VERSION 1 STATS 330THE UNIVERSITY OF AUCKLANDSEMESTER ONE 2018Campus: CitySTATISTICSAdvanced Statistical Modelling(Time allowed: THREE hours)INSTRUCTIONSSECTION A: Multiple Choice (24 marks) Answer ALL 12 questions on the coloured teleform sheet provided. To answer, fill in the appropriate box on the teleform sheet. Use pencil only. To change an answer, erase the original answer completely andfill in a new answer. If you give more than one answer to any question, you will receive zero marksfor that question. All questions carry the same mark value. All questions have a single correct answer. Incorrect answers are not penalised.SECTION B (76 marks) Answer all questions.Total for both parts: 100 marks.Page 1 of 24VERSION 1 STATS 330SECTION A1. Suppose we fit a model and calculate a 95% confidence interval and a 95%prediction interval for an observation. The confidence interval is (1.1, 2.1) andthe prediction interval is (?0.8, 1.8). Which of these statements is TRUE?(1) If we calculate the confidence interval for a number of successive samplesfrom the same population, it will contain the true mean 90% of the time.(2) If we calculate the confidence interval for a number of successive samplesfrom the same population, it will contain the future observation 95% of thetime.(3) If we calculate the prediction interval for a number of successive samplesfrom the same population, it will contain the future observation 90% of thetime.(4) If we calculate the prediction interval for a number of successive samplesfrom the same population, it will contain the true mean 95% of the time.(5) These intervals are not consistent with statistical theory; they have beencalculated incorrectly.2. Consider diagnostics for a linear model. If the constant variance assumptionfails, which of the following options is a potential solution?(1) Transform the response and fit the model again, using a Box-Cox plot tochoose the transformation.(2) Delete influential points and fit the model again.(3) Use the backwards elimination method.(4) Use weighted least squares, where the weights are the inverse of the coeffi-cients.(5) Delete the variables that are collinear and fit the model again.Page 2 of 24VERSION 1 STATS 330The next two questions are based on the following scenario.Suppose we have a response variable Y and an explanatory factor X, which hasfour levels. We ran the following code in R:> mymodel > plot(mymodel, which = 1:6)3. We want to test whether the effect of all levels of X is the same. What is thecorrect code to do this in R?(1) anova(submodel, mymodel)(2) plot(mymodel)(3) anova(mymodel)(4) t.test(mymodel)(5) summary(mymodel)4. Which plots are produced by the code?(1) Residuals vs Fitted, Normal Q-Q, Scale-Location, Cook’s distance, Residualsvs Leverage, Fitted values.(2) Residuals, Normal Q-Q, Scale-Location, Cook’s distance, Leverage, Fittedvalues.(3) Residuals vs Fitted, Normal Q-Q, Scale-Location, Cook’s distance, Residualsvs Leverage, Cook’s distance vs Leverage.(4) Residuals vs Fitted, Normal Q-Q, Scale-Location, Residuals vs Leverage.(5) Residuals vs Fitted, Normal Q-Q, Scale-Location vs Cook’s distance, Residualsvs Fitted, Cook’s distance vs Leverage.Page 3 of 24VERSION 1 STATS 330The next two questions are based on the following analysis.Blackburn Rovers is a football club based in Lancashire, England. One of theirfans cross-classified all their league matches over the last five seasons by result(Win, Draw, or Loss) and match location (Home or Away). A ‘Home’ matchis played at Ewood Park in Blackburn, while an ‘Away’ match is played at theopposition’s football ground. The data are shown in the following contingencytable:Win Draw LossHome 53 35 27Away 35 40 40The following code was used to analyse these data:> blackburn.dfresult location count1 Win Home 532 Draw Home 353 