STATS 201编程写作、辅导Data Analysis编程

” STATS 201编程写作、辅导Data Analysis编程STATS 201/208THE UNIVERSITY OF AUCKLANDFirst SEMESTER 2020Campus: CitySTATISTICSData Analysis/Data Analysis for CommerceINSTRUCTIONS Time Allowed: This Final Assessment has been designed so that a well-prepared studentcould complete it within two hours. From the 1pm release time you will have 24hours to complete and submit your assessment. No marks will be deducted for takinglonger than two hours within that 24-hour period, but you must submit before thedeadline. This assessment is open book, you are permitted to access your course manuals andother written material including online resources. Calculators are permitted. Answers should be written using the R Markdown file provided. It is your responsibility to ensure your assessment is successfully submitted on time. We recommend you aim to submit a couple of hours in advance of the deadline, toallow time to deal with any technical issues that might arise. We STRONGLY recommend you download your submitted document from Canvas,after submitting it, to verify you have uploaded the correct document. Attempt all questions. There are 70 marks in total for this examination.Page 1 of 32STATS 201/208STATS 201作业写作、辅导Data Analysis作业 If you have any concerns regarding your Final Assessment, please call the ContactCentre for advice, rather than your instructors. The Contact Centre can be reached on these numbers: Auckland: 09 373 7513 Outside Auckland: 0800 61 62 63 International: +64 9 373 7513 For any Canvas issues, please use 24/7 help on Canvas by chat or phone. If any corrections are announced during the 24 hours of the final assessment, youwill be notified by a Canvas Announcement. Please ensure your notifications areturned on during this period.Question Interpretation:Please note that during the final assessment period you cannot contact your instructorsfor clarification on how to interpret the wording of any specific questions or to verify thatyour answer is correct. Interpreting wording and making appropriate assumptions is partof what is being assessed. You will need to interpret the question yourself and check yourown answers.If you believe there is a typo, first re-read the question to check you have not misunderstoodthe question, as it is very common for students to misread questions. If you stillbelieve there is a typo, please phone the Contact Centre.Page 2 of 32STATS 201/208Academic Honesty Declaration:By completing this assessment, I agree to the following declaration:I understand the University expects all students to complete coursework with integrityand honesty. I promise to complete all online assessment with the same academic integritystandards and values. Any identified form of poor academic practice or academicmisconduct will be followed up and may result in disciplinary action.As a member of the Universitys student body, I will complete this assessment in a fair,honest, responsible and trustworthy manner. This means that: I declare that this assessment is my own work. I will not seek out any unauthorised help in completing this assessment. I am aware the University of Auckland may use plagiarism detection tools to checkmy content. I will not discuss the content of the assessment with anyone else in any form,including, Canvas, Piazza, Facebook, Twitter or any other social media or onlineplatform within the assessment period. I will not reproduce the content of this assessment anywhere in any form at anytime. I declare that I generated the calculations and data in this assessment independently,using only the tools and resources defined for use in this assessment. I will not share or distribute any tools or resources I developed for completing thisassessment.Page 3 of 32STATS 201/2081. This question refers to the Job Prestige Data in Appendix A. [Total 18 marks]a. Using the pairs plot, comment on the most important features of the data.