辅导STATS 330编程设计、写作R程序

” 辅导STATS 330编程设计、写作R程序STATS 330 Revision1. The selling prices (in British pounds) at auction of 32 antique grandfather clocks wererecorded. The age of each clock (in years) and the number of people who participated inthe bidding were also recorded.Age Bidders Price Age Bidders Price Age Bidders Price127 13 1235 115 12 1080 127 7 845150 9 1522 156 6 1047 182 11 1979156 12 1822 132 10 1253 137 9 1297113 9 946 137 15 1713 117 11 1024137 8 1147 153 6 1092 117 13 1152126 10 1336 170 14 2131 182 8 1550162 11 1884 184 10 2041 143 6 854159 9 1483 108 14 1055 175 8 1545108 6 729 179 9 1792 111 15 1175187 8 1593 111 7 785 115 7 744194 5 1356 168 7 1262(a) A conditional plot of Price versus Age given the level of Bidders was created using coplot.This plot suggests that Age and Bidders interact.i. What is meant by the statement Age and Bidders interact.ii. Explain what feature of the conditional plot indicates that there is an interactionbetween Age and Bidders (your explanation should clearly identify how the plot wouldbe different if the two variables did not interact).[5 marks](b) The output for the Regression model containing the Age:Bidders interaction is:Estimate Std. Error t value Pr(|t|)(Intercept) 320.4580 295.1413 1.086 0.28684Age 0.8781 2.0322 0.432 0.66896Bidders -93.2648 29.8916 -3.120 0.00416 **Age:Bidders 1.2978 0.2123 6.112 1.35e-06 ***—Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1Residual standard error: 88.91 on 28 degrees of freedomMultiple R-squared: 0.9539,Adjusted R-squared: 0.9489F-statistic: 193 on 3 and 28 DF, p-value: 2.2e-16i. Consider the following line of this output.F-statistic: 193 on 3 and 28 DF, p-value: 2.2e-16What does indicate about the fitted model?ii. Consider the following statement: As the fitted coefficient for Bidders is negative,this model indicates that E(Price) decreases as the number of bidders increases. Doyou agree with this statement? Explain your answer.[5 marks](c) Suppose that a seller has a clock that is 150 years old.i. Based on our fitted model, write down the relationship between E(Price) and Biddersfor this clock.ii. How does this relationship change as the value of Age is increased?[5 marks]2. The Stat2Data package in R contains a data set that records the prices of horses advertised forsale on the internet. The data was collected by a group of students from California PolytechnicState University. The information they collected for each horse included:Price Price in US dollars.Age Age of the horse in years.Height Height of the horse in hands.Sex f=female and m=male.This data was used to Create a data frame horse.df in R. head(horse.df)Price Age Height Sex1 38000 3 16.75 m2 40000 5 17.00 m3 12000 8 16.00 f4 25000 4 16.25 m5 35000 8 16.25 f6 35000 5 16.50 m summary(horse.df)Price Age Height SexMin. : 1100 Min. : 0.500 Min. :14.25 f:181st Qu.:15750 1st Qu.: 5.000 1st Qu.:16.00 m:29Median :25000 Median : 7.000 Median :16.50Mean :27957 Mean : 7.489 Mean :16.333rd Qu.:40000 3rd Qu.: 8.500 3rd Qu.:16.75Max. :60000 Max. :20.000 Max. :17.25The following regression model was fitted using this data: horse.lm=lm(log(Price)~Age*Sex+Height,data=horse.df) summary(horse.lm)Coefficients:Estimate Std. Error t value Pr(|t|)(Intercept) 4.19038 2.14289 1.955 0.057201 .Age -0.14882 0.02116 -7.035 1.3e-08 ***Sexm -0.53174 0.29584 -1.797 0.079464 .Height 0.41017 0.13818 2.968 0.004926 **Age:Sexm 0.13585 0.03253 4.176 0.000146 ***Residual standard error: 0.4666 on 42 degrees of freedomMultiple R-squared: 0.6884, Adjusted R-squared: 