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Unformatted text preview: STAT 326 LAB 4 DUE IN LAB 1. NAME: — Group#: 2. NAME: — Group#: 3. NAME: Group#: 4. NAME: — Group#: — LEGOs. In this example we consider a linear model aimed at predicting the Original Listing price for different LEGO sets based on the number of pieces in the LEGO set and the number of figures (not including animals) in the LEGO set. This dataset consists of information taken from 37 randomly selected LEGO sets. 0 J MP column label reference — Number: This is the LEGO box reference number — Original Price (y): This is the Orginal Listing Price of the LEGO sets. Notice the prices are in multiple of 5 dollars minus 1 cent. Example: 999, 14.99, 19.99 etc. — Number of Pieces (.731): This is the number of LEGO Pieces in the set as recorded on the LEGO box. — Figures (:52): This is the number of human LEGO figures (not animals) in a set as seen on the LEGO box. For all your final calculations, please round to the nearest hundredth (cents). 1. (Model 1) First fit a simple linear regression line predicting the original price based on the number of pieces in the LEGO set. Record the estimated equation in the space below. Also record the R2 value and interpret this value in the context of this example. 0 Estimated equation: 3) = 1.227 + 0.1033; c x is the number of pieces 0 g is the predicted price given the number of pieces. Interpretation of the R2: 85.16 percent of the variability in the price of LEGO sets is explained by the linear model with explanatory variable number of pieces. 2. (Model 2) Next fit a simple linear regression line predicting the original price based on the number of figures in the LEGO set. Record the estimated equation in the space below. Also record the R2 value and interpret this value in the context of this example. 0 Estimated equation: 17 = 10.905 + 9.2973: c x is the number of figures 0 g) is the predicted price given the number of figures. Interpretation of the R2: 36.9 percent of the variability in the price of LEGO sets is explained by the linear model with explanatory variable number of figures. 3. Create a scatterplot of the original price (y axis) by the number of pieces. We have already created symbols for you. An A is displayed for sets with 1 figure, a B is displayed for sets with 2 figures a C is displayed for sets with 3 figures and a D is displayed for sets With 4 or more figures. Save this plot for your own records. In the space provided discuss any characteristics you observe. You should discuss the 4 features of the scatter plot (between original price and number of pieces). Further you should discuss what you observe in the original price point for sets with different numbers of figures (specifically your discussion should relate to differences in prices even for LEGO sets with a similar number of pieces). There is a strong positive linear relationship between the number of pieces and the price of LEGO sets. There are no obvious outliers in this example. After accounting for the number of pieces there appears to additionally be higher prices in sets with more Figures. For example the sets with 4 or more figures (coded with a D in the plot) tend to have higher prices compared to sets with 1 or 2 figures (coded with an A or B in the plot) even for sets with a similar number of pieces. 4. For the remainder of the lab we will use a multiple regression model to describe the original price of LEGO sets. Let y be the original Price of LEGO sets, an the Number of pieces in the LEGO set and $2 be the number figures in the LEGO set. Here we will assume the population model: My = 50 + fllxl + 52352 In the space below provide an interpretation of the population intercept in the context of this problem. Include the interpretation even if the interpretation doesn’t make much sense from a practical perspective. 50 is the mean price of LEGO set with no pieces and no figures. You should not have any numbers related to the estimated model here with your solution. Remember this is a population parameter interpretation. 