[3 marks]b. The model fit.prestige1 uses both the variables income and log10incomewhile the other models use only log10income. Give TWO reasons why incomewas dropped instead of log10income. [2 marks]c. What would happen to the P-value for income if we dropped log10incomefrom fit.prestige1 instead of dropping income? Briefly justify your answer.[2 marks]d. According to the model fit.prestige3, which job has greater prestige:gov.administrators or physicists? Briefly justify your answer. [2 marks]Reminder: The data contains information collected from a large number ofpeople averaged for each of 102 different professions.e. Give a brief Executive Summary of the main conclusions of theanalysis of the Job Prestige Data in APPENDIX A. [9 marks]Note: To describe the effect of log10income it will be extremely helpfulto note that a 0.30 increase in log10income is equivalent to doubling income.2. This question refers to the mms Data in Appendix B.[Total 18 marks]a. Comment on what the interaction plot at the start of Appendix B reveals.[2 marks]b. Write brief Methods and Assumption Checks of the analysis of themms Data in Appendix B. [6 marks]c. Give a brief Executive Summary of the main conclusions of themms Data in Appendix B. [10 marks]Page 4 of 32STATS 201/2083. This question refers to the Viral pock data in Appendix C.[Total 10 marks]a. Comment on what the plot at the start of Appendix C reveals. [2 marks]b. Write down an equation of the final model fitted to the data, just as you wouldfor a Method and Assumption Checks section. [2 marks]c. Give a brief Executive Summary of the main conclusions of theanalysis of the Viral pock data in Appendix C. [6 marks]4. This question refers to the Turbines data in Appendix D.[Total 10 marks]a. Write brief Methods and Assumption Checks notes of the analysis of theTurbines data in Appendix D. [4 marks]b. Give a brief Executive Summary of the main conclusions of theanalysis of the Turbines data in Appendix D. [6 marks]Page 5 of 32STATS 201/2085. This question refers to the Nitrous Oxide Data in Appendix E.[Total 14 marks]a. Comment on the time series plot at the start of Appendix E (Global atmosphericconcentration of Nitrous Oxide). [2 marks]b. Comment on the Holt-Winters coefficients. [2 marks]c. Is the Holt-Winters model a good model of the N2O data? Justify your answer.[1 mark]d. Using the Holt-Winters model, provide a 95% prediction interval for theN2O level in May, 2020. [2 marks]e. Comment on the diagnostic plots for the model SF.fit1. Why was this modelrejected? [2 marks]f. Comment on the diagnostic plots for the model SF.fit2. Are the underlyingassumptions satisfied? Briefly explain. [2 marks]g. Using model SF.fit2, write an equation to calculate the estimated N2O forNovember, 2019 . This is the estimated equation, not the model equation, souse the estimated values for the coefficients and substitute appropriate valuesfor variables. You DO NOT need to calculate the final value. [3 marks]APPENDICES BOOKLET FOLLOWSPage 6 of 32STATS 201/208APPENDICES BOOKLETCONTENTSAppendix Name PagesA Job Prestige Data 813B mms Data 1418C Viral pock data 1921D Turbines data 2225E Nitrous Oxide Data 2632Page 7 of 32STATS 201/208Appendix A Job Prestige DataEmployment