Loss Home 274 Win Away 355 Draw Away 406 Loss Away 40> blackburn.fit data = blackburn.df)> anova(blackburn.fit, test = Chisq)Analysis of Deviance TableModel: poisson, link: logResponse: countTerms added sequentially (first to last)Df Deviance Resid. Df Resid. Dev Pr(>Chi)NULL 5 9.49result 2 2.91 3 6.58 0.234location 1 0.00 2 6.58 1.000result:location 2 6.58 0 0.00 0.037 *---Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1Page 4 of 24VERSION 1 STATS 330> summary(blackburn.fit)Call:glm(formula = count ~ result * location, family = poisson,data = blackburn.df)Deviance Residuals:[1] 0 0 0 0 0 0Coefficients:Estimate Std. Error z value Pr(>|z|)(Intercept) 3.69e+00 1.58e-01 23.33 resultDraw -2.34e-16 2.24e-01 0.00 1.000resultWin -1.34e-01 2.31e-01 -0.58 0.564locationHome -3.93e-01 2.49e-01 -1.58 0.115resultDraw:locationHome 2.60e-01 3.40e-01 0.76 0.445resultWin:locationHome 8.08e-01 3.31e-01 2.44 0.015 *---Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1(Dispersion parameter for poisson family taken to be 1)Null deviance: 9.4884e+00 on 5 degrees of freedomResidual deviance: -1.4655e-14 on 0 degrees of freedomAIC: 44.81Number of Fisher Scoring iterations: 3> coef(blackburn.fit)(Intercept) resultDraw resultWin3.6889e+00 -2.3424e-16 -1.3353e-01locationHome resultDraw:locationHome resultWin:locationHome-3.9304e-01 2.5951e-01 8.0799e-01> exp(coef(blackburn.fit))(Intercept) resultDraw resultWin40.0000 1.0000 0.8750locationHome resultDraw:locationHome resultWin:locationHome0.6750 1.2963 2.2434Page 5 of 24VERSION 1 STATS 3305. Which of the following statements is FALSE?(1) We estimate that the log-odds of Blackburn Rovers winning rather thanlosing if they are playing at home are approximately 0.81 higher than thelog-odds of them winning rather than losing if they are playing away.(2) We estimate that the odds of Blackburn Rovers winning are approximately0.88 times the odds of them losing.(3) We estimate that the odds of Blackburn Rovers winning rather than losingare approximately 124% higher if they are playing at home than they areif they are playing away.(4) We estimate that the odds of Blackburn Rovers winning rather than losingif they are playing at home are approximately 2.24 times the odds of themwinning rather than losing if they are playing away.(5) There is evidence of an association between the result and the location ofa Blackburn Rovers match.6. Which of the following statements is TRUE?(1) We should drop the result:location interaction because the p-value forresultDraw:locationHome is large.(2) The residual deviance of the model blackburn.fit is zero because we havefitted the saturated model.(3) The null deviance of the model blackburn.fit is smaller than the residualdeviance.(4) If we dropped the result:location interaction, the resulting model wouldprobably have a smaller residual deviance than the model blackburn.fit.(5) We probably should not trust this analysis, because some of the cells havecounts that are too small.Page 6 of 24VERSION 1 STATS 3307. Which of the following statements about logistic regression with a logit linkfunction is FALSE?(1) By default, R uses the logit link function when a logistic regression modelis fitted.(2) We assume that the observations are independent.(3) We assume that the variance of the response variable is constant across allobservations.(4) We assume that the response variable comes from a binomial distribution.(5) We assume that the log-odds of a trial being successful is a linear combinationof the explanatory variables.8. A Poisson regression model with a log link funcion was fitted to a responsevariable Y , using only a single numeric expanatory variable, X. Estimates ofthe linear predictor’s intercept, β0, and slope, β1, were obtained. Which of thefollowing is an appropriate interpretation?