longevity of men. However, the latter implies that, mother and father can cover some of the information in gmothers and gfathers, and thus gmothers and gfathers are insignificant when mother and father are known. This is understandable because the the ages at death of the mother and father are highly related to the age at death of the grandparents. 3. Problem 17 (4.14) (0) We run a forward stepwise regression model where we set the probability to enter to be 0.05. Our results indicate that we should include all three variables, ERA, BA, an HR, into predicting WINS (aka our Y) Summary of Fit RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) 0.896931 0.885039 5.00226 80.83333 30 Analysis of Variance Source Model Error C. Total DF 3 26 29 Sum of Squares 5661.5790 650.5876 6312.1667 Mean Square 1887.19 25.02 F Ratio 75.4195 Prob > F <.0001* Parameter Estimates Term Intercept HR BA ERA Estimate -21.88065 0.0975914 606.30599 -16.89736 Std Error 28.92622 0.035722 100.7689 1.758246 t Ratio -0.76 2.73 6.02 -9.61 Prob>|t| 0.4562 0.0112* <.0001* <.0001* The model tells us that to be a successful baseball team, you should have higher home runs, higher average and batting, have low average ERA. (a) Our R2 is 0.896931, which is very high. This indicates that the regression model we built is a good fit to the data and be an excellent predictor of WINS (b) The predicted number of wins for Team A, A, is 68.2285357 while the predicted number of wins for Team B, B, is 91.4148184. The predicted difference is simply 91.4148184 68.2285357 = 23.18628. The prediction interval is simply B A +/- 2 * RMSE * sqrt(2) = 23.18628 +/- 2*5.00226 *sqrt(2) = (9.03775, 37.33481) OR (slightly more accurate one where we actually obtain the critical t value and use Sp of each team instead of approximating Sp by RMSE) B A +/- 2.055529 * sqrt(Sp(A)2 + Sp(B)2) = 23.18628 +/- 2.055529 *7.580458 = (7.604429, 38.76813) 4. Problem 4.8 (pg 169) Log10(Cost/SqFt) = 1.3050734 - 0.0206945*Log10(SqFt) + 0.0319822*(Log10(SqFt)-4.18431)^2 Summary of Fit RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) 0.14286 0.135067 0.039938 1.227443 223 Analysis of Variance Source Model Error C. Total DF 2 220 222 Sum of Squares 0.05848530 0.35090492 0.40939021 Mean Square 0.029243 0.001595 F Ratio 18.3337 Prob > F <.0001* Parameter Estimates Term Intercept Log10(SqFt) (Log10(SqFt)-4.18431)^2 Estimate 1.3050734 -0.020695 0.0319822 Std Error 0.022416 0.005164 0.008855 t Ratio 58.22 -4.01 3.61 Prob>|t| <.0001* <.0001* 0.0004* (a) For a house with 2,500 sqFt, we get \$17.96933/SqFt with a 95% prediction interval of (1.17514488, 1.33391885) in log scale. After unlogging it, we get (\$14.9635/SqFt, \$21.57341/SqFt). The 50% prediction interval is +/- t * RMSE = 1.25453186 +/- (0.6756066)*0.039938 = (1.227549, 1.281514) After unlogging, we get (\$16.88686/SqFt, \$19.12115/SqFt) as our interval. OR (slightly more accurate one where we use Sp instead of approximating Sp by RMSE) +/- t * sqrt(Sp) = 1.25453186 +/- 0.6757066 * 0.04028149 = (1.227313, 1.281750) (b) There are two possible residual plots. Both are correct for the purposes of studying normality and homoscedasticity. Based on the plots, the residuals are normally distributed because it fits inside the butterfly lines and the regression looks homoscedastic because there is no spreading in the residual plots. Residual by Predicted Plot Log10(Cost/ SqFt) Res idual 0.15 0.10 0.05 0.00 -0.05 -0.10 1.1 1.15 1.2 1.25 1.3 1.35 1.4 R es id ual Log10(Cost/SqFt) Predicted 0.10 0.00 -0.10 10 20 Log10(SqFt)^2 2.33 0.98 1.64 0.95 1.28 0.9 0.67 0.0 -0.67 0.8 0.5 0.2 -1.28 0.1 -1.64 0.05 0.02 -2.33 -0.1 -0.05 (c) (Optional) 0 0.05 0.1 0.15 No rm al Qu a ntil e Pl ot And the QQ plot is As the number of square feet increases, the marginal utility of renting the office space decreases. Marginal utility, which is the change in utility for every Sqft increase in office space, goes down and using a quadratic term takes into account the presence of marginal utility. 