0.6587F-statistic: 23.2 on 4 And 42 DF, p-value: 3.582e-10(a) The fitted model uses log(Price) as the response. The assumptions for the ordinaryregression model are linearity, independence, constant variance and Normality. Which ofthese assumptions are affected by using log(Price) rather than Price as the response?2 marks(b) For the fitted model, Sex is a factor and its indicator variable is set up using f as thebaseline level. An explanatory variable Age:Sexm which represents the interaction be-tween Age and Sex has also been set up. For the first six observations of horse.df (givenon the previous page) write down the values of Sexm and Age:Sexm. 2 marks(c) The output for the anova command is: anova(horse.lm)Response: log(Price)Df Sum Sq Mean Sq F value Pr(F)Age 1 7.1303 7.1303 32.745 9.955e-07 ***Sex 1 6.9854 6.9854 32.080 1.207e-06 ***Height 1 2.2938 2.2938 10.534 0.0023039 **Age:Sex 1 3.7970 3.7970 17.437 0.0001465 ***Residuals 42 9.1456 0.2178The p-value for Sex in this table is much smaller than the p-value for Sexm in the summaryoutput on the previous page. However the p-value for Age:Sex in this table and that forAge:Sexm in the summary output are the same. Explain why this occurs. 3 marks(d) Use the fitted model to compare the impact Age, Sex and Height have on Price.5 marks3. An advertisement for diamonds was included in the February 18, 2000 edition of The BusinessTimes. This advertisement gave the following characteristics for 308 diamonds:price Price in Singapore dollars.carats Size measured in carats.colour Colour rating on the GIA colour scale which goes from D (colourless) through Z (lightcolour). The diamonds listed all had ratings of D, E, F, G, H or I.clarity Clarity rating on the GIA clarity scale. The diamonds listed all had ratings of IF(internally flawless), VVS1, VVS2, VS1 or VS2. VS1 and VS2 indicate very slightlyflawed diamonds whereas VVS1 and VVS2 indicate very very slightly flawed diamonds.For VS1 and VVS1 the flaw is only visible from the pavilion (bottom) and for VS2 andVVS2 the flaw is only visible from the crown (top).cert The certification Agency: Gemmological Institute of America (GIA), International Gem-mological Institute (IGI) or Hoge Raad Voor Diamant (HRD).This data was read into a data frame named diamond.df in R and the following outputobtained: summary(diamond.df)carats colour clarity cert priceMin. :0.1800 D:16 IF :44 GIA:151 Min. : 6381st Qu.:0.3500 E:44 VS1 :81 HRD: 79 1st Qu.: 1625Median :0.6200 F:82 VS2 :53 IGI: 78 Median : 4215Mean :0.6309 G:65 VVS1:52 Mean : 50193rd Qu.:0.8500 H:61 VVS2:78 3rd Qu.: 7446Max. :1.1000 I:40 Max. :16008(a) The diamond data was used to construct a regression model that relates the price ofa diamond to the remaining characteristics. The following set of diagnostic plots wasobtained for the linear model: diamond1.lm-lm(price~carats+colour+clarity+cert,data=diamond.df) plot(diamond1.lm)0 2000 6000 10000Based on these plots discuss how well this model fits the data. Give reasons for anyconclusions you reach. (5 marks)(b) It was decided to modify the regression model by usingprice as the response. Considerthe following output: diamond2.lm-lm(price^.5~carats+colour+clarity+cert,data=diamond.df) new.df=data.frame(carats=1, colour = F, clarity=VS2, cert=HRD) predict(diamond2.lm,new.df,interval = confidence)fit lwr upr1 98.00119 97.06094 98.94144 predict(diamond2.lm,new.df,interval = prediction)fit lwr upr1 98.00119 93.27767 102.7247i. Use this output to find a 95% confidence interval and a 95% prediction interval forthe price of a 1 