(1) When x = 0, we estimate that the odds of success are equal to exp(β0)/(1+exp(β0)).(2) For every one-unit increase in X, we estimate that the expected value ofY increases by β1.(3) When x = 0, we estimate that the expected value of Y is β0.(4) For every one-unit increase in X, we estimate that the odds of success aremultiplied by exp(β1).(5) For every one-unit increase in X, we estimate that the expected value ofY is multiplied by exp(β1).Page 7 of 24VERSION 1 STATS 3309. Which of the following statements about the use of offsets in generalised linearmodels is FALSE?(1) While offsets are most common for Poisson regression models, we can usethem for logistic or standard linear regression models, too.(2) We estimate a coefficient for an offset, which allows us to interpret therelationship between the offset and the response variable.(3) An offset adds a fixed value to the linear predictor for each observation.(4) When we fit a Poisson regression model with a log link function in R, wecan use the argument offset = log(t) if we think the expected value ofthe response is directly proportional to the variable t.(5) An explanatory variable’s estimated coefficient is very close to 1. The fittedvalues from our model are unlikely to change much if we use the variableas an offset instead.10. Which of the following statements about estimated coefficients of linear models(LMs) and generalised linear models (GLMs) is FALSE?(1) The estimated coefficients of a GLM maximise the deviance.(2) The estimated coefficients of a GLM minimise the sum of the squareddeviance residuals.(3) The estimated coefficients of a GLM maximise the likelihood.(4) The estimated coefficients of a GLM maximise the log-likelihood.(5) The estimated coefficients of a LM minimise the residual sum of squares.Page 8 of 24VERSION 1 STATS 33011. Consider a logistic regression model fitted to ungrouped data. The observedresponses from the first two observations are given by y1 and y2. The first wasobserved as a success (y1 = 1), and has a fitted probability under the modelof p1 = 0.8. The second was observed as a failure (y2 = 0), and has a fittedprobability under the model of p2 = 0.7. Which of the following statements isTRUE?(1) Under the fitted model, the expected value of the first observation is 0.2,and the expected value of the second observation is 0.3.(2) The Pearson residual of the second observation is larger in magnitude (i.e.,further from zero) than the Pearson residual of the first observation.(3) The deviance residual of the first observation is negative, and the devianceresidual of the second observation is positive.(4) It is possible for an observation to have a positive deviance residual, but anegative deviance residual.(5) The further an observed value is from its expected value, the closer to zerothe deviance residual is.12. Which of the following statements is FALSE?(1) The null deviance is always equal to or smaller than the residual deviance.(2) The residual deviance is equal to twice the difference between the loglikelihoodof the saturated model and the log-likelihood of the fitted model.(3) The residual deviance of the saturated model is always equal to zero.(4) The residual deviance is equal to the sum of the squared deviance residuals.(5) The log-likelihood of the saturated model is always equal to or larger thanthe log-likelihood of the fitted model.Page 9 of 24VERSION 1 STATS 330SECTION B13. [8 marks] Guess the analysis: choose the most appropriate model to fit foreach of the scenarios described below. Different scenarios may have the sameanswer.For each scenario, select one of these three possible answers:(1) Linear regression model(2) Logistic regression model(3) Poisson regression model(a) TVNZ has just released a new TV show. Their market analyst wishes tobuild a model to predict whether or not specific individuals will enjoy theshow. They conduct a survey, collecting variables from participants suchas gender, age, income, and occupation. They also asked participants ifthey enjoyed a pilot episode of the show.