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Semester One Final Examinations, 2021 STAT1201 Analysis of Scientific Data
Exam information
Course code and
title
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
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A Casio fx-82 series or UQ approved calculator may be used.
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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.
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of the exam.
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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
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
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.
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
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
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)
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
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
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
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).
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.
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
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
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
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
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)
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
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
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)