carat diamond certified by HRD that has colour F and clarityVS2.ii. In the context of this example, clearly explain why the prediction interval is widerthan the confidence interval. (5 marks)(c) The factor cert was included in the model since it was thought that the reputation ofthe certification agency (GIA, IGI or HRD) may have an impact on price. Consider usingthe anova function to evaluate the relevance of cert in the model using an F-test: diamond2.lm-lm(price^.5~carats+colour+clarity+cert,data=diamond.df) anova(diamond2.lm)Analysis of Variance TableDf Sum Sq Mean Sq F value Pr(F)carats 1 177237 177237 32036.7488 2.2e-16 ***colour 5 5227 1045 188.9561 2.2e-16 ***clarity 4 2950 737 133.3014 2.2e-16 ***cert 2 107 54 9.6783 8.49e-05 ***Residuals 295 1632 6If the regressors Are ordered differently in the model, the value of the F-statistic for certchanges from 9.6783 to 5617.27 in the ANOVA table. diamond2A.lm-lm(price^.5~cert+carats+colour+clarity,data=diamond.df) anova(diamond2A.lm)Analysis of Variance TableDf Sum Sq Mean Sq F value Pr(F)cert 2 62153 31076 5617.27 2.2e-16 ***carats 1 115217 115217 20826.30 2.2e-16 ***colour 5 5356 1071 193.62 2.2e-16 ***clarity 4 2795 699 126.28 2.2e-16 ***Residuals 295 1632 6Both tests result in very small p-values but they are testing different hypotheses.i. Carefully explain the difference between the tests for cert in these two tables.ii. Which test is more relevant in investigating whether the reputation of the certifica-tion agency has an impact on price? Explain your answer. (5 marks)(d) The output from summary for theprice model used in parts (b) and (c) is: summary(diamond2A.lm)Call:lm(formula = price^0.5 ~ cert + carats + colour + clarity,data = diamond.df)Coefficients:Estimate Std. Error t value Pr(|t|)(Intercept) 25.62462 0.84435 30.348 2e-16 ***certHRD 0.03969 0.35509 0.112 0.911certIGI -1.80848 0.42466 -4.259 2.77e-05 ***carats 91.47532 0.62917 145.391 2e-16 ***colourE -5.52416 0.68862 -8.022 2.47e-14 ***colourF -8.17357 0.64641 -12.645 2e-16 ***colourG -10.71430 0.66350 -16.148 2e-16 ***colourH -14.12727 0.67184 -21.028 2e-16 ***colourI -18.23773 0.70429 -25.895 2e-16 ***clarityVS1 -8.45895 0.52868 -16.000 2e-16 ***clarityVS2 -10.96488 0.56679 -19.346 2e-16 ***clarityVVS1 -2.69994 0.52951 -5.099 6.12e-07 ***clarityVVS2 -5.95314 0.49254 -12.087 2e-16 ***Consider the Coefficients for the four indicator (dummy) variables used for the factorclarity.i. What does the estimated coefficient for clarityVS1 (-8.45895) represent in terms ofthe levels of clarity?ii. What is the estimated difference in the expectedprice for very very slightly flaweddiamonds where the flaw is only visible from the pavilion (VVS1) and where the flawis only visible from the crown (VVS2)?iii. Suppose we want to formally test (i.e. find a p-value) the hypothesis that there isno difference in the average price of VVS1 and VVS2 diamonds (given that all othercharacteristics are the same). Suggest a simple method of adjusting the regressionmodel so that the required p-value is calculated for us. (5 marks)4. Sahoo and H.S. Pandalai (Natural Resources Research, vol. 8, 1999) used logistic regressionto predict the probability of a gold deposit being present based the concentrations of Arsenic(As) and Antimony (Sb) in ground water samples. Ground water samples were taken from atotal of 64 sites of these, 28 had a gold deposit present. The concentrations of As and Sbin the samples were measured as parts per million (ppm).