[2 marks](b) A STATS 330 lecturer is interested to see if the number of questions postedon Piazza is related to the closeness of an assignment deadline. Each day,they count the number of Piazza questions that were posted, and recordthe number of days until the next assignment deadline.[2 marks](c) A detective wishes to determine whether or not there is an association betweena serial killer’s gender (female or male) and their preferred method(poisoning, strangulation, and so on). They cross-classify a sample of convictedserial killers using these two variables.[2 marks](d) A University of Auckland empolyee wishes to determine the quickest wayto get to work. Each day, they randomly select a transportation method(bus, train, or walk) and a departure time (8:00am, 8:15am, or 8:30am).They record how long their journey took.[2 marks]Page 10 of 24VERSION 1 STATS 33014. [4 marks] Suppose we wish to predict the expenditure on cancer treatmentfor a patient at the Auckland hospital. We have collected the following variablesfrom a sample of cancer patients: Stage of cancer Type of cancer Cancer treatment expenditure Age Gender Ethnicity Marital status(a) Is it possible to build a predictive model with this information? Explainyour answer.[2 marks](b) We also want to build an explanatory model to investigate if dietary habitscause cancer. Can we fit a model to do so using only the variables above?Explain your answer.[2 marks]Page 11 of 24VERSION 1 STATS 33015. [17 marks] The data for this question were collected from a sample of 44male and 51 female athletes at the Australian Institute of Sport. The data setcontains the following variables:sex The athlete’s sex, either female or male.sport The athlete’s sport, either basketball (BBal), rowing (Row),swimming (Swim), or tennis (Tennis).BMI The athletes body mass index, calculated by dividing theirweight (in kg) by their height (in m) squared.X.Bfat The athlete’s body fat percentage.Printed below are the first three observations, and summary statistics for eachof the代做STATS 330留学生作业、代写R程序语言作业、代做VERSION作业、R实验作业代写 调试Matlab程序|帮做 variables:> head(sport.df, 3)sex sport BMI X.Bfat1 female BBall 20.56 19.752 female BBall 20.67 21.303 female BBall 21.86 19.88> summary(sport.df)sex sport BMI X.Bfatfemale:51 BBall :25 Min. :17.1 Min. : 6.16male :44 Row :37 1st Qu.:21.3 1st Qu.: 8.92Swim :22 Median :22.7 Median :12.20Tennis:11 Mean :22.8 Mean :13.913rd Qu.:24.0 3rd Qu.:18.62Max. :26.8 Max. :28.83We analysed these data using the following code:> sport.fit > reg > regAIC BIC CV BMI Row Swim Tennis male Row:male Swim:male Tennis:male1 161.12 166.23 0.470 0 0 0 0 1 0 0 02 130.05 137.71 0.384 1 0 0 0 1 0 0 03 112.36 122.58 0.332 1 0 1 0 1 0 0 04 107.89 120.66 0.318 1 0 1 0 1 0 1 05 106.14 121.46 0.316 1 0 1 1 1 0 1 06 102.21 120.09 0.308 1 0 1 1 1 0 1 17 102.84 123.28 0.310 1 1 1 1 1 0 1 18 104.00 126.98 0.313 1 1 1 1 1 1 1 1Page 12 of 24VERSION 1 STATS 330> sport.fit.2 > anova(sport.fit.2)Analysis of Variance TableResponse: log(X.Bfat)Df Sum Sq Mean Sq F value Pr(>F)sport 3 1.87 0.625 3.91 0.011 *Residuals 91 14.54 0.160---Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1> summary(sport.fit.2)Coefficients:Estimate Std. Error t value Pr(>|t|)(Intercept) 2.5956 0.0799 32.47 sportRow 0.0752 0.1035 0.73 0.469sportSwim -0.2835 0.1168 -2.43 0.017 *sportTennis -0.1043 0.1446 -0.72 0.472---Residual standard error: 0.4 on 91 degrees of freedomMultiple R-squared: 0.114,Adjusted R-squared: 0.085F-statistic: 3.91 on 3 and 91 DF, p-value: 0.0112(a) Write down the mathematical formula for the model sport.fit. Thisshould describe the relationship between the explanatory variables and theresponse, and show its assumptions.