data was collected from a large number of people working in Canada in1971 across 102 different professions. For each profession, the data was combined andaverage education and income levels were calculated. The prestige score and percentageof women was also recorded for each profession.We want to build a model to explain the perceived prestige of a profession in terms ofthe other information collected. In particular, is there any evidence that professions witha higher proportion of women are either more or less prestigious?The variables are:job The professionprestige Pineo-Porter prestige score for the profession. A numeric measure ofprestige, the higher the score the more prestigious the profession.education Average education (years) for people in the professionincome Average income ($) for people in the professionlog10income The logarithm to base 10 of incomewomen Percentage of women in the professionPage 8 of 32STATS 201/208 pairs20x(Prestige.df[, c(1:5)]) fit.prestige1 = lm(prestige ~ income + log10income + education+ + women, data = Prestige.df) plot(fit.prestige1,which=1)20 30 40 50 60 70 8020 100 10 20Fitted valuesResidualsResiduals vs Fitted summary(fit.prestige1)Call:lm(formula = prestige ~ income + log10income + education + women,data = Prestige.df)Residuals:Min 1Q Median 3Q Max-17.3357 -4.3855 -0.0915 4.3120 19.1733Coefficients:Estimate Std. Error t value Pr(|t|)(Intercept) -1.097e+02 2.227e+01 -4.925 3.46e-06 ***income 2.936e-05 3.736e-04 0.079 0.938log10income 3.056e+01 6.557e+00 4.661 1.00e-05 ***education 3.724e+00 3.669e-01 10.150 2e-16 ***women 4.706e-02 3.012e-02 1.562 0.121—Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1Residual standard error: 7.129 on 97 degrees of freedomMultiple R-squared: 0.8351,Adjusted R-squared: 0.8283F-statistic: 122.8 on 4 and 97 DF, p-value: 2.2e-16Page 10 of 32STATS 201/208 fit.prestige2 = lm(prestige ~ log10income + education + women,data=Prestige.df) summary(fit.prestige2)Call:lm(formula = prestige ~ log10income + education + women, data = Prestige.df)Residuals:Min 1Q Median 3Q Max-17.364 -4.429 -0.101 4.316 19.179Coefficients:Estimate Std. Error t value Pr(|t|)(Intercept) -110.9658 14.8429 -7.476 3.27e-11 ***log10income 30.9427 4.4066 7.022 2.90e-10 ***education 3.7305 0.3544 10.527 2e-16 ***women 0.0469 0.0299 1.568 0.12—Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1Residual standard error: 7.093 on 98 degrees of freedomMultiple R-squared: 0.8351,Adjusted R-squared: 0.83F-statistic: 165.4 on 3 and 98 DF, p-value: 2.2e-16 fit.prestige3 = lm(prestige ~ log10income + education, data = Prestige.df) summary(fit.prestige3)Call:lm(formula = prestige ~ log10income + education, data = Prestige.df)Residuals:Min 1Q Median 3Q Max-17.0346 -4.5657 -0.1857 4.0577 18.1270Coefficients:Estimate Std. Error t value Pr(|t|)(Intercept) -95.1940 10.9979 -8.656 9.27e-14 ***log10income 26.3357 3.3090 7.959 2.94e-12 ***education 4.0020 0.3115 12.846 2e-16 ***—Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1Residual standard error: 7.145 on 99 degrees of freedomMultiple R-squared: 0.831,Adjusted R-squared: 0.8275F-statistic: 243.3 on 2 and 99 DF, p-value: 2.2e-16Page 11 of 32STATS 201/208 plot(fit.prestige3,which=1)20 30 40 50 60 70 8020 100 10 20Fitted valuesResidualsResiduals vs Fitted674682 cooks20x(fit.prestige3)observation number0 20 40 60 80 1000.00 0.05 0.10 0.15 0.20Cooks Distance plotCooks distance6367 20Page 12 of 32STATS 201/208 normcheck(fit.prestige3,main=histogram)Theoretical QuantilesSample QuantileshistogramResiduals from lm(prestige ~ log10income + education) confint(fit.prestige3)2.5 % 97.5 %(Intercept) -117.016298 -73.371676log10income 19.769848 32.901591education 3.383824 4.620081 head(Prestige.df)prestige education women income log10income job1 68.8 13.11 11.16 12351 4.091702 gov.administrators2 69.1 12.26 4.02 25879 4.412947 general.managers3 63.4 12.77 15.70 9271 3.967127 accountants4 56.8 11.42 9.11 8865 3.947679 purchasing.officers5 73.5 14.62 11.68 8403 3.924434 chemists6 77.6 15.64 5.13 11030 4.042576 physicists head(predict(fit.prestige3, interval = confidence))fit lwr upr1 65.02953 62.71844 67.340622 70.08810 65.99493 74.181273 60.38808 58.51342 62.262744 54.47327 52.71791 56.228635 66.66736 64.20611 69.128616 73.86069 70.95756 76.76381Page 13 of 32STATS 201/208Appendix B mms DataIn this study researchers were interested in whether low-fat nutrition labels increase theactual consumption of chocolate candies (mms) for overweight and normal weightconsumers.The researchers set up the experiment as follows. They asked forty adult family membersparticipating in a university open house to serve themselves unusual colours of mms(gold, teal, purple, and white), which were clearly labelled either as New Colors ofRegular mms (regular-label) or as New Low-Fat mms (low-fat-label).They then measured how many calories of mms the participants ate. The participantwas also assessed as being normal weight or overweight based on their body mass index.The variables in the data set are:Calories the number of calories consumed per helping by the participant,BMI whether the participant had a BMI that was normal or ow for overweight,MM whether mms were labelled regular or lowfat.The researchers were interested in whether the designations of low-fat resulted ingreater calorific consumption and, if so, did the effect of the label on the mmsdependon whether the respondent was of normal weight or deemed overweight.They also want the following estimated: the average number of calories consumed per helping by normal weight people eatingregular mms. the difference(s) in average number of calories consumed per helping between normaland overweight people – for each type of mms. the difference(s) in average number of calories consumed per helping between when themmsare labelled regular and lowfat – for both weight groups.Reference: Can Low-Fat Nutrition Labels Lead to Obesity?, Wansink, B Chandon,P., Journal of Marketing Research, November 2006.Page 14 of 32STATS 201/208We wish to order the factor MM so that regular is first. MM.df=within(MM.df,{MM=factor(MM,levels=c(regular,lowfat))}) interactionPlots(Calories~BMI+MM,ylab=Calories consumed,data =MM.df)Plot of Calories consumedby levels of BMI and MMBMICalories consumed100 150 200 250 300 350normal owlowfaregul MM.fit1 = lm(Calories~BMI*MM, data = MM.df) eovcheck(MM.fit1)200 220 240 260 28080 400 20 60Fitted valuesResidualsPage 15 of 32STATS 201/208 normcheck(MM.fit1)Theoretical QuantilesSample QuantilesResiduals from lm(Calories ~ BMI * MM) cooks20x(MM.fit1)observation number0 10 20 30 400.00 0.10 0.20Cooks Distance plotCooks distance28 2634Page 16 of 32STATS 201/208 anova(MM.fit1)Analysis of Variance TableResponse: CaloriesDf Sum Sq Mean Sq F value Pr(F)BMI 1 11465 11465 12.3800 0.001195 **MM 1 35593 35593 38.4339 3.765e-07 ***BMI:MM 1 9030 9030 9.7507 0.003528 **Residuals 36 33339 926—Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 summary(MM.fit1)Call:lm(formula = Calories ~ BMI * MM, data = MM.df)Residuals:Min 1Q Median 3Q Max-77.50 -14.82 2.56 15.10 56.89Coefficients:Estimate Std. Error t value Pr(|t|)(Intercept) 187.790 