Assignment 1 TM5515 Face-to-Face Block 2018 Epidemiology for Public Health (TM5515) ASSIGNMENT 1 Block, 2018 Due: 2nd October, 11:59pm AEST Assignment marks: The marks allocated to each question are...

Assignment 1 TM5515 Face-to-Face Block 2018
Epidemiology for Public Health
(TM5515)
ASSIGNMENT 1
Block, 2018
Due: 2nd October, 11:59pm AEST
Assignment marks: The marks allocated to each question are indicated beside the question. Some questions have parts, and marks assigned to part questions are also indicated. Note that the questions are not equally weighted.
In addition, ten (10) marks are assigned for presentation. Your assignment should have numbered pages, with your name and student number on each page.
In total, this assignment will contribute 25% towards the assessment for this subject.
Total marks: 190
Format of the assignment: Your assignment should be typed with adequate space left between questions. Handwritten and scanned assignments will be accepted, if handwriting is legible and presentation neat. No answers need to be longer than a few short sentences or short paragraphs. Where you are required to perform a calculation, you will be marked on the associated process, and not simply on obtaining the co
ect numerical answer. Take care in your calculations. The final answer should be reported and co
ect to two (2) decimal places, unless indicated otherwise. Leave rounding to the very end to avoid rounding e
ors. Please show all relevant formulas and working.
Method of submission: Assignments should be submitted as a Word or PDF document viaSubmission viarequires you to read and understand the plagiarism statement, and attest that the work you submit is entirely your own.
Late assignments will not be accepted without prior approval. Assignments for which an extension has not been sought prior to the due date may attract a penalty of 5% per day of the total marks available for the assignment, as per
Important: While we have attempted to make these questions relevant to health, some liberties have been taken with the finer details. Questions may not reflect reality as you know it.
Many questions are based on real data and published research. However, many of the epidemiological measures and concepts have been simplified. Please base your answers on the data and information provided in these assignment questions and instructions. References to articles that have been used in questions for this assignment will be provided later, in the worked answers, for your interest. For this assignment (Assignment1) you are not required to read or research beyond the core course materials and prescribed textbook. Of course you are welcome to read beyond the core materials, but you will not receive extra marks for doing so.
Referencing: You are not expected to read or research beyond the core course materials and prescribed text. Therefore no or minimal referencing is required (reference the text book if appropriate, referencing of lecture/tutorial materials not required). If you do source and include other information, then you should reference as appropriate, using any suitable referencing style.
Question 1 [26 marks]
Researchers conducted a study to investigate frequency of dental decay in school children in a Pacific country. Children at five community schools in a region were enrolled. At commencement of the study in April 2015, 554 children were enrolled. The 554 children were assessed, at baseline, and 123 were found to have decayed or missing permanent teeth. These same children were examined in April 2016. (For this study, the researchers were interested in only the first case of dental decay, in permanent teeth, per child. They did not count subsequent cases of dental decay in different teeth). 35 more children were found to have new first dental decay. 15 of the original children were no longer attending these schools. The children were examined again in April 2017 and 2018. In April 2017, 5 further children were no longer attending the school, and there were 22 new cases of dental decay. In April 2018, 12 further children were not attending the school, and there were 15 new cases of dental decay. There were no new missing permanent teeth during the course of the study.
· What was the prevalence of dental decay in students at this school at the start of the study?            
                
Total number of children = 554
Number of dental decay cases initially = 123
Prevalence of dental decay at the start of the study
= Number of cases/ Number of children
= 123/554            
= 0.22%                                [3 marks]
· What is the cumulative incidence of first dental decay in this group of school students, during the entire course of the study?                            
Total number of new cases= 35+22+15= 72
Number of 'at risk’ children
= 554 - 123 = 431    
    
Cumulative incidence
= No. of new reported cases/ No. of 'at risk’ individuals
=72/431
= 0.16 % over the course of four years
    [3 marks]
· One of the reasons that the researchers were conducting this study was to decide if there was a need for increased health promotion measures to reduce dental decay, such as education on diet and tooth
ushing techniques. Based on the values calculated in parts and b, the researchers concluded that the amount of students developing dental decay was decreasing, so there was no need for new health interventions. Do you agree with this conclusion? In your answer, compare the measures in part a) and b), making sure to fully define each of these measures.
I don't agree with the researchers because of the stagnant reduction in the number of cases in the consecutive years. Prevalence refers to the individuals already diseased during the specific period while incidence is the proportion of individuals having developed the condition during the time.
At the start of the study, there were 123 affected students (0.22%) who could have developed dental decay over the course of many years while there was an increase of 0.16% in the next four years which is a matter of concern. Improved interventions could disseminate more information resulting in enhanced oral hygiene practices.
[3 marks]

· What is the incidence rate of dental decay in this group of school students?
No. of years = 4
Total number of cases = 195
During calculation, we assume that the number of children who were newly diagnosed and those who lost follow-ups, remained healthy for the initial half of the respective year.
Person-years of observation
= (431 + 1/2×123)
+ (381 + ½ × 35 + ½×15)
+ (354 + 1/2×22 + ½×5)
+ (327 + 1/2×15 + 1/2×12)
= 1606.5 person-years
Incidence rate
= No. of cases reported/ No. of person-years of follow up
= 195/ 1606.5
= 0.12 cases per person-yea
= 12 cases per 100 person-yea