(a) The fitted model logistic for this data is: gold.glm-glm(gold~As+Sb,family=binomial,data=gold.df) summary(gold.glm)Coefficients:Estimate Std. Error z value Pr(|z|)(Intercept) -4.9664 1.3675 -3.632 0.000281 ***As 1.2490 0.3777 3.307 0.000943 ***Sb 0.9235 0.4486 2.059 0.039518 *—Null deviance: 87.720 on 63 degrees of freedomResidual deviance: 18.306 on 61 degrees of freedomAIC: 24.306i. Let pi represent the probability that a gold deposit is present. Write down the logisticform of the fitted model (i.e. pi = . . .).ii. Find a 95% confidence interval for the coefficient for As.iii. Interpret the coefficient for As in terms of the impact of the concentration of As onthe odds of a gold Deposit being present.(5 marks)(b) The following output was obtained using the anova function. anova(gold.glm,test=Chisq)Terms added sequentially (first to last)Df Deviance Resid. Df Resid. Dev Pr(Chi)NULL 63 87.720As 1 65.117 62 22.603 7.057e-16 ***Sb 1 4.297 61 18.306 0.03818 *i. What hypothesis is being tested by the lineAs 1 65.117 62 22.603 7.057e-16 ***in this table? What do you conclude from this test?ii. The output for summary(gold.glm) also provides a test for As.Estimate Std. Error z value Pr(|z|)As 1.2490 0.3777 3.307 0.000943 ***How does this test differ from the test in (i)?(5 marks)i. Which points are identified as being usual by these plots? For each of these pointsdescribe what makes it unusual. Note that using HMD and deviance changes asdiagnostics were not covered this semester so you can ignore those two plots andjust look at the Cooks distance plot.ii. Briefly describe how you would further investigate the unusual points you identified.(5 marks)(d) Suppose the fitted model is to be used to predict the presence of gold deposits for othersites (it is not known whether gold is present at these sites). A gold deposit is predicted tobe present if the estimated probability is more than one half and is otherwise predicted tobe absent. The following table summarizes the performance of this prediction procedurewhen it is applied to the 64 observations in the data set.PredictionActual gold present gold absentgold present 26 2gold absent 1 35i. Based on this table, estimate the sensitivity and the specificity for this predictionmodel. Explain why these estimates are too optimistic.ii. Suggest a different method for estimating the sensitivity and specificity that stilluses the original data but should give more realistic estimates.(5 marks)5. The data for this question come from an ecological study conducted by Dr Rick Linhurst(1979, PhD thesis, North Carolina State). The study focussed on the relationship betweenthe biomass (weight per unit area) of spartina (a type of grass found in coastal salt marshes)and the properties of the soil at different sites in the Cape Fear estuary of North Carolina.Data was collected from three different locations: Oak Island (OI), Smith Island (SI) andSnows Marsh (SM). At each location, there were areas where the spartina vegetation wasshort (SHRT), other areas where it was tall (TALL) and still other areas where it had diedand re-vegetated (DVEG). For each combination of location and vegetation type, five siteswere selected and the biomass of spartina was measured.Consider the following conditional plot for this data.