Biostatistics Assignment 2 - University of Eastern Sydney

This is a solved assignment for Biostatistics course (401077) at University of Eastern Sydney. The assignment covers topics like sign test, confidence interval, hypothesis testing, Chi-square test, and sample size calculation. The assignment also includes questions related to drug use and well-being score of students.

Running head: 401077 INTRODUCTION TO BIOSTATISTICS, AUTUMN 2018
401077 Introduction to Biostatistics, Autumn 2018
Assignment 2
Due Sunday April 29, 2018
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401077 INTRODUCTION TO BIOSTATISTICS, AUTUMN 2018
Question 1
The research question being considered is that whether the given data shows evidence
of there being any change in drinking habits from the population in University of Eastern
Sydney. Let p denote the proportion of “more” or “+”. “No change” is equivalent to
proportion of “more” or “+” being equal to proportion of “less” or “-”. Thus the hypotheses
to be tested could be defined as follows:
H0: p =0.5 (Null hypothesis)
H1: p ≠ 0.5 (Alternate hypothesis)
Using sign test, the statistic of interest is X where it represents the number of “+” or
more and follows Binomial (10, p) .The test is thus reduced to test for binomial proportion.
The p-value is used for the decision making at 0.05 level of significance. The p-value was
observed to be 0.3438 which is greater than 0.05. Thus there is not enough evidence to reject
null and claim that drinking habits of the sample has changed from the one at University of
Eastern Sydney.

401077 INTRODUCTION TO BIOSTATISTICS, AUTUMN 2018
Question 2
a.
The 95% confidence interval for the WEMWBS value was found to be (43.73318,
44.42872).
b.
The point estimate of mean well-being score for the student was found to be 44.081.
Additionally, it can be said with 95% confidence that the interval (43.733, 44.429) would
contain the mean value of WEMWBS.
Question 3
a.
The given research question, that whether drug users in University of Eastern Sydney
has lower WEMWBS than non-users, focuses on the case that where WEMWBS score is less

401077 INTRODUCTION TO BIOSTATISTICS, AUTUMN 2018
for drug users than non-users only and in such as case the consideration that it might be more
is moot. The alternative thus would have the condition that the mean is greater for non-users
failing to gather evidence for which would determine answer to the research question. Thus is
should be addressed with a left sided one tailed test.
b.
Let muser be mean of WEMWBS for self-reported drug user in the University of
Eastern Sydney and mnon-user be the mean of WEMWBS for self-reported drug non-user. Then
the hypotheses are:
H0: m user = m non-user (Null Hypothesis)
H1: m user < m non-user (Alternate Hypothesis)
c.
i.
The mean WEMWBS for drug users was found to be 43.93.

401077 INTRODUCTION TO BIOSTATISTICS, AUTUMN 2018
ii.
The mean WEMWBS for drug users was found to be 44.20.
iii.
The p-value for the test for the hypothesis defined in part b was found to be 0.2624.
d.
The p-value was found to be greater than 0.05 and thus there is not enough evidence
to suggest that null ought to be rejected in favor of the alternate at 5% level of significance.
Thus it is concluded that there is no difference between the WEMWBS of students who are
drug users from the non-drug users.
Question 4
Approaching the research question specified in part 3 using non-parametric method,
the statistic of interest is taken to be the median WEMWBS of drug users and non-drug users
respectively. Let med user be median of WEMWBS for self-reported drug user in the
University of Eastern Sydney and med non-user be the median of WEMWBS for self-reported
drug non-user.

401077 INTRODUCTION TO BIOSTATISTICS, AUTUMN 2018
The hypotheses can be defined as:
H0: med user = med non-user (Null Hypothesis)
H1: med user < med non-user (Alternate Hypothesis)
Then Wilcoxon Mann Whitney test was applied to test for the alternative. The p-value
was found to be 0.3058. Therefore sufficient evidence was not found which supports
rejection of null in favour of the alternative. Thus the conclusion is that drug users do not
have lesser WEMWBS than non-drug users.
Question 5
The calculation for sample size was conducted through the online calculator available
via clinical.com (http://clincalc.com/stats/samplesize.aspx ). When testing for whether
population mean 44 with standard deviation being 4 is at the least half of the standard
deviation greater than the average. It was found that minimum of 42 sample points are
required for the specified test parameters of alpha being 0.05 and power of test, or 1-beta
value, being 90%. The following figure shows the parameters used and results as obtained via
the calculator online.

401077 INTRODUCTION TO BIOSTATISTICS, AUTUMN 2018

401077 INTRODUCTION TO BIOSTATISTICS, AUTUMN 2018
Question 6
a.
The following table gives the contingency table with column total and percentages as
computed in R.
Drug
Sex
user non-user
male 68.90 48.05
female 31.09 51.95
Total 100.00 100.00
Sum(Frequency
)
119.00 154.00

401077 INTRODUCTION TO BIOSTATISTICS, AUTUMN 2018
b.
The conditions for Chi-square test are,
1. Variables “sex” and “drug” are independent
2. Frequency of no cell in the contingency table is less than 5.
Therefore, comparing it to the contingency table above, although second found to be
satisfied, first condition does not hold since Probability(user, male) does not equal
Probability(user) x Probability(male).
c.
The hypothesis under testing can be written as:
H0: There exists no association between sex and drug use (Null Hypothesis)
H1: There exists association between sex and drug use (Alternate Hypothesis)
The p-value for the test was found to be 0.0008698 which is less than 0.05. Thus there
exists sufficient evidence to support that null hypothesis ought to be rejected in favour of the
alternative, that is, drug use does depend on and differ by gender of the student.
Question 7
The minimum sample size needed for a 95% confidence interval for proportion of
illicit drug users with margin of error being ± 4 % is given by the formula,

401077 INTRODUCTION TO BIOSTATISTICS, AUTUMN 2018
n= p∗(1− p)
( marginof error
1.96 )2
Since the proportion is unknown take p= 0.5. “n” was found to be 600.25. Thus 601
would be the minimum required sample size for the given specifications.