MAS183 Statistical Data Analysis Final Examination – Semester 1, 2020Page 1 of 51. [8 marks]A tyre manufacturer tested the braking performance of one of its tyre models on a test track. Newtyres were fitted to 10 different cars. Each car was run at 100 kilometres per hour for 15 minutes, andthen stopping distance was measured from start of braking to full stop. The test was run on both dryand wet road (continuous wetting equivalent to 2 mm of rain per hour). The company wants to knowhow braking distance differs between wet and dry road conditions.(a) Complete the following statement by choosing ONE option from each of columns 1-4 below. Ifyou think more than one answer may be possible, choose the best answer. Write your statement in your answer book.Statement: “The research question concerns the…”[4] 1comparison ofdistribution ofrelation betweenvalue of 2multipleonetwo 3categoricalnumericalpopulationsample 4mean(s)median(s)proportion(s)variable(s) (b) Write appropriate hypothesis statements for the relevant hypothesis test. [4]2. [7 marks]In a study of patients with stable coronary artery disease, patients were randomly allocated to twotreatment groups. One group stayed with their old medications, while the other group was started ona new medication. Researchers performed several tests to check for differences between the groupsat the start of the study. One characteristic of interest was the proportion of patients reporting ahistory of congestive heart failure. Summary data are reported below: TreatmentgroupNo. ingroupNo. reportingcongestive heart failureOld medsNew med147 48153 28 (a) Calculate a 99% confidence interval for the difference in the proportions reporting congestiveheart failure in the (notional) populations from which the two groups were selected. [5](b) Use your result from part (a) to assess whether the treatment groups differ significantly withrespect to reported congestive heart failure. [2]3. [12 marks]For certain parent genotypes, offspring will have genotypes characterised as ΑA (healthy), Αa (carrier),and aa (has disease symptoms). These genotypes are expected to occur in the ratios 1:2:1 respectively.For ninety five offspring of parents with the relevant genotypes, the following genotype distributionwas observed: GenotypeAA Aa aaCount22 55 18 Evaluate the evidence that the observed distribution of genotypes differs from the expecteddistribution. If you conclude that there is a difference, describe it.MAS183 Statistical Data Analysis Final Examination – Semester 1, 2020Page 2 of 54. [18 marks]Gingko biloba, commonly known as gingko, is an ancient tree species native to China. It is marketed ashaving the ability to improve memory. In 2002 a study was conducted to test this claim. Elderlypeople in generally good health were recruited via newspaper advertisements. They were screened toensure that all were at least 60 years old, in good health, living in the general community andindependent in their daily activities. Any with a history of mental illness or any major illness in the last5 years were excluded, as were those who had taken any psychoactive medication in the last 60 days.In all, 98 men and 132 women participated in the study. At the outset, they were randomly allocatedbetween a treatment group (taking an over-the-counter gingko supplement according to themanufacturer’s instructions) and a placebo group. The medications were similar in appearance.Neither participants nor those treating them knew who was in which group. During the study someparticipants withdrew consent or were excluded for non-compliance with the medication regime.Participants did a baseline memory test before starting to take medication, and were tested again aftertaking it for 6 weeks. The data presented here are the differences between baseline and final memoryscores, calculated as final score – baseline, and given the variable name “Improv”.> describeBy(Improv, Group, mat=TRUE,skew=FALSE, ranges=FALSE, digits=3)item group1 vars n mean sd seX11 1 Gingko 1 104 4.231 5.032 0.493X12 2 Placebo 1 99 5.222 4.117 0.414> t.test(Improv ~ Group, conf.level = .9) # Independent samples t-testWelch Two Sample t-testdata: Improv by Groupt = -1.5395, df = 196.61, p-value = 0.1253alternative hypothesis: true difference in means is not equal to 090 percent confidence interval:-2.05573526 0.07282928sample estimates:mean in group Gingko mean in group Placebo4.230769 5.222222(a) Provide a 95% confidence interval estimate for the mean memory score improvement amonghealthy elderly people taking gingko according to manufacturer’s instructions. [3](b) Formally evaluate the evidence that ginko and placebo differ in their effect on memory score. [12](c) The gingko supplement manufacturer claimed memory improvement would be noticeable after4 weeks. Using your results in (a) and (b), comment on the validity of this claim. [3]Gingko Placebo-10-55 010Memory score improvementTreatmentMemory score improvement while takingGingko or PlaceboMAS183 Statistical Data Analysis Final Examination – Semester 1, 2020Page 3 of 55. [16 marks]Most climate scientists claim that changes in mean global temperature are related to changes inatmospheric carbon dioxide (CO2) concentrations. The data presented here are from two publiclyavailable data sets covering the period 1959 to 2016. The variables are – GTAGlobal Temperature Anomaly (oC), is a sophisticated measure reflecting the globally-averageddifference between mean temperature in a particular year and the average temperature overthe twentieth century. (source https://www.ncdc.noaa.gov/cag/data-info/global)Atmospheric carbon dioxide concentration in parts per million (ppm) measured at Mauna LoaCO2 Observatory, Hawaii. (source ftp://aftp.cmdl.noaa.gov/products/trends/co2/co2_annmean_mlo.txt)Call:lm(formula = GTA ~ CO2)Residuals:Min 1Q Median 3Q Max-0.19302 -0.06951 -0.01530 0.07302 0.17683Coefficients:Estimate Std. Error t value Pr(>|t|)(Intercept) -3.1793300 0.1584344 -20.07 summary(CB.model)Df Sum Sq Mean Sq F value Pr(>F)Treatment 2 716.2 358.1 7.415 0.00139 **Residuals 56 2704.4 48.3> leveneTest(CB.model,center=median)Levene’s Test for Homogeneity of VarianceDf F value Pr(>F)group 2 3.0492 0.0553456> describeBy(VolChg, Treatment, mat = TRUE, skew=FALSE, range=FALSE, digits=3)item group1 vars n mean sd seX11 1 Bath 1 22 4.545 7.763 1.655X12 2 Bath+Exer 1 23 8.000 7.039 1.468X13 3 Exercise 1 14 -1.071 5.181 1.385 (a)What assumptions are required for the above statistical analysis?[3](b) As far as possible, assess the extent to which the assumptions and conditions have been met.[5] (c) Using a 5% significance level, evaluate the evidence that the treatments have differing effects onthe average change in hand volume. Do not repeat matters already addressed in parts (b) and (c).[5](d) Briefly assess whether contrast baths help to reduce hand volume in post-operative CTS patients.[2]END OF PAPERBath Bath+Exer Exercise-10-55 0101520Change in hand volume during treatmentTreatmentChange in hand volume (ml)

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