9.623 19.514 2e-16 ***BMIow 3.810 13.609 0.280 0.78112MMlowfat 29.610 13.609 2.176 0.03622 *BMIow:MMlowfat 60.100 19.247 3.123 0.00353 **—Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1Residual standard error: 30.43 on 36 degrees of freedomMultiple R-squared: 0.6272,Adjusted R-squared: 0.5961F-statistic: 20.19 on 3 and 36 DF, p-value: 7.604e-08 confint(MM.fit1)2.5 % 97.5 %(Intercept) 168.272946 207.30705BMIow -23.791283 31.41128MMlowfat 2.008717 57.21128BMIow:MMlowfat 21.065891 99.13411Page 17 of 32STATS 201/208 summary2way(MM.fit1,page=nointeraction)Tukey multiple comparisons of means95% family-wise confidence levelFit: aov(formula = fit)$BMIdiff lwr upr p adjow-normal 33.86 14.34295 53.37705 0.0011955$MMdiff lwr upr p adjlowfat-regular 59.66 40.14295 79.17705 4e-07 summary2way(MM.fit1,page=interaction)Tukey multiple comparisons of means95% family-wise confidence levelFit: aov(formula = fit)$`Comparisons within BMI`diff lwr upr p adjnormal:lowfat-normal:regular 29.61 -7.043392 66.26339 0.1494601ow:lowfat-ow:regular 89.71 53.056608 126.36339 0.0000007$`Comparisons between BMI`diff lwr upr p adjow:regular-normal:regular 3.81 -32.84339 40.46339 0.9922076ow:lowfat-normal:lowfat 63.91 27.25661 100.56339 0.0002140Page 18 of 32STATS 201/208Appendix C Viral pock dataWhen creating vaccines and medications for treating viruses, researchers need to be ableto produce large amounts of samples of the virus in petri dishes. A set quantity of a viralmedium is placed in a petri dish and left for the virus to grow. One question researchershad was what was the effect of diluting the viral medium on the viral Activity.They set up an experiment as follows. A supply of viral medium was produced. This wasthen divided into 5 batches, each with a different level of dilution applied to them. Petridishes were randomly assigned to the batches and filled. After a fixed amount of time,the viral activity in each petri dish was assess by counting the number of pock marksthat appeared. The dilution levels are actually in increasing powers of 2, so each increaseof 1 in dilution level doubles the dilution and therefore halves the concentration of theviral medium.The variables are:Count a count of the number of pock marks recorded in the petri dishDilution the dilution level used for viral medium in the petri dishThe researcher has two questions they want answered. What is the estimated number ofpock marks you would expect to see at dilution level 0 and what is the effect of doublingthe dilution level? plot(Count~Dilution, data=Pock.df)0 1 2 3 40 50 100 150 200 250DilutionCountPage 19 of 32STATS 201/208 fit.glm- glm(Count ~ Dilution,family = poisson, data=Pock.df ) summary(fit.glm)Call:glm(formula = Count ~ Dilution, family = poisson, data = Pock.df)Deviance Residuals:Min 1Q Median 3Q Max-6.058 -2.031 -0.462 1.373 5.357Coefficients:Estimate Std. Error z value Pr(|z|)(Intercept) 5.26793 0.02255 233.60 2e-16 ***Dilution -0.68094 0.01544 -44.09 2e-16 ***—Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1(Dispersion parameter for poisson family taken to be 1)Null deviance: 2729.74 on 47 degrees of freedomResidual deviance: 290.44 on 46 degrees of freedomAIC: 562.42Number of Fisher Scoring iterations: 4 1-pchisq(deviance(fit.glm), df.residual(fit.glm)) fit.quasi=glm(Count ~ Dilution,family = quasipoisson, data=Pock.df ) summary(fit.quasi)Call:glm(formula = Count ~ Dilution, family = quasipoisson, data = Pock.df)Deviance Residuals:Min 1Q Median 3Q Max-6.058 -2.031 -0.462 1.373 5.357Coefficients:Estimate Std. Error t value Pr(|t|)(Intercept) 5.26793 