[3 marks](b) Based on the results from the allposregs() function, which model willreturn the lowest prediction error? Write down its mathematical formulaand the R code that could be used to fit this model.[3 marks](c) Consider the model sport.fit.2 and the output from the anova() function,shown above. What is the null hypothesis, or what are the nullhypotheses, associated with the p-value(s) in this output?[3 marks]Page 13 of 24VERSION 1 STATS 330(d) What do you conclude about the hypothesis (or hypotheses) from the outputof the anova() function?[2 marks](e) Consider the model sport.fit.2 and the output from the summary() function,shown above. What is the null hypothesis, or what are the null hypotheses,associated with the p-value(s) in this output?[3 marks](f) What do you conclude about the hypothesis (or hypotheses) from the outputof the summary() function?[3 marks]Page 14 of 24VERSION 1 STATS 33016. [18 marks] The data for this question are related to a sample of 1599Portugese red wines. Various physiochemical properties of the wines were measured.Additionally, a panel of judges decided whether or not each wine was of‘good quality’. The data set contains the following variables:good.quality This variable takes the value 1 if the wine is of ‘good quality’,and 0 otherwise.fixed.acidity The fixed concentration of tartaric acid (g per dm3).volatile.acidity The volatile concentration of tartaric acid (g per dm3).residual.sugar The concentration of residual sugars (g per dm3).chlorides The concentration of sodium chloride (g per dm3).f.sulfur.dioxide The concentration of free sulfur dioxide (mg per dm3).density The density of the wine (g per cm3).sulphates The concentration of potassium sulphate (g per dm3).alcohol The alcohol level of the wine (percentage alcohol by volume).The following final model was fitted in R:> wine.fit residual.sugar + chlorides + t.sulfur.dioxide +I(t.sulfur.dioxide^2) + density + sulphates +I(sulphates^2) + alcohol + I(alcohol^2),family = binomial, data = wine.df)> summary(wine.fit)Call:glm(formula = good.quality ~ fixed.acidity + volatile.acidity +Coefficients:Estimate Std. Error z value Pr(>|z|)(Intercept) 2.44e+02 1.01e+02 2.42 0.01554 *fixed.acidity 2.68e-01 8.48e-02 3.17 0.00154 **volatile.acidity -2.28e+00 6.66e-01 -3.43 0.00061 ***residual.sugar 2.24e-01 7.87e-02 2.85 0.00436 **chlorides -6.89e+00 3.57e+00 -1.93 0.05378 .t.sulfur.dioxide -2.85e-02 7.65e-03 -3.73 0.00020 ***I(t.sulfur.dioxide^2) 1.17e-04 5.07e-05 2.31 0.02104 *density -2.93e+02 1.02e+02 -2.88 0.00399 **sulphates 2.22e+01 5.06e+00 4.39 1.1e-05 ***I(sulphates^2) -1.10e+01 3.25e+00 -3.39 0.00070 ***alcohol 5.67e+00 1.49e+00 3.81 0.00014 ***I(alcohol^2) -2.20e-01 6.51e-02 -3.38 0.00072 ***---(Dispersion parameter for binomial family taken to be 1)Null deviance: 1269.92 on 1598 degrees of freedomResidual deviance: 825.17 on 1587 degrees of freedomPage 15 of 24VERSION 1 STATS 330> ROC.curve(wine.fit)Area under ROC curve = 0.87660.0 0.2 0.4 0.6 0.8 1.00.0 0.2 0.4 0.6 0.8 1.0False positive rateTrue positive rate●AUC = 0.8766The following code was used to predict which wines in the sample were ‘goodquality’ wines. A probability cutoff of c = 0.5 was used.> wine.pred > wine.predcode The results of this code have been reformatted to give the following confusionmatrix:PredictedGood Not goodObserved Good 85 132Not good 41 1341Page 16 of 24VERSION 1 STATS 330(a) Interpret the effects of the following variables on wine quality under themodel wine.fit:(i) Concentration of residual sugars.[2 marks](ii) Volatile concentration of tartaric acid.[2 marks](b) Calculate the following:(i) In-sample sensitivity.[2 marks](ii) In-sample specificity.[2 marks](iii) In-sample error rate.