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Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 1 of 21 Exam information Course code and title r> STAT1201 Analysis of Scientific Data Semester Semester 1, 2021 Exam type Online, non-invigilated, final examination Exam technology Exam accessed through the Exam Tool in Blackboard Exam date and time Your examination will begin at the time specified in your personal examination timetable. If you commence your examination after this time, the end for your examination does NOT change. The total time for your examination from the scheduled starting time will be: 2 hours 10 minutes (including 10 minutes reading time during which you should read the exam paper and plan your responses to the questions). A 15-minute submission period is available for submitting your examination after the allowed time shown above. If your examination is submitted after this period, late penalties will be applied unless you can demonstrate that there were problems with the system and/or process that were beyond your control. Exam window You must commence your exam at the time listed in your personalised timetable. You have from the start date/time to the end date/time listed in which you must complete your exam. Permitted materials During the exam you may access your own notes and any of the material on the STAT1201 Blackboard and Edge sites. A Casio fx-82 series or UQ approved calculator may be used. You may not make use of any other material. This includes web sites,books, or other software Required materials Ensure the following materials are available during the exam: You will need access to RStudio to complete the exam. Instructions The exam involves a total of 50 multiple-choice questions split into four scenarios. You may attempt the scenarios and questions in any order and return to make changes if needed. The Exam is accessed through the STAT1201 Blackboard page under Examinations. Answers are automatically recorded so there is no submission process at the end of the exam. Who to contact If you have any concerns or queries about a particular question, or need to make any assumptions to answer the question, state these using the ‘Assumptions’ link at the bottom of the exam. If you experience any technical difficulties during the exam, join Zoom 835 0875 5861 (via app or by dialling +61 2 8015 2088), or email
[email protected]
with details of the issue. Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 2 of 21 Late or incomplete submissions Answers are automatically recorded so there is no submission process at the end of the exam. Important exam condition information Academic integrity is a core value of the UQ community and as such the highest standards of academic integrity apply to all examinations, whether undertaken in- person or online. This means: • You are permitted to refer to the allowed resources for this exam, but you cannot cut-and-paste material other than your own work as answers. • You are not permitted to consult any other person – whether directly, online, or through any other means – about any aspect of this examination during the period that it is available. • If it is found that you have given or sought outside assistance with this examination, then that will be deemed to be cheating. Undertaking this online exam deems your commitment to UQ’s academic integrity pledge as summarised in the following declaration: “I certify that I have completed this examination in an honest, fair and trustworthy manner, that my submitted answers are entirely my own work, and that I have neither given nor received any unauthorised assistance on this examination”. Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 3 of 21 MDMA and post-traumatic stress disorder A study investigated if psychotherapy combined with limited administration of 3,4 Methylenedioxymethamphetamine (MDMA) can reduce symptoms of post-traumatic stress disorder. Severity of symptoms was measured via the CAPS-IV score with higher scores indicating more severe symptoms. Subjects recruited to the study were randomly allocated to one of the three dosage levels (Low – 40 mg, Medium – 100 mg, High – 125 mg). The primary outcome was the reduction in CAPS-IV score one month after the end of treatment. A secondary outcome was whether the subject experience a drop of 20% or more in CAPS-IV score. Download the csv file using the link below and read it into RStudio: MDMA.csv The data contains the following variables: • Before - CAPS-IV scores before