0.05678 92.78 2e-16 ***Dilution -0.68094 0.03888 -17.51 2e-16 ***—Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1(Dispersion parameter for quasipoisson family taken to be 6.338981)Null deviance: 2729.74 on 47 degrees of freedomResidual deviance: 290.44 on 46 degrees of freedomAIC: NANumber of Fisher Scoring iterations: 4 confint(fit.quasi)2.5 % 97.5 %(Intercept) 5.1548909 5.3775069Dilution -0.7583492 -0.6058599 exp(confint(fit.quasi))2.5 % 97.5 %(Intercept) 173.2769014 216.4819000Dilution 0.4684391 0.5456051 100*(exp(confint(fit.quasi))-1)2.5 % 97.5 %(Intercept) 17227.69014 21548.19000Dilution -53.15609 -45.43949Page 21 of 32STATS 201/208Appendix D Turbines dataAn experiment was set up to test the life of turbine wheels. Of interest was the likelihoodof the turbine wheels developing fissures (narrow cracks) that would require maintenanceto prevent more serious damage.In an experiment, turbine wheels were randomly allocated to be run for a number of hours.Each wheel was then examined to see whether or not it had developed any fissures.The data collected contains the variables:Hours the number of hours the turbines were run, a numeric vector,Turbines the number of turbine wheels that where tested at that number of hours,Fissures the number of turbine wheels that developed fissures at that number of hours.The researchers wanted to know the effects of time on the likelihood of turbine wheelsdeveloping fissures. They also wanted to estimate how many hours the turbine could runbefore the probability of the wheel developing fissures exceeded 25%. Turbines.dfHours Turbines Fissures plot(Fissures/Turbines~Hours,ylab=probability,data=Turbines.df,+ main=Probability of having a fissure by number of hours)1000 2000 3000 40000.0 0.1 0.2 0.3 0.4 0.5 0.6Probability of having a fissure by number of hoursHoursprobability turbines.glm – glm( Fissures/Turbines ~ Hours, family=binomial,+ weights=Turbines, data=Turbines.df) plot(turbines.glm,which=1)3 2 1 010 1 2Predicted valuesResidualsResiduals vs Fitted917Page 23 of 32STATS 201/208 summary(turbines.glm)Call:glm(formula = Fissures/Turbines ~ Hours, family = binomial, data = Turbines.df,weights = Turbines)Deviance Residuals:Min 1Q Median 3Q Max-1.5055 -0.7647 -0.3036 0.4901 2.0943Coefficients:Estimate Std. Error z value Pr(|z|)(Intercept) -3.9235966 0.3779589 -10.381 2e-16 ***Hours 0.0009992 0.0001142 8.754 2e-16 ***—Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1(Dispersion parameter for binomial family taken to be 1)Null deviance: 112.670 on 10 degrees of freedomResidual deviance: 10.331 on 9 degrees of freedomAIC: 49.808Number of Fisher Scoring iterations: 4 1-pchisq(deviance(turbines.glm), df.residual(turbines.glm))[1] 0.3243236 confint(turbines.glm)2.5 % 97.5 %(Intercept) -4.704479229 -3.219091395Hours 0.000783381 0.001231879 exp(100*confint(turbines.glm))2.5 % 97.5 %(Intercept) 4.864778e-205 1.572668e-140Hours 1.081488e+00 1.131097e+00 100*(exp(100*confint(turbines.glm))-1)2.5 % 97.5 %(Intercept) -100.000000 -100.00000Hours 8.148825 13.10969Page 24 of 32STATS 201/208 Pred.data = data.frame(Hours = seq(400,5000,by=100)) Turbine.pred = predict(turbines.glm, Pred.data) Prob=round(exp(Turbine.pred) / (1 + exp(Turbine.pred)),3) names(Prob)=seq(400,5000,by=100) plot(Fissures/Turbines~Hours,ylab=probability, main=Probability of+ turbine having a fissure by number of hour,data=Turbines.df) lines(seq(400,5000,by=100),Prob)1000 2000 3000 40000.0 0.2 0.4 0.6Probability ofturbine having a fissure by number of hourHoursprobability Prob400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 