401077 Introduction to Biostatistics For Null and Alternate Hypothesis

A random sample of 10 former University of Eastern Sydney students was ask whether they drink more or less since they graduated from University. Their answers were:
more, less, less, less, more, less, less, less, less, more

Conduct a sign test to see whether or not this provides evidence of a change in drinking habits of the wider population of graduates from University of Eastern Sydney. Use the 5-step method and show all working. Use the Table 6 from the textbook (below) to determine the decision rule or use R Commander to calculate the p-value.
 

Using sign test, the statistic of interest is 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. 

Question 2

The 95% confidence interval for the WEMWBS value was found to be (43.73318, 44.42872). 

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.

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 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.

Let m_user be mean of WEMWBS for self-reported drug user in the University of Eastern Sydney and m_non-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) 

The mean WEMWBS for drug users was found to be 43.93.

The mean WEMWBS for drug users was found to be 44.20.

The p-value for the test for the hypothesis defined in part b was found to be 0.2624.

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. 

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.

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.

The calculation for sample size was conducted through the online calculator available via clinical.com (https://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. 

The following table gives the contingency table with column total and percentages as computed in R.

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).

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.

The minimum sample size needed for a 95% confidence interval for proportion of illicit drug users with margin of error being is given by the formula, 

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.