treatment. • After - CAPS-IV scores at the one month after the second treatment session. • Change - Reduction in CAPS-IV scores (Before – After). • Drop20 - Was a 20% reduction in CAPS-IV score achieved (TRUE or FALSE). • Dose – Dosage level of MDMA given (Low – 40 mg, Medium – 100 mg, High – 125 mg). Question 1 Based on the information above, which of the following best describes the study? Answer: Randomised comparative experiment Distractors: ❑ Observational study ❑ Randomised complete block design ❑ Randomised comparative double-blind experiment. Question 2 Mean CAPS-IV score before treatment in low dose group Answer: 78.5 Distractors: ❑ 78 ❑ 77.75926 ❑ 62.2963 Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 4 of 21 Question 3 Researchers conduct a one-way ANOVA to determine if there is a difference in the mean reduction in CAPS-IV for the three dosage levels of MDMA. Which of the following describes the null hypothesis that is test in a one-way ANOVA? Answer: The mean change in CAPS-IV score is the same for all dosage levels. Distractors: ❑ The mean change in CAPS-IV score is zero for all dosage levels. ❑ The mean change in CAPS-IV score is different for all dosage levels. ❑ The mean change in CAPS-IV score is different for at least one dosage level. Question 4 What is the residual degrees of freedom in the one-way ANOVA to compare dosage level? Answer: 51 Distractors: ❑ 2 ❑ 52 ❑ 53 Question 5 What is the Total Sum of Squares for the one-way ANOVA to compare dosage levels? Answer: 297.43 Distractors: ❑ 56.26 ❑ 241.17 ❑ 303.04 Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 5 of 21 Question 6 Based on the F-test for the one-way ANOVA to compare the dosage levels, you can conclude there is Answer: strong evidence to suggest that there is an effect of MDMA dosage level on the mean change in patient’s CAPIS-IV score (p < 0.01) Distractors: ❑ no evidence to suggest that there is an effect of MDMA dosage level on the mean change in patient’s CAPIS-IV score (p >0.1) ❑ weak evidence to suggest that there is an effect of MDMA dosage level on the mean change in patient’s CAPIS-IV score (p < 0.1) ❑ moderate evidence to suggest that there is an effect of MDMA dosage level on the mean change in patient’s CAPIS-IV score (p < 0.05) Question 7 What is the R2 value for this one-way ANOVA? Answer: 0.1892 Distractors: ❑ 0.1574 ❑ 0.8108 ❑ 0.4349 Question 8 Which of the following is NOT an assumption of the one-way ANOVA Answer: Linear relationship between mean change in CAPS-IV score and dosage level. Distractors: ❑ The subjects recruited to the study are independent. ❑ The variability of the change in CAPS-IV scores does not depend on the dosage level. ❑ The change in CAPS-IV score has a normal distribution. Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 6 of 21 Question 9 Which of the following plots would be useful in assessing model assumptions? Answer: Normal probability plot of the residuals from the one-way ANOVA Distractors: ❑ Normal probability plot of the changes in CAPS-IV score ❑ Scatterplot of the fitted values from the one-way ANOVA against dosage level. ❑ Scatterplot of Before against After CAPS-IV scores. Question 10 The researchers would like to understand which (if any) dosage levels of MDMA result in different mean changes in CAPS-IV score. Using Tukey’s Honestly Significant Difference, the researchers are able to conclude that the mean change in CAPS-IV score are different at the 5% (familywise) significance level for the following dosage levels: Answer: Low – Medium, and Low – High Distractors: ❑ Only Low – High ❑ Low – High, and Medium – High ❑ Low – Medium, Low – High, and Medium – High Question 11 A secondary outcome for the experiment was whether subjects experienced a 20% drop in CAPS-IV scores at one month after the second treatment session. How many subjects in the study experienced a 20% drop in their CAPS-IV score? Answer: 26 Distractors: ❑ 18 ❑ 22 ❑ 29 Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 7 of 21 Question 12 What proportion of participants received a low dose of MDMA and experienced a 20% drop in their CAPS-IV score? Answer: 0.074 Distractor: ❑ 0.154 ❑ 0.222 ❑ 0.160 Question 13 If there was no association between experiencing a 20% drop in CAPS-IV score and dosage level, what is the expected count for subjects in the low dose group experiencing a 20% drop in CAPS-IV score? Answer: 8.667 Distractors: ❑ 8.308 ❑ 12 ❑ 9 Question 14 What is the value of the 2 statistic used to test for an association between