As a practicing health professional you are quite concerned about the lack of sun smart behavior among your community members; you have even heard some people boasting about not using sunscreen and proudly showing off their sun burns. You are quite aware of the beneficial effects of sunscreen and would like to conduct your own research project to demonstrate to others the benefits of using sun screen in terms of prevention and reduction of the cases of melanoma. It has been few years since you passed EPID1000 and are a bit rusty on different study designs so you immediately email your ex-unit coordinator to send you all the materials (especially lecture recordings and tutorial notes) but you don’t get any reply. Eventually,on an old USB you found everything.
Please give us a short description of how each design can be used to demonstrate the benefits of using sun screen. You should use the points provided in the review exercise (tutorial 11) to cover all the features of each design.
a) Cross-sectional study.
b) Prospective Cohort study.
c) Retrospective Cohort study.
d) Case Control study.
e) Quasi-experimental study.
f) Randomized controlled trial
(No need to worry about the ethics for e and f parts); this is a hypothetical exercise to assess your understanding of different study designs).
Note: Please be specific about the specified exposure and outcome given for this question. No marks for writing general design features from the lectures. We want you to ‘apply’ your knowledge of study designs in light of the given scenario.
Formatting &Line limit Requirements:For each of the six parts please use Times New Roman size 12 font, normal page borders and 10 lines maximum for each part,(10 lines DONOT mean ten sentences or ten statements). These formatting requirements are for Q1 ONLY, rest of the questions are free from any formatting requirements.
For Q2, Q3, Q4, Q5, Q6, Q7 and Q8you will conduct ALL analyses using your own random sample of 50(Video & instructions posted separately).
NO MARKS will be awarded:
If you used the entire dataset(N=700) instead of your own sample of 50.
If you do not provide the relevant SPSS output for all questions.NO SPSS Output = No Mark
You can choose a different sample of 50 for every question or use the same sample of 50 for all questions; it is up to you (Please read FAQs for more questions).
Write null and a non-directional alternative hypotheses that can be tested with anIndependent Samples t test. Briefly describe, and report on the assumption at the analysis stage.Provide complete interpretation of your results including their statistical and practical significance.
H0: population means of resting energy expenditure are equal for male and female
H1: population means of resting energy expenditure are not equal for male and female
The hypothesis can be testing using an independent sample t test.
Group Statistics |
|||||
|
Gender |
N |
Mean |
Std. Deviation |
Std. Error Mean |
Resting Energy Expenditure |
Males |
31 |
1370.35 |
350.910 |
63.025 |
Females |
19 |
1268.58 |
279.237 |
64.061 |
Independent Samples Test |
||||||||||
|
|
Levene's Test for Equality of Variances |
t-test for Equality of Means |
|||||||
|
|
F |
Sig. |
t |
df |
Sig. (2-tailed) |
Mean Difference |
Std. Error Difference |
95% Confidence Interval of the Difference |
|
|
|
Lower |
Upper |
|||||||
Resting Energy Expenditure |
Equal variances assumed |
.979 |
.327 |
1.072 |
48 |
.289 |
101.776 |
94.949 |
-89.133 |
292.684 |
Equal variances not assumed |
|
|
1.133 |
44.624 |
.263 |
101.776 |
89.867 |
-79.267 |
282.819 |
The assumption of normality has been met for male population. However, normality is slightly deviated for female population. The p-value of the F test suggests that the assumption of equality of variances has been met.
The p-value of the independent sample t-test is 0.289. As the p-value is greater than the level of significance 0.05, we can accept the null hypothesis. This concludes that there is no such evidence of significant difference in population means of resting energy expenditure between male and female populations.
a) Are people more willing to help strangers who ask for money than those who ask for a cigarette?
Choose a suitable test to answer this question and provide a short description of your data analysis including statistical significance of your findings. No need to list or test any assumptions
H0: population proportion of peoples willing to help strangers ask for money equal to population proportion of peoples willing to help strangers ask for a cigarette
H1: population proportion of peoples willing to help strangers ask for money is greater than population proportion of peoples willing to help strangers ask for a cigarette
Stranger help (Money) |
|||||
|
|
Frequency |
Percent |
Valid Percent |
Cumulative Percent |
Valid |
Definitely not |
5 |
10.0 |
10.0 |
10.0 |
Probably Not |
2 |
4.0 |
4.0 |
14.0 |
|
Possibly yes |
24 |
48.0 |
48.0 |
62.0 |
|
Definitely yes |
19 |
38.0 |
38.0 |
100.0 |
|
Total |
50 |
100.0 |
100.0 |
|
Stranger help (Cigarettes) |
|||||
|
|
Frequency |
Percent |
Valid Percent |
Cumulative Percent |
Valid |
Definitely Not |
36 |
72.0 |
72.0 |
72.0 |
Probably Not |
6 |
12.0 |
12.0 |
84.0 |
|
Possibly Yes |
4 |
8.0 |
8.0 |
92.0 |
|
Definitely Yes |
4 |
8.0 |
8.0 |
100.0 |
|
Total |
50 |
100.0 |
100.0 |
|
The two population proportion test is suitable for this problem.