[7 marks]
· Compare the measures you have calculated in parts b and d. [Hint: make sure you compare measures across similar time periods]. Which of these measures of incidence is larger, and why? Which is most accurate and why? Generally, under what circumstances will the CI and IR be numerically more similar, and under what circumstances more different? What are the benefits and limitations of each measure?     
Larger value            
IR has a larger value as in its formula, the denominator represents varying observation time when the same for CI considers the observation time to be equal for the entire population under study.         
For example: In 2016;
CI = 35/ (554-123) = 0.08%
IR = 35/ (381 + ½ × 35 + ½×15)
= 0.086 per person-yea
Hence, it may be concluded that CI is the bigger value.    

Accuracy    
Incidence rate is a more accurate value than cumulative incidence as it is more instintive that the latter in determining the health results. Instead of considering the risk factors, it calculates the rate at which new cases are added with respect to unit time.
Similarity and difference
When rate and time period is low, CI and IR have numerically similar values. With increase in the duration of study, more different will be the values.
            
Advantages and disadvantages
Cumulative incidence represents the probability of the risk of developing a particular disease or condition. It is also easy to calculate. However, it does not take into account the other risk factors which may cause the disease or say, death during the time period. It also do not consider the persons lost to follow-up as well as the time at which the individuals developed the disease.
Incidence rate, on the other hand, considers the individuals lost to follow-ups. It also considers the time of the study when the disease was detected. Nevertheless, it's demerit is that it considers the disease probability to be constant throughout the study. Also, it is difficult to be understood by a layman.
[4 marks]
· Many of the cases of dental decay were not treated promptly, and progressed to causing severe toothache. On average, each student with dental decay experienced toothache for six weeks. On any one school day, what is the prevalence of severe tooth ache in these children? In a group of 500 students, how many would have tooth ache on any particular day? For the purposes of this question, consider that only first episode of dental decay causes pain [we know that in real life pain would occur with decay in any teeth! - but make this assumption for the purposes of this question].
Considering a day post April 2018, there are 195 cases in a population of 450 students.
Therefore, prevalence of tooth ache on any particular day = 195/450
= 0.43%
Number of tooth ache cases out of 500 children = IR × 500 = 0.12× 500
=60 person-years
[6marks]
Question 2 [16 marks]
Internationally, ship
eaking workers are typically exposed to a wide range of hazardous chemicals. Researchers investigated deaths in male Taiwanese ship
eaking workers compared to the general male population of Taiwan. The following table presents data on the observed and expected deaths from various causes among these male ship
eaking workers, as well as (age-) standardised mortality ratios.
a) State the type of standardisation used in this study and define this term. Explain the likely reason for choosing this type of standardisation. Additionally, why are standardised rather than crude rates used?                             Indirect standardisation is used here.
When age-specific rates are unavailable, it applies a set of rates from the standard population to each of the study population to estimate the mo
idity/ mortality ratios.
Crude rates considers study population as a whole which may often mislead health statistics while standardised rates considers specific characteristic (s) as standard, say age and sex. As a result, standardised rates are prefe
ed over crude rates.
                                    [4 marks]
) Define the terms ‘Obs’, ‘Exp’, and ‘SMR’, and how they are derived. In your answer, provide a general description of the steps of this standardisation process.     
‘Obs’ or 'observed number of deaths’ is the actual number of deaths during a time period.
‘Exp’ or ‘expected number of deaths’ is the age- and sex- dependent deaths in the standard population to that of the study population during the same time period.

SMR or Standardized Mortality Ratio is the ratio between the observed number of deaths in the population under study and the number of expected deaths which if greater than one indicate death abundance. It compares the mortality risk of the study population to that of the standard population.
Steps of the standardisation process:
· Calculate age- and sex-dependent death rates in the standard population (a)
· Consider the population size estimate for each group (b)
· Calculate number of expected deaths
= a×
100,000
· Add together all the age- and sex-specific dependent death rates calculated as above
· Consider the observed number of deaths during the same time period.
· Calculate SMR using the formula = Obs/ Exp
· To ascertain the statistical importance of SMR,95% Confidence Interval is calculated which is equal to 1.96 times the standard e
or of the estimate.
        [4 marks]
c) Interpret the findings in the table in relation to liver cancer and cancer of the Esophagus (separately). Include a comment on the relevant 95% confidence intervals. [4 marks]
Liver cancer:
SMR = Obs/ Exp
= 65/14
= 4.64
95% C.I. = 1.96 × s.e. SMR
s.e. SMR = √Obs/ Exp
= √65 / 14 = 0.58
1.96 × s.e. SMR = 1.96 × 0.58
= 1.14 (95% C.I.)
Lower limit of 95% C.I. = SMR - 1.14
= 4.64 - 1.14
= 3.5
Upper limit of 95% C.I. = SMR + 1.14
= 5.78
We can interpret that the 95% C.I do not include the value ‘1.0’ which means the observed number of...