experiencing a 20% drop in CAPS-IV score and dosage level? Answer: 9.049 Distractors: ❑ 4.352 ❑ 5.217 ❑ 7.273 Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 8 of 21 Question 15 What is the degrees of freedom for the 2 test in Question 12? Answer: 2 Distractors: ❑ 3 ❑ 5 ❑ 6 Question 16 Based on this analysis, you can conclude that there is Answer: moderate evidence to suggest that there is an association between dosage level and experiencing a 20% drop in CAPS-IV score (p = 0.0108) Distractors: ❑ no evidence to suggest that there is an association between dosage level and experiencing a 20% drop in CAPS-IV score (p = 0.1135) ❑ weak evidence to suggest that there is an association between dosage level and experiencing a 20% drop in CAPS-IV score (p = 0.0736) ❑ moderate evidence to suggest that there is an association between dosage level and experiencing a 20% drop in CAPS-IV score (p = 0.0263) Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 9 of 21 Age estimation from blood cells DNA methylation is a known biomaker for age and has been applied in forensic investigations. The aim is to construct a model which accurately predicts a person’s age from DNA methylation data obtained from blood samples. The blood samples were taken from people aged between 1 year and 90 years old. The percentage of DNA methylation was measured for two genes ELOVL2 and PDE4C. Download the csv file using the link below and read it into RStudio: DNA.csv The data contains the following variables: • Age - Chronological age (years) • Gender – (Female/Male) • ELOVL2 - % of DNA methylation at ELOVL2 • PDE4C - % of DNA methylation at PDE4C Question 1 What is the interquartile range of the ELOVL2 variable? Answer: 42.69 Distractors: ❑ 46.00 ❑ 29.55 ❑ 37.69 Question 2 Based on a box plot of ELOVL2, how would you describe the shape of its distribution? Answer: Approximately symmetric Distractors: ❑ Symmetric with outliers ❑ Skewed to the right ❑ Skewed to the left Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 10 of 21 Question 3 Which of the following best describes the relationship between Age and ELOVL2? Answer: A strong positive linear relationship (r = 0.96) Distractors: ❑ A positive non-linear relationship ❑ A strong negative linear relationship (r = -0.96) ❑ A weak positive linear relationship (r=0.32) Question 4 What is the slope of the least-squares line for the relationship between Age and ELOVL2? Answer: 0.7571 Distractors: ❑ 2.2376 ❑ 0.8117 ❑ 0.8731 Question 5 What is the margin of error in a 95% confidence interval for the slope of the relationship between Age and ELOVL2? Answer: 0.0547 Distractors: ❑ 0.0536 ❑ 0.1124 ❑ 0.0273 Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 11 of 21 Question 6 Based on the linear model, what is the estimated Age of a person whose blood sample has 60% DNA methylation at ELOVL2? Answer: 47.662 Distractors: ❑ 50.121 ❑ 45.948 ❑ 63.236 Question 7 Can we improve the model of Age by including the percentage of DNA methylation at both ELOVL2 and PDE4C? We can address this question using a multiple regression model of Age where the explanatory variables are ELOVL2 and PDE4C. Assuming there is no interaction term, the PDE4C coefficient in the multiple regression model is Answer: 0.2446 Distractors: ❑ 0.5261 ❑ 0.1112 ❑ 0.7683 Question 8 What is the line sum of squares (or explained sum of squares) for the multiple regression model in question 7? Answer: 19101 Distractors: ❑ 18989 ❑ 20450 ❑ 17365 Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 12 of 21 Question 9 Based on the normal probability plot of residuals for the multiple regression model fitted in Question 7 we can conclude: Answer: The distribution of the residuals appears roughly normal. Distractors: ❑ The distribution of the residuals is skewed to the right. ❑ The distribution of the residuals is skewed to the left. ❑ The residuals have a non-constant variance. Question 10 Which of the following plots could not be used to detect a violation of the linearity assumption for the multiple regression model: Answer: Normal probability plot of residuals Distractors: ❑ Scatter plot of residuals against fitted values ❑ Scatter plot of residuals against values of ELOVL2 ❑ Scatter plot of residuals against values of PDE4C Question 11 Assume that the model assumptions are satisfied. Based on the multiple regression model, you can conclude that there is Answer: moderate evidence to suggest an association between Age and percentage of DNA methylation at PDE4C, after taking into percentage of DNA methylation at ELOVL2 (p < 0.05). Distractors: ❑ no evidence to suggest an association between Age and percentage of DNA methylation at PDE4C, after taking into percentage of DNA methylation at ELOVL2 (p > 0.1). ❑ weak evidence to suggest an association between Age and percentage of DNA methylation at PDE4C, after taking into percentage of DNA methylation at ELOVL2 (p < 0.1). ❑ strong evidence to suggest an association between Age and percentage of DNA methylation at PDE4C, after taking into percentage of DNA methylation at ELOVL2 (p < 0.01). Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 13 of 21 Question 12 Based on the multiple regression model, what is the residual for the person Aged 51 years whose blood sample has 71.63% DNA methylation at ELOVL2 and 62.06% DNA methylation at PDE4C? Answer: -3.49 Distractors: ❑ -5.46 ❑ 3.49 ❑ -2.76 Question 13 Based on the multiple regression model, construct a 95% prediction interval for the age of a person whose blood sample has 40% DNA methylation at ELOVL2 and 50% DNA methylation at PDE4C. The upper limit for this prediction interval is Answer: 44.87 (years) Distractors: ❑ 24.93 (years) ❑ 48.16 (years) ❑ 43.27 (years) Question 14 Overall, the multiple regression model suggests that old age is associated with Answer: a high percentage of DNA methylation at both ELOVL2 and PDE4C. Distractors: ❑ a high percentage of DNA methylation at ELOVL2 and low percentage of DNA methylation at PDE4C. ❑ a low percentage of DNA methylation at ELOVL2 and high percentage of DNA methylation at PDE4C. ❑ a low percentage of DNA methylation at both ELOVL2 and PDE4C. Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 14 of 21 Obstructive sleep apnea A study was conducted to determine if there is a difference in the efficacy of two treatments of obstructive sleep apnea (OSA). Thirty patients (20 male, 10 female) were recruited to the study with fifteen patients being randomly allocated each of the mandibular advancement splint (MAS) treatment group and tongue stabilizing device (TSD) treatment group. The efficacy of the treatments were compared in terms of the reduction in the apnea-hypopnea index (AHI). The thirty patients at the commencement of the study had an average AHI of 28.32 with a sample standard deviation of 16.94. After two months the AHI of each patient was again measured. The MAS treatment group experienced an average reduction in AHI of 11.28 with a sample standard deviation of 9.23. The TSD treatment group experienced an average reduction in AHI of 13.45 with a sample standard deviation of 10.66. Question 1 The standard error for the sample mean AHI of patients at commencement is Answer: 3.0928 Distractors: ❑ 3.1457 ❑ 5.1705 ❑ 0.5646 Question 2 The margin of error for a 95% confidence interval of the population mean AHI of patients at commencement is Answer: 6.3255 Distractor: ❑ 6.3163 ❑ 5.2493 ❑ 5.2550 Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 15 of 21 Question 3 Let M be the population mean decrease in AHI for the MAS treatment and let T be the population mean decrease in AHI for the TSD treatment. The researchers would like to test Answer: H0: M = T versus H1: M ≠ T Distractors: ❑ H0: M = T versus H1: M < T ❑ H0: M = T versus H1: M > T ❑ H0: M = 0 versus H1: M > 0 Question 4 The standard error for the difference in the sample means for the two treatment groups is Answer: 3.6408 Distractors: ❑ 1.1515 ❑ 1.3260 ❑ 7.1360 Question 5 The t statistic used to test the hypotheses in Question 2 is Answer: -0.5960 Distractors: ❑ -1.8844 ❑ -2.1239 ❑ -0.9447 Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 16 of 21 Question 6 To do a test by hand, we can use the minimum degrees of freedom from the two samples. Here the degrees of freedom to be used is Answer: 14 Distractors: ❑ 29 ❑ 15 ❑ 16 Question 7 Based on the test, using your degrees of freedom from Question 6, you can conclude that there is Answer: no evidence to suggest the mean decrease in AHI is different between the two treatments (p = 0.56). Distractors: ❑ no evidence to suggest the mean decrease in AHI is different between the two treatments (p = 0.28). ❑ weak evidence to suggest the mean decrease in AHI is different between the two treatments (p = 0.08). ❑ moderate evidence to suggest the mean decrease in AHI is different between the two treatments (p = 0.02). Question 8 Assuming the data was available, which of the following tests would provide an alternative nonparametric test to test the hypotheses of Question 3? Answer: Rank Sum test Distractors: ❑ Sign test ❑ Signed rank test ❑ Chi-squared test Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 17 of 21 Question 9 A secondary measure (“response to treatment”) was defined as resolution of all symptoms or an improvement in symptoms and ≥ 50% reduction in AHI. In the MAS group 10 patients experienced a “response to treatment”. The standard error for the proportion of patients in the MAS group that experience a “response to treatment” is Answer: 0.1217 Distractors: ❑ 0.0574 ❑ 0.0905 ❑ 0.2381 Question 10 The margin of error for constructing a 95% confidence interval by hand for the population proportion of patients that would experience a “response to treatment” with the MAS is Answer: 0.2385 Distractors: ❑ 0.2831 ❑ 0.5164 ❑ 0.2610 Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 18 of 21 Question 11 In addition to the 10 patients in the MAS group who experienced a “response to treatment”, 7 patients in the TSD group also experienced a “response to treatment”. The standard error for the difference in the proportion of patients experiencing a “response to treatment” between the two groups is Answer: 0.1772 Distractors: ❑ 0.0539 ❑ 0.1232 ❑ 0.3655 Question 12 The researchers would like to test if there is a difference in the population proportion of patients that experience a “response to treatment” between the two treatment groups. Carrying out this test by hand, you can conclude there is Answer: no evidence to suggest a difference in the population proportion of patients that experience a “response to treatment” (p > 0.1) Distractors: ❑ weak evidence to suggest a difference in the population proportion of patients that experience a “response to treatment” (p < 0.1) ❑ moderate evidence to suggest a difference in the population proportion of patients that experience a “response to treatment” (p < 0.05) ❑ strong evidence to suggest a difference in the population proportion of patients that experience a “response to treatment” (p < 0.01) Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 19 of 21 Parasites The fish Rutilus rutilus is one of the hosts of the parasitic worm Ligula intestinalis. The distribution of the number (X) of Ligula intestinalis worms infecting a randomly selected host from the Rutilus rutilus population is as follows: x 0 1 2 3 P(X = x) 0.21 0.45 0.23 0.11 Question 1 The expected number of Ligula intestinalis in a fish from the Rutilus rutilus population is Answer: 1.24 Distractors: ❑ 0.79 ❑ 1 ❑ 2.24 Question 2 The standard deviation of Ligula intestinalis in a fish from the Rutilus rutilus population is Answer: 0.9069 Distractors: ❑ 0.8224 ❑ 1.3281 ❑ 0.5238 Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 20 of 21 Question 3 We are often interested in prevalence of parasitic infection, that is the probability a randomly selected individual from the host species has at least one parasite. Based on the above probabilities, what is the prevalence of Ligula intestinalis in the Rutilus rutilus population? Answer: 0.79 Distractors: ❑ 0.21 ❑ 0.89 ❑ 0.45 Question 4 Ligula intestinalis can sometimes be difficult to identify in hosts. Suppose we correctly identify an infected host as having one or more parasite only 80% of the time. What is the probability that a randomly selected fish from the Rutilus rutilus population is identified as being infected by Ligula intestinalis? Answer: 0.632 Distractors: ❑ 0.36 ❑ 0.712 ❑ 0.11 Question 5 A randomly selected fish from the Rutilus rutilus population is inspected and no Ligula intestinalis parasites are identified. What is the probability that this fish has no Ligula intestinalis parasites? Answer: 0.5706 Distractors: ❑ 0.368 ❑ 0.158 ❑ 0.834 Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data Page 21 of 21 Question 6 A sample of 10 fish from the Rutilus rutilus population is taken. The distribution of the number of fish infected with Ligula intestinalis in the sample is Answer: Binomial(10, 0.79) Distractors: ❑ Binomial(10, 0.8) ❑ Normal(0.79, 0.13) ❑ Normal(7.9, 1.3) Question 7 The probability that six or fewer fish are infected with Ligula intestinalis in the sample from Question 4 is Answer: 0.1391 Distractors: ❑ 0.0993 ❑ 0.6177 ❑ 0.2508 Question 8 A sample of 100 fish from the Rutilus rutilus population is taken. The distribution of the number of infected hosts is approximately Answer: Normal(79, 4.1) Distractors: ❑ Normal(0.79, 0.041) ❑ Normal(79, 8.89) ❑ Binomial(79,0.79) END OF EXAMINATION
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