16000.029 0.032 0.035 0.038 0.042 0.046 0.051 0.056 0.062 0.068 0.074 0.081 0.0891700 1800 1900 2000 2100 2200 2300 2400 2500 2600 2700 2800 29000.098 0.107 0.117 0.127 0.139 0.151 0.164 0.179 0.194 0.210 0.227 0.245 0.2643000 3100 3200 3300 3400 3500 3600 3700 3800 3900 4000 4100 42000.284 0.304 0.326 0.348 0.371 0.395 0.419 0.444 0.468 0.493 0.518 0.543 0.5684300 4400 4500 4600 4700 4800 4900 50000.592 0.616 0.639 0.662 0.684 0.705 0.726 0.745Page 25 of 32STATS 201/208Appendix E Nitrous Oxide DataThe following data are the monthly average global atmospheric concentrations of N2Ofrom January 2008 to October 2019 in parts per billion (ppb). N2O.ts = ts(N2O.df$N2O,start=2008,frequency=12) N2O.tsJan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec2008 321.3 321.4 321.4 321.4 321.4 321.5 321.4 321.4 321.5 321.6 321.8 322.02009 322.2 322.3 322.3 322.1 322.0 322.0 322.0 322.1 322.3 322.5 322.7 322.82010 322.9 322.9 322.9 323.0 323.0 323.0 323.0 323.1 323.3 323.5 323.7 323.92011 324.0 324.1 324.1 324.1 324.1 324.0 324.1 324.1 324.2 324.4 324.6 324.82012 324.9 325.0 325.0 324.9 324.9 324.9 324.9 325.0 325.1 325.3 325.4 325.52013 325.6 325.6 325.7 325.7 325.8 325.9 326.0 326.0 326.1 326.2 326.4 326.52014 326.6 326.7 326.7 326.8 326.8 326.9 327.0 327.2 327.3 327.5 327.7 327.92015 328.0 328.1 328.0 328.0 327.9 328.0 328.0 328.1 328.2 328.4 328.6 328.82016 328.9 328.9 328.9 328.9 328.9 328.9 328.9 328.8 328.9 329.0 329.2 329.42017 329.5 329.5 329.5 329.5 329.5 329.6 329.7 329.8 329.9 330.0 330.2 330.32018 330.4 330.6 330.7 330.7 330.7 330.7 330.7 330.9 331.1 331.3 331.5 331.72019 331.8 331.7 331.7 331.6 331.6 331.7 331.9 331.9 331.9 332.1 plot(N2O.ts,xlab=month,ylab=ppb,main=Global atmospheric concentration+ of Nitrous Oxide – Jan 2008 to Oct 2019)Global atmospheric concentrationof Nitrous Oxide Jan 2008 to Oct 2019monthppb2008 2010 2012 2014 2016 2018 2020322 324 326 328 330 332Page 26 of 32STATS 201/208 HW.fit = HoltWinters(N2O.ts) HW.fitHolt-Winters exponential smoothing with trend and additive seasonal component.Call:HoltWinters(x = N2O.ts)Smoothing parameters:alpha: 0.8527811beta : 0.01102914gamma: 1Coefficients:[,1]a 332.17880227b 0.07663861s1 0.01925176s2 0.10884812s3 0.18119490s4 0.22432467s5 0.23304180s6 0.14142845s7 0.01506802s8 -0.08650202s9 -0.14807969s10 -0.18645879s11 -0.16506783s12 -0.07880227Page 27 of 32STATS 201/208 HW.pred = predict(HW.fit,n.ahead=12,prediction.interval=T) plot(HW.fit,HW.pred,main=Global atmospheric concentration of Nitrous Oxide)Global atmospheric concentration of Nitrous OxideTimeObserved / Fitted2010 2012 2014 2016 2018 2020322 326 330 334 HW.predfit upr lwrNov 2019 332.2747 332.4768 332.0726Dec 2019 332.4409 332.7077 332.1741Jan 2020 332.5899 332.9096 332.2702Feb 2020 332.7097 333.0756 332.3437Mar 2020 332.7950 333.2028 332.3873Apr 2020 332.7801 333.2266 332.3336May 2020 332.7303 333.2132 332.2475Jun 2020 332.7054 333.2227 332.1881Jul 2020 332.7205 333.2707 332.1703Aug 2020 332.7587 333.3406 332.1768Sep 2020 332.8568 333.4693 332.2442Oct 2020 333.0197 333.6620 332.3773Page 28 of 32STATS 201/208Note: Define Month as a factor for month of the year coded as 1 for January, 2 forFebruary, etc. Define Time as a numeric variable with values 1 for January 2008, 2 forFebruary 1980, …, 142 for October 2019. Time = 1:142 Month = Factor(c(rep(1:12,11),(1:10))) SF.fit1 = lm(N2O.ts~Time+Month) plot.ts(residuals(SF.fit1),main=Residual Series)Residual SeriesTimeresiduals(SF.fit1)0 20 40 60 80 100 120 1400.3 0.1 0.1 0.2 0.3Page 29 of 32STATS 201/208 acf(residuals(SF.fit1))0 5 10 15 20\r\”