5. Problem 4.8 (pg 169) (a) H0: %<=20 = %>=50 = 0 Ha: at least one of the two betas F=36.618, p-value<0.0001, so we reject the null hypothesis and conclude that class sizes are useful to predict Grad.Rate. (b) Bs GradRate will be higher than As by 1.2095*0.15=0.181425. (c) The 95% confidence will be (d) Grad.Rate=-0.973+1.325Fresh.Ret_0.000397SAT75 (e) Ha: at least one of the two betas F=501.22, p-value<0.0001, so we reject the null hypothesis and conclude that the overall model under Analysis II useful for predicting Grad.Rate. (f) So we reject the null hypothesis and conclude that SAT75 is a useful predictor of Grad.Rate after controlling for Fresh.Ret. (g) Analysis of Variance Source DF Model 4 Error 168 C. Total 172 Sum of Squares F-Ratio 4.684 270.975 0.72 6 5.410 SST=5.410 from Analysis II SST; SSR=SST-SSE=5.410-0.726=4.684 F=MSR/MSE=(4.684/4)/(0.726/168)=270.975 (h) Ha: at least one the two betas So we reject the null hypothesis and conclude that there be statistically significant information about class sizes contained in the two variables %<=20 and %>=50 after controlling for Fresh.Ret and SAT75. (i) We choose the three variables based on how backward stepwise regression choose these three variables. Backward stepwise regression chooses the t with the lowest ratio to go from a four variable model to a three variable model. Hence, we get Fresh.Ret, SAT75, %<=20 (j) Analysis of Variance Source DF Sum of Squares F-Ratio Model 3 4.68397 363.43 Error 169 0.72603 C. Total 172 5.410 SST=5.410 from Analysis II; From ANOVAIII, for %>=50 t-ratio=0.08 If we consider two regression models: Analysis III and Analysis IV Partial F ratio= But partial F ratio=(t ratio)^2 where the t ratio comes from the variable we take out. We solve the equation and get SSEr=0.72603, which is SSE in Analysis IV SSR=SST-SSE=5.410-0.72603=4.68397 F=MSR/MSE=(4.68397/3)/(0.72603/169)=363.43 Note that F is very high (probably higher than the critical value of F) and hence, the overall model is useful. (k) Three-variable model R^2=4.68397/5.410=0.8658 The Best Two-variable model R^2=4.625/5.410=0.8549 (Fresh Ret. And SAT 75) To test if %<=20 is significant in the three-variable model So we reject the null hypothesis and conclude that three-variable model is significantly better than the two-variable model.
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Chapter 8Competitive Firmsand MarketsThe love of money is the root of allvirtue.George Bernard ShawCopyright 2011 Pearson Addison-Wesley. All rights reserved.Chapter 8 Outline8.18.28.38.4Perfect CompetitionProfit MaximizationCompetition in t
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Chapter 9Properties &amp;Applications of theCompetitive ModelNo more good must be attemptedthan the public can bear.Thomas JeffersonCopyright 2011 Pearson Addison-Wesley. All rights reserved.Chapter 9 Outline9.1 Zero Profit for Competitive Firms in t
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Chapter 11MonopolyI think its wrong that only onecompany makes the game Monopoly.Steven WrightCopyright 2011 Pearson Addison-Wesley. All rights reserved.Chapter 11 Outline11.111.211.311.511.611.711.8Monopoly Profit MaximizationMarket Power
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Chapter 12Pricing andAdvertisingEverything is worth what itspurchaser will pay for it.Publilius Syrus(first century BC)Copyright 2011 Pearson Addison-Wesley. All rights reserved.Chapter 12 Outline12.112.212.312.412.512.6Why and How Firms Pr