5. Explain why the interpretation of the population intercept is not very practical for this example. Hopefully it is clear that a LEGO set would not exist if there were no figures and no LEGO pieces. Again you should not have any output numbers referenced in your answer here. This discussion is reacted to the population intercept. 6. (Model 3) Use the Fit Model utility in JMP to fit the estimated multiple regression model. Save the column of residuals and the column of predicted values. Record the estimated model in the space below. :17 = —5.5267 + 0.0907031 + 4.1279132 7. 10. 11. (Model 3) Interpret the estimate for the slope associated with Number of Pieces. For every addition piece we predict the price of a LEGO set to increase by 9.07 (or just 9) cents assuming the number of LEGO figures does not change. . (Model 3) Interpret the estimate for the slope associated with Figures. For each additional Figure we predict the price of a LEGO set to increase by $4.13 assuming the number of LEGO pieces remains the same. NOTE: In practice a regular LEGO piece would technically have to be replaced with a Figure piece for the number of pieces in the set to actually remain constant. (Model 3) Assume that LEGO is considering offering a new LEGO set with 5 Figures and 400 pieces. Use your estimated model to help propose a good price for the LEGO set. 3] = —5.5267 + 0.0907 * 400 + 4.127 * 5 = 51.39 This prediction suggests a proposed price of $51.39. As a side note: The original Prices of the sets appear to be grouped by $5 minus a penny. If we took this into account it is likely a marketer would suggest a price of either 49.99 or 54.99. (Model 3) Assume two LEGO sets have the same number of Figures. How much more would you expect to pay for the set with 200 additional pieces? 200*0.0907 = 18.14. We would predict to pay $18.14 more for a set with 200 additional pieces. (Model 3) As a consumer would you prefer to pay for a LEGO set with a positive residual or a negative residual? Explain your answer. I would prefer to pay for a LEGO set with a negative residual. This would mean the actual price I paid would be lower than the predicted price based on the number of pieces and the number of figures. This would mean I would get more pieces for my money. NOTE: One may argue they would hope to get better pieces by paying more per piece (with a positive Residual). As long as the argument supports the direction you choose this is reasonable. Fit Group Bivariate Fit of Original Price By Number of Pieces Bivariate Fit of Original Price By Figures 100 100 80 .5 .5 60 E E E" E" 31° .20 o o 40 20 0 0 1 00 200 300 400 500 600 700 800 0 1 2 3 4 5 6 7 8 9 Number of Pieces Figures Linear Fit Linear Fit Original Price = 1.2276593 + 0.1033135"Number of Pieces Original Price = 10.905033 + 9.2973856*Figures Summary of Fit Summary of Fit RSq uare 0.851632 RSquare 0.368942 RSquare Adj 0.847393 RSquare Adj 0.350911 Root Mean Square Error 8.598023 Root Mean Square Error 17.73225 Mean of Response 35.53054 Mean of Response 35.53054 Observations (or Sum Wgts) 37 Observations (or Sum Wgts) 37 Sum of Sum of Source DF Squares Mean Square F Ratio Source DF Squares Mean Square F Ratio Model 1 14851.779 14851.8 200.9006 Model 1 6434.042 6434.04 20.4624 Error 35 2587.410 73.9 Prob > F Error 35 11005.147 314.43 Prob > F C. Total 36 17439.189 < .0001 * C. Total 36 17439.189 <.0001* Parameter Estimates Parameter Estimates Term Estimate Std Error tRatio Prob> |t| Term Estimate Std Error tRatio Prob> |t| Intercept 1 .2276593 2.802688 0.44 0.6641 Intercept 1 0.905033 6.1 75259 1 .77 0.0861 Number of Pieces 0.1033135 0.007289 14.17 < .0001* Figures 9.2973856 2.055336 4.52 < .0001* Response Original Price Summary of Fit RSquare 0.911706 RSquare Adj 0.906512 Root Mean Square Error 6.729599 Mean of Response 35.53054 Observations (or Sum Wgts) 37 Sum of Source DF Squares Mean Square F Ratio M odel 2 15899.414 7949.71 175.5386 Error 34 1539.775 45.29 Prob > F C. Total 36 17439.189 < .0001* Parameter Estimates Term Estimate Std Error tRatio Prob>|t| Intercept -5.526762 2.604655 -2.12 0.0412* Number of Pieces 0.0907332 0.006276 14.46 <.0001* Figures 4.127177 0.858099 4.81 <.0001* ...
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