[2 marks](c) Based on your calculations, Comment on the model’s predictive powerusing a probability cutoff of c = 0.5. How well does it predict wines thatare good quality? How well does it predict wines that are not good quality?[3 marks]Page 17 of 24VERSION 1 STATS 330(d) A chemist wishes to use this model to predict which wines are good quality.However, they want the model to correctly predict 80% of the good-qualitywines in the sample. They adjust their probability cutoff c accordingly.Using the ROC curve above, approximately what proportion of wines inthe sample that are not of good quality will they correctly predict usingthis adjusted cutoff? Give your answer to one decimal place.[2 marks](e) Specificity and sensitivity can also be calculated via crossvalidation usingthe R function cross.val(). Would you expect the sensitivity and speci-ficity calculated by cross.val() to be higher or lower than your in-samplecalculations above? Explain your answer.[3 marks]Page 18 of 24VERSION 1 STATS 33017. [29 marks] In 2015, New Zealand’s National Institute of Water and AtmosphericResearch (NIWA) sampled 486 sites on rivers around the country. Ateach site, they recorded whether or not various freshwater species were present.Each site was cross-classified based on these presence/absence data to form acontingency table. The data set fishy.df has the following variables:eel Presence (1) or absence (0) of the longfin eel.koura Presence (1) or absence (0) of the koura, a type of crayfish.bully Presence (1) or absence (0) of the upland bully.trout Presence (1) or absence (0) of the brown trout.count The number of sites with a particular combination of theabove variables.Freshwater biologists were interested in how the species interact. Is it commonto find species at the same site? Or do some species avoid one another?The data are shown below:> fishy.dfeel koura bully trout count1 0 0 0 0 2332 0 0 0 1 413 0 0 1 0 354 0 0 1 1 125 0 1 0 0 126 0 1 0 1 27 0 1 1 0 28 0 1 1 1 19 1 0 0 0 5210 1 0 0 1 3411 1 0 1 0 912 1 0 1 1 1313 1 1 0 0 1614 1 1 0 1 815 1 1 1 0 416 1 1 1 1 12For example, from the first row, there were 233 sites that had none of the speciespresent. From the fourth row, there were 12 sites at which the upland bully andthe brown trout present, but the longfin eel and koura were not present.> fishy.fit.1 family = poisson, data = fishy.df)>> fishy.fit.2 family = poisson, data = fishy.df)Page 19 of 24VERSION 1 STATS 330> summary(fishy.fit.2)Call:glm(formula = count ~ (eel + koura + bully + trout)^2, family = poisson,Coefficients:Estimate Std. Error z value Pr(>|z|)(Intercept) 5.4632 0.0642 85.04 eel -1.5204 0.1449 -10.49 koura -3.0946 0.2668 -11.60 bully -1.9837 0.1711 -11.59 trout -1.7869 0.1595 -11.21 eel:koura 1.8526 0.3251 5.70 1.2e-08 ***eel:bully 0.2525 0.2749 0.92 0.35833eel:trout 1.3429 0.2347 5.72 1.1e-08 ***koura:bully 0.7185 0.3367 2.13 0.03286 *koura:trout 0.0911 0.3280 0.28 0.78115bully:trout 0.8875 0.2627 3.38 0.00073 ***---(Dispersion parameter for poisson family taken to be 1)Null deviance: 850.0360 on 15 degrees of freedomResidual deviance: 3.0544 on 5 degrees of freedom> fishy.fit.3 eel:bully + eel:trout + koura:bully + bully:trout,family = poisson, data = fishy.df)> summary(fishy.fit.3)Call:glm(formula = count ~ eel + koura + bully + trout + eel:koura +Coefficients:Estimate Std. Error z value Pr(>|z|)(Intercept) 5.4628 0.0643 85.02 eel -1.5290 0.1421 -10.76 koura -3.0821 0.2626 -11.73 bully -1.9869 0.1710 -11.62 trout -1.7838 0.1591 -11.22 eel:koura 1.8773 0.3127 6.00 1.9e-09 ***eel:bully 0.2463 0.2743 0.90 0.36920eel:trout 1.3623 0.2241 6.08 1.2e-09 ***koura:bully 0.7363 0.3305 2.23 0.02588 *bully:trout 0.8959 0.2609 3.43 0.00059 ***---(Dispersion parameter for poisson family taken to be 1)Null deviance: 850.0360 on 15 degrees of