The two proportions:
p1 = 43/50 = 0.86
p2 = 8/50 = 0.16
The overall sample proportion: p = (43 + 8) / (50 + 50) = 0.51.
The test statistic formula:
Z = (p1 – p2)/Ö{(p(1-p))(1/n1+1/n2)}
= (0.86 – 0.16)/Ö{(0.51(1-0.53))(1/50+1/50)}
= 9.81
p-value = 0.000
The p-value of the two sample proportion test is 0.000. As the p-value is less than the level of significance 0.05, we can reject the null hypothesis. Therefore, we can conclude that people are more willing to help strangers who ask for money than those who ask for a cigarette.
b) Are females more willing than males to help strangers when they ask for money?
Choose a suitable test to answer this question and provide a short description of your data analysis including statistical significance of your findings. No need to list or test any assumptions
H0: population proportion of females willing to help strangers ask for money equal to population proportion of males willing to help strangers ask for money
H1: population proportion of female willing to help strangers ask for money is greater than population proportion of males willing to help strangers ask for money
Stranger help (Money) * Gender Crosstabulation |
||||
Count |
|
|
|
|
|
|
Gender |
Total |
|
|
|
Males |
Females |
|
Stranger help (Money) |
Definitely not |
5 |
0 |
5 |
Probably Not |
2 |
0 |
2 |
|
Possibly yes |
14 |
10 |
24 |
|
Definitely yes |
10 |
9 |
19 |
|
Total |
31 |
19 |
50 |
The two population proportion test is suitable for this problem.
The two proportions:
p1 = 19/19 = 1.00
p2 = 24/31 = 0.61
The overall sample proportion: p = (19 + 24 / (19 + 31) = 0.86.
The test statistic formula:
Z = (p1 – p2)/Ö{(p(1-p))(1/n1+1/n2)}
= (0.1.00 – 0.61)/Ö{(0.86(1-0.86))(1/19+1/31)}
= 3.01
p-value = 0.003
The p-value of the two sample proportion test is 0.003. As the p-value is less than the level of significance 0.05, we can reject the null hypothesis. Therefore, we can conclude that females are more willing than males to help strangers when they ask for money.
Choose only those who are27 years and younger and test the hypothesis that they come from a population in which mean exercise time is 6 hours per week.
Write null and alternative hypotheses.
Choose a suitable test and carry out the appropriate analyses and write a short summary of your results including their statistical significance No need to list or test any assumptions.
H0: population mean exercise time is 6 hours per week
H1: population mean exercise time is different from 6 hours per week
A one sample t test wtll be suitable to answer this question.
One-Sample Test |
||||||
|
Test Value = 6 |
|||||
|
t |
df |
Sig. (2-tailed) |
Mean Difference |
95% Confidence Interval of the Difference |
|
|
Lower |
Upper |
||||
Exercise hours per week |
-.419 |
23 |
.679 |
-.124 |
-.74 |
.49 |
The p-value of the one sample t test is 0.679 which is greater than the level of significance 0.05. We can accept the null hypothesis. Therefore, we can conclude that those who are 27 years and younger have mean exercise time of 6 hours per week.
Is there any significant difference in the Academic performance between those who Strongly Agree, Agree, Disagree and Strongly Disagree to the statement It is great not to have an invigilated final exam?
Choose a suitable test to answer this question. Carry out the appropriate analyses and write a short summary of your results and conclusion.Report on the assumption at the analysis stage.
H0: Population means of academic performance between those who Strongly Agree, Agree, Disagree and Strongly Disagree are equal
H1: Population means of academic performance between those who Strongly Agree, Agree, Disagree and Strongly Disagree are different for at least one
A one way ANOVA is suitable to answer this question.
Test of Homogeneity of Variances |
|||
Academic Performance University |
|
||
Levene Statistic |
df1 |
df2 |
Sig. |
12.406 |
3 |
46 |
.000 |
ANOVA |
|||||
Academic Performance University |
|
|
|
|
|
|
Sum of Squares |
df |
Mean Square |
F |
Sig. |
Between Groups |
23.712 |
3 |
7.904 |
11.373 |
.000 |
Within Groups |
31.968 |
46 |
.695 |
|
|
Total |
55.680 |
49 |
|
|
|
The histogram with normal curve suggests that the assumption of normality is violated in this case. Also the assumption of equality of variances has not been met as the p-value of the Levene Statistic is less than the level of significance 0.05.