(a) Consider creating a regression model that relates biomass (BIO) to location (Loc) andvegetation type (Type). The plot on the previous page indicates that it may be appro-priate toi. use log(BIO) as the response andii. include the Loc:Type interaction in the model.Explain how the plot indicates that each of these actions may be appropriate. [4 marks](b) Consider a regression model that uses the logged biomass (BIO) as the response andvegetation type (Type) and location (Loc) as regressors. model1.lm-lm(log(BIO) ~ Loc*Type,data=spartina.df) summary(model1.lm)Coefficients:Estimate Std. Error t value Pr(|t|)(Intercept) 6.6805 0.1466 45.575 2e-16 ***LocSI -0.8659 0.2073 -4.177 0.000179 ***LocSM 0.6189 0.2073 2.985 0.005066 **TypeSHRT -0.3688 0.2073 -1.779 0.083668 .TypeTALL 0.4512 0.2073 2.176 0.036171 *LocSI:TypeSHRT 0.2311 0.2932 0.788 0.435751LocSM:TypeSHRT -0.7450 0.2932 -2.541 0.015496 *LocSI:TypeTALL 1.0486 0.2932 3.577 0.001015 **LocSM:TypeTALL -0.4358 0.2932 -1.486 0.145882—Residual standard error: 0.3278 on 36 degrees of freedomMultiple R-squared: 0.8195,Adjusted R-squared: 0.7794F-statistic: 20.43 on 8 and 36 DF, p-value: 3.142e-11Use this model to estimate each of the following:i. the biomass of spartina for short vegetation at Smith Island.ii. the difference in the biomass of spartina between re-vegetated sites (DVEG) at SnowsMarsh and at Oak Island.iii. the difference in the biomass of spartina short vegetation and tall vegetation at SmithIsland. [6 marks](c) Recall that for each Combination of location and vegetation type, five sites were selectedand the biomass of spartina was measured. Characteristics of the soil were also measuredat each site. One of these characteristics was the pH which indicates the acidity oralkalinity of the soil. The following search procedure was used to see whether it wouldbe useful to include pH (or interactions involving pH) in the model.model2.lm-lm(log(BIO) ~ Loc+Type+pH,data=spartina.df)model3.lm-lm(log(BIO) ~ Loc*Type*pH,data=spartina.df)spartstepln.lm-step(object=model2.lm,scope=formula(model3.lm),direction=both)Briefly, describe the following aspects of this search procedure.i. The starting model.ii. The set of regressors being considered.iii. The procedure used at each step to select the next model. [4 marks](d) Consider the model selected by this search procedure.Call:lm(formula = log(BIO) ~ Loc + Type + pH + Loc:Type + Loc:pH,data = spartina.df)Estimate Std. Error t value Pr(|t|)(Intercept) 6.92216 0.92315 7.498 1.27e-08 ***LocSI -12.65295 2.75000 -4.601 5.95e-05 ***LocSM -0.62059 1.74134 -0.356 0.723821TypeSHRT -0.37689 0.17288 -2.180 0.036491 *TypeTALL 0.39757 0.26485 1.501 0.142840pH -0.05055 0.19148 -0.264 0.793424LocSI:TypeSHRT 0.37952 0.24460 1.552 0.130294LocSM:TypeSHRT -0.55440 0.36236 -1.530 0.135559LocSI:TypeTALL -13.14536 3.20872 -4.097 0.000255 ***LocSM:TypeTALL -0.49372 0.35519 -1.390 0.173828LocSI:pH 3.55980 0.80948 4.398 0.000107 ***LocSM:pH 0.25336 0.35515 0.713 0.480610—Residual standard error: 0.269 on 33 degrees of freedomMultiple R-squared: 0.8885,Adjusted R-squared: 0.8514F-statistic: 23.91 on 11 and 33 DF, p-value: 1.315e-12Use this model to describe the impact that pH has on the biomass of spartina. [6 marks]6. The number of Different mussel species was recorded in 41 rivers along the east coast of theUSA. Nine possible explanatory variables were recorded and described as follows:area – the area of drainage basin.sAC, sAP, sSL, sSV – the number of intermediate rivers to 4 major species-source riversystems: Alabama-Coosa (sAC), Apalachicola (sAP) St. Lawrence (sSL), and Savannah(sSV).nitrate – nitrate concentration.hy – hydronium ion concentration.solres the amount of solid residue present in the water.logarea – it was thought that taking the log of the area might improve the model fit so thiswas also included as a potential regressor.Reference: J.J. Sepkoski, Jr., M.A. Rex (1974). Distribution of Freshwater Mussels: Coastal Rivers asBiogeographic Islands, Systematic Zoology, Vol. 23(2), pp. 165-188.The first six rows of the data frame for the mussels data: head(mussels.df)species area sAC sAP sSV sSL nitrate solres hy logareaPenobscot 9 8440 33 28 21 4 0.8 57 4.0 9.0407Kennebec 8 5960 32 27 20 5 0.4 31 3.2 8.6928Androscoggin 7 3510 31 26 19 6 0.6 65 2.5 8.1634Saco 6 1730 30 25 18 7 0.8 33 2.5 7.4559Merrimac 11 5020 29 24 17 8 2.6 78 6.3 8.5212Blackstone 8 425 26 21 14 11 8.4 120 20.0 6.0521The step function was used to conduct two searches for a suitable model using the followingR code: null.glm- glm(species~1,family = poisson, data=mussels.df) full.glm-glm(species~.,family = poisson, data=mussels.df) formula(full.glm)species ~ area + sAC + sAP + sSV + sSL + nitrate + solres + hy + logarea step1.glm=step(null.glm,scope=formula(full.glm),trace=0,direction=both) summary(step1.glm)Estimate Std. Error z value Pr(|z|)(Intercept) 0.9483756 0.4701266 2.017 0.043667 *logarea 0.2495191 0.0513711 4.857 1.19e-06 ***sAC -0.0258976 0.0056666 -4.570 4.87e-06 ***solres -0.0022974 0.0006491 -3.540 0.000401 ***hy -0.0329310 0.0112529 -2.926 0.003429 **nitrate 0.0514423 0.0285902 1.799 0.071971 .Null deviance: 127.527 on 43 degrees of freedomResidual deviance: 46.426 on 38 degrees of freedomAIC: 240.45 step2.glm=step(full.glm,scope=formula(full.glm),trace=0,direction=both) summary(step2.glm)Estimate Std. Error z value Pr(|z|)(Intercept) 0.765412 0.487918 1.569 0.116711sAP -0.034419 0.007187 -4.789 1.67e-06 ***sSV 0.016718 0.011265 1.484 0.137800nitrate 0.049525 0.028270 1.752 0.079799 .solres -0.002328 0.000655 -3.554 0.000379 ***hy -0.031616 0.011363 -2.782 0.005394 **logarea 0.257352 0.051853 4.963 6.94e-07 ***Null deviance: 127.527 on 43 degrees of freedomResidual deviance: 44.303 on 37 degrees of freedomAIC: 240.33(a) Consider the two searches that were performed.i. Briefly describe the Procedure used for the first search (step1.glm).ii. How does the second search differ from the first search?iii. Consider the models step1.glm and step2.glm found by the two searches. Is eithermodel clearly better than the other one? Explain your answer. (5 marks)(b) Consider the impact of hydronium concentration (hy) on the mean number of musselspecies found in a river.i. Based on the output for model step1.glm, explain the impact on the mean numberof mussel species if the hydronium concentration is increased by 1 unit (and allremaining explanatory variables are fixed).ii. Use the following output to produce a 95% confidence interval for your estimate ofthis impact. confint(step1.glm)2.5 % 97.5 %(Intercept) 0.020585439 1.864031858logarea 0.149254036 0.350694327sAC -0.037083119 -0.014862806solres -0.003625416 -0.001075839hy -0.056618776 -0.012315184nitrate -0.006685696 0.105584270iii. Estimate the impact of increasing hydronium concentration by 3 units on the meannumber of mussel species found in a river. (6 marks)(c) If a goodness of fit test is performed for model step1.glm, the p-value is equal to 0.19.Explain how the p-value is calculated. What do you conclude about model step1.glmfrom this test? (4 marks)(d) Consider the following set of diagnostic plots for model step1.glmBased on these plots describe any concerns or issues you have with the fitted model.ii. Briefly explain how you would address each of the concerns/issues you have identi-fied. (5 marks)请加QQ：99515681 或邮箱：99515681@qq.com WX：codehelp

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