freedomResidual deviance: 3.1312 on 6 degrees of freedomPage 20 of 24VERSION 1 STATS 330> fishy.fit.4 eel:trout + koura:bully + bully:trout,family = poisson, data = fishy.df)> summary(fishy.fit.4)Call:glm(formula = count ~ eel + koura + bully + trout + eel:koura +Coefficients:Estimate Std. Error z value Pr(>|z|)(Intercept) 5.4569 0.0641 85.09 eel -1.4937 0.1355 -11.02 koura -3.1160 0.2630 -11.85 bully -1.9363 0.1596 -12.13 trout -1.8106 0.1583 -11.44 eel:koura 1.9013 0.3109 6.12 9.7e-10 ***eel:trout 1.3934 0.2213 6.30 3.0e-10 ***koura:bully 0.8300 0.3132 2.65 0.00804 **bully:trout 0.9636 0.2494 3.86 0.00011 ***---(Dispersion parameter for poisson family taken to be 1)Null deviance: 850.0360 on 15 degrees of freedomResidual deviance: 3.9273 on 7 degrees of freedom> confint(fishy.fit.4)Waiting for profiling to be done...2.5 % 97.5 %(Intercept) 5.32863 5.5801eel -1.76581 -1.2340koura -3.66910 -2.6328bully -2.26073 -1.6339trout -2.13076 -1.5094eel:koura 1.30797 2.5336eel:trout 0.96218 1.8307koura:bully 0.20057 1.4340bully:trout 0.47172 1.4517> AIC(fishy.fit.1, fishy.fit.2, fishy.fit.3, fishy.fit.4)df AICfishy.fit.1 16 101.936fishy.fit.2 11 94.990fishy.fit.3 10 93.067fishy.fit.4 9 91.863Page 21 of 24VERSION 1 STATS 330(a) What are the assumptions of a Poisson regression model?[2 marks](b) Write an equation to calculate the expected number of sites with a particularcombination of the presence/absence variables under the modelfishy.fit.4. Define any notation you use that is not obvious.[2 marks](c) Consider the following code and output:> anova(fishy.fit.2, fishy.fit.1, test = Chisq)Analysis of Deviance TableModel 1: count ~ (eel + koura + bully + trout)^2Model 2: count ~ eel * koura * bully * troutResid. Df Resid. Dev Df Deviance Pr(>Chi)1 5 3.052 0 0.00 5 3.05 0.69What is the null hypothesis being tested here? Refer to effects estimatedby the model fishy.fit.1 in your answer.[3 marks](d) What can you conclude from the hypothesis test conducted in question(c)?[2 marks](e) The model fishy.fit.2 was simplified to fishy.fit.3, and then furthersimplified to fishy.fit.4. Briefly state why you think that these weresensible decisions.[2 marks]Page 22 of 24VERSION 1 STATS 330(f) Sketch the association graph for the model fishy.fit.4.[3 marks](g) Describe the relationship between the following pairs of factors under themodel fishy.fit.4. For each pair, select one of these three possible answers:(1) Independent(2) Conditionally independent given other factors(3) DependentIf you select option (2), state which other factor(s) the independence isconditional upon. Both pairs may have the same answer.(i) bully and trout[2 marks](ii) bully and eel[2 marks](h) Assume the model fishy.fit.4 is the correct model. A freshwater biologistis interested in the association between presence of the koura andpresence of the brown trout. They wish to simplify the contingency tableby collapsing over another factor.(i) Is it appropriate to collapse over the factor eel? Briefly explain youranswer.[2 marks](ii) Is it appropriate to collapse over the factor bully? Briefly explainyour answer.[2 marks]Page 23 of 24VERSION 1 STATS 330(i) The freshwater biologist believes that the longfin eel and the brown troutavoid one another. In other words, holding all other variables constant,sites with longfin eel are less likely to have brown trout present than siteswithout longfin eel. Does the analysis above suggest that the biologist’sbelief is correct? Explain your answer.[3 marks](j) Provide a 95% confidence interval for the odds ratio that quantifies theassociation between the presence of koura and the presence of the uplandbully, holding all other variables constant.[2 marks](k) Write a sentence interpreting your confidence interval from question (j).[2 marks]转自：http://ass.3daixie.com/2019011930482420.html