The p-value of the ANOVA test is 0.000. This suggests that we can reject the null hypothesis at 5% level of significance. Therefore, we can conclude that there is significant difference in the Academic performance between those who Strongly Agree, Agree, Disagree and Strongly Disagree to the statement It is great not to have an invigilated final exam?
Is there any significant difference in text messaging behavior while driving among people with different blood groups?
Choose a suitable test to answer this question. Carry out the appropriate analyses and write a short summary of your results and conclusion.No need to list or test any assumptions
H0: Population means of test massaging behavior while driving among people with different blood groups are equal
H1: Population means of test massaging behavior while driving among people with different blood groups are different for at least one
A one way ANOVA is suitable to answer this question.
Test of Homogeneity of Variances |
|||
Texts while Driving |
|
|
|
Levene Statistic |
df1 |
df2 |
Sig. |
2.975 |
3 |
46 |
.041 |
ANOVA |
|||||
Texts while Driving |
|
|
|
|
|
|
Sum of Squares |
df |
Mean Square |
F |
Sig. |
Between Groups |
1.768 |
3 |
.589 |
.484 |
.695 |
Within Groups |
56.012 |
46 |
1.218 |
|
|
Total |
57.780 |
49 |
|
|
|
The histogram with normal curve depicts the violation of assumption of normality. Also the assumption of equality of variances has not been met as the p-value of the Levene Statistic is less than the level of significance 0.05.
The p-value of the ANOVA test is 0.695. As this p-value is greater than 0.05, the level of significance, this suggests that we can accept the null hypothesis at 5% level of significance. Therefore, we can conclude that there is no significant difference in text messaging behavior while driving among people with different blood groups.
Is there any significant difference in academic achievement at school (based on ATAR score School) and later at university (based on Academic Performance Uni score)?
Choose a suitable test to answer this question. Carry out the appropriate analyses and write a short summary of your results and conclusion.No need to list or test any assumptions
H0: Population means of academic achievement at school and academic achievement at university groups are equal
H1: Population means of academic achievement at school and academic achievement at university groups are different
An independent sample t test is suitable to answer this question.
Independent Samples Test |
||||||||||
|
|
Levene's Test for Equality of Variances |
t-test for Equality of Means |
|||||||
|
|
F |
Sig. |
t |
df |
Sig. (2-tailed) |
Mean Difference |
Std. Error Difference |
95% Confidence Interval of the Difference |
|
|
|
Lower |
Upper |
|||||||
Academic_performance |
Equal variances assumed |
132.009 |
.000 |
11.740 |
98 |
.000 |
3.880 |
.330 |
3.224 |
4.536 |
Equal variances not assumed |
|
|
11.740 |
73.086 |
.000 |
3.880 |
.330 |
3.221 |
4.539 |
The histogram with normal curve shows assumption of normality holds for academic performance at University. However, a non-normal behaviour has been seen for academic performance at school. Also the assumption of equality of variances has not been met as the p-value of the F Statistic is less than the level of significance 0.05.
The p-value of the independent sample t test is 0.000. As this p-value is less than 0.05, the level of significance, this suggests that we can reject the null hypothesis at 5% level of significance. Therefore, we can conclude that there is significant difference in academic achievement at school (based on ATAR score School) and later at university (based on Academic Performance Uni score).
Participants were asked to name one positive aspect, if possible, about COVID-19 experience. (Variable Corona_Positives). Unexpectedly all participants had at least one positive aspect to report. You will see that top five responses have been coded as A, B, C, D and Ebut no labels or descriptions. Use your discretion (and creativity or personal experience) to assign each code a hypothetical response or description, make up any whatever you think these are/can/or should be, and provide a suitable graph showing labels/description of these five responses.
A = Environment
B = Better Hygiene
C= Innovation of connectedness
D = Peace
E = Digitized Education
(Relevant SPSS output not provided with your written answer = No mark).
This applies to Qs2-8
‘FAQs’ posted will be extremely useful so please refer to it before you ask any question.
Thank you everyone for posting your questions on the Discussion Board or taking them to Open Collaborate sessionslast time. Same rules apply to this assignment.
No emails,otherwise it will not be fair to others.
Remember, at the center of any academic work, lies clarity and evidence. Should you need further assistance, do look up to our Statistics Assignment Help
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