Part Three: Question 3 of Chapter 5(a) Show that the coefficient of the interaction term in model (5.10) is statisti- cally significant.(b) Estimate the number of days of delay to the end of harvest it takes to decrease the quality rating by 1 point when there is:Part Four: QUESTION 5 of Chapter 6(a) A statistician from Australia has recommended to the analyst that they not transform any of the predictor variables but that they transform Y using the log transformation. Do you agree with this recommendation? Give reasons to support your answer.(b) Develop a valid full regression model containing all seven potential predictor variables listed above. Ensure that you provide justification for your choice of full model, which includes scatter plots of the data, plots of standardized residuals, and any other relevant diagnostic plots.(c)Identify any points that should be investigated. Give one or more reasons to support each point chosen.(d) Describe any weaknesses in your model.(e) The golf fan wants to remove all predictors with insignificant t -values from the full model in a single step. Explain why you would not recommend this approach.stats101A hw6Yuh-Tzer Lin11/14/2016Part Three: Question 3 of Chapter 5(a) Show that the coefficient of the interaction term in model (5.10) is statisti- cally significant.Based on the model output, the coefficient of the interaction termˆβ3, has a correspondingtstatistic of0.0120<0.05 =α, therefore, we should reject the null hypothesis thatβ3= 0 and conclude that the coefficientof the interaction term is significant.(b) Estimate the number of days of delay to the end of harvest it takes to decrease the qualityrating by 1 point when there is:(i) No unwanted rain at harvestRain = 0 When (End of Harvest) = 0, Quality = 5.16122. Then Quality - 1 = 4.16122 Quality = b0 + b1(End of Harvest) + e 4.16122 = 5.16122 - 0.03145 (End of Harvest) (End of Harvest) = (4.16122 - 5.16122) /(-0.03145) = 31.80 = 32 days(ii) Some unwanted rain at harvestRain = 1 When (End of Harvest) = 0, Quality = 5.16122 + 1.78670 = 6.94792. Then Quality -1 =5.94792Quality = b0 + b1 (End of Harvest) + b2 + b3 (End of Harvest) 5.94792 = 5.16122 - 0.03145 (End ofHarvest) + 1.78670 - 0.08314 (End of Harvest) End of Harvest = (5.94792 - 5.16122 - 1.78670) / (-0.03145-0.08314) = 8.73 = 8 daysPart Four: QUESTION 5 of Chapter 6(a) A statistician from Australia has recommended to the analyst that they not transform anyof the predictor variables but that they transform Y using the log transformation. Do youagree with this recommendation? Give reasons to support your answer.I would not agree with the statistcian from Australia because the model he uses only two of the predictorvariables are coming out significant as contributors to PrizeMoney, that being GIR and BirdieConversion.But after 2B, it seems the model is okay to use: Residuals look random and there is no particular pattern tothem, QQ Plot looks good, vast majority of the points line on the line if not really close to it, with that said,data is said to be consistent with that from a normal distribution, square root Standardized Residuals lookgood as well, we have a relatively flat line, suggesting we have constant variance.(b) Develop a valid full regression model containing all seven potential predictor variableslisted above. Ensure that you provide justification for your choice of full model, which includesscatter plots of the data, plots of standardized residuals, and any other relevant diagnosticplots.After looking at the data untransformed and performing a PowerTransform of only the predictor variables, ittells us to log all the predictor variables, but there are clearly still problems with the model. Looking atthe invResPlot for the log(predictor variables) model, it suggests us to use a lambda of 0.05 but 0 almost1follows the 0.05 lambda line just as well. As a result, to make our lives easier, we’ll log PrizeMoney as well.After using a log transform for both Y and all the X variables, the model seems a bit better. Looking at thismodel, there doesn’t seem to be any particular differences between this one and the Australian one stated in2a. I would not have any issue the Australian and the log everything model interchangeably.(c)Identify any points that should be investigated. Give one or more reasons to support eachpoint chosen.Observation 185 should be looked at. It seems to have a fairly high leverage and have fairly high residualvalue as well. It seems to be a one of those points that benefits the model though. Point 63 should beinvestigated as well for the same reasons as 185(d) Describe any weaknesses in your model.The ultimate weakness of my model is the fact that 5 of the predictor variables are insignificant to predictingour Y, PrizeMoney. It seems that GIR and BirdieConversion are the main contributors to prediction andthat was the same pitfalls of the Australian model in 2a. Our R-squared value is 0.5585 for the transformedmodel and that is not particularly strong.(e) The golf fan wants to remove all predictors with insignificant t -values from the full modelin a single step. Explain why you would not recommend this approach.That is not a particularly good idea because maybe certain predictors are correlated with one another.Although one predictor variable may not contribute to Y, there is a good chance it is related to another X.By removing it, it may cause weird, counter-intuitive