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Question 1: Case-Control Study on BRCA7 and Ovarian Cancer
a) Presentation of Results
The results of the study are summarised in the 2-by-2 table below:
Figure 1: Summary Table
(Source: Self-calculated in Stata)
b) Hypotheses
- Null Hypothesis (H0): That is, the BRCA1 risk variant is not a risk factor for ovarian cancer. An odds ratio is one means there is no association between the independent variable and the dependent variable.
- Alternative Hypothesis (H1): It has been identified that there is a relationship between the risk variant of the BRCA1 gene and ovarian cancer (Herdea et al. 2023). This condition states that the odds ratio is not equal to one.
c) Statistical Test
Chi-square test of independence is the most suitable statistical test when it comes to situations where the null hypothesis has to be tested. This test measures the extent of the relationship between two non-continuous variables or variables that can be categorised (Kedir et al. 2022). The Chi-Square test is appropriate here since the comparison is made about the frequencies of the BRCA1 risk variant positivity between women with ovarian cancer and women without it.
Figure 2: Chi-square test
(Source: Self-calculated in Stata)
d) Risk Estimate and Summary
The estimated odds ratio is 1.6 (95% CI: These are a summation of the different components as seen on the SENTAN chart: 0.636 to 3)153. This shows that women with the BRCA1 risk variant had an elevated risk of ovarian cancer as compared to women who did not possess this genetic risk factor (Almoznino et al. 2020). However, the p-value equals t. 321, which means that this correlation is not statistically significant at 5% level.
Figure 3: Risk Estimate
(Source: Self-calculated in Stata)
Question 2: Dietary Fiber Intake and Health Promotion Campaign
a) Stating the Null Hypothesis and Alternative Hypothesis
- Null Hypothesis (H0): Based on the results obtained from the assessment of the campaign intervention, the average fibre intake data show that there is no change. Again, it means that the mean fibre intake as shown in the table below was the same both before the campaign and after the campaign.
- Alternative Hypothesis (H1): The Prevailing study highlighted the contrast in the general fibre consumption when a health promotion campaign was in progress (Vahid et al. 2020). There exists a mean difference in fibre intake before the campaign and after the campaign.
b) Proper Test and Justification
Figure 4: Data list
(Source: Self-calculated in Stata)
The paired t-test is used for this analysis because:
Design: This means that the data collected was engaged and therefore paired because the same participants were assessed before and after the campaign.
Data Type: Thus, due to the research question and hypotheses, as well as the objectives of the study, comparing the fibre intake before and after the campaign involves a paired t-test because it is a suitable test for comparing the means of two related groups.
Normality Assumption: The paired t-test requires the differences between the two paired observations to be roughly normally distributed (Eshete Tadesse, Chane Mekonnen, & Adane, 2020). In view of the fact that the sample used was 47, this assumption is reasonably met.
c) Carry out the test and write of conclusion
The dependent t-test was used with a view to test the hypothesis on the difference in mean fibre intake before and after the health promotion campaign.
Results: “The mean fibre intake before the campaign was 19.” From before the campaign, 02 grams usage, it rose to 21 grams usage after the campaign. 41 grams. The mean difference between pre-campaign and post-campaign fibre intake was given as -2. 39 grams (Ghoreishy, et al. 2021). The t-statistic was -6. Regarding the cognitive status of the patients, the mean of the test statistic was 6171, and the p-value was 0.0000.
Conclusion: The paired t-test revealed a statistically significant change with regard to the second research question on the increase in fibre intake per day after the health promotion campaign. The mean fibre intake rose from 19. The consumption of chop increased from 21 |02 grams before the campaign to 21. 41 grams after the campaign means that there is little awareness created by the campaign. The mean difference of the pre-post analysis was -2. 39 grams with a t-statistic of -.6 All of these results can easily be interpreted in their respective metrics to possibly signify a link between the two independent variables. 6171 as well as data from a t-test with a p value of: 0. It is significant when the value is 0000. Thus, it can be concluded that the execution of the health promotion campaign enhanced the daily fibre consumption.
Figure 5: The paired t-test
(Source: Self-calculated in Stata)
Question 3: Cross-Sectional Study on Bone Mineral Density
a) Relationship Between Bone Mineral Density and Age
The following scatter plot shows the comparative analysis of BMD and age of the patients.
Description: From the scatter plot, it can be deduced that there is no relationship between age and Bone Mineral Density. The values obtained are separated without any seeming link between them.
Figure 6: Scatter Plot
(Source: Self-created in Stata)
b) Correlation Coefficient
The regression analysis of BMD and age is:
bmd = -0. 51+0. 10*age;
r = 0.0278.
Interpretation: The actual value of this positive coefficient is as close to zero as possible, and this means that there is almost no linear relationship between the age of the subjects and their bone mineral density in this data set (Altun et al. 2021). This statistic proves that there is a low degree of relationship between age and bone mineral density, as it is close to zero.
Figure 7: Correlation
(Source: Self-calculated in Stata)
c) Simple Linear Regression Model
The simple linear regression analysis results are as follows:
Regression Equation:
bmd = 1. 061 + 0. 000249 * age
Model Summary:
Intercept (Constant):. 061
The coefficient for Age: 0 final. 000249
R-squared: 0 0008
p-value for Age: 0.735
Interpretation: From the regression model, it is evident that the value of age is very small, and the p-value associated is 0.735. Age was considered a, nd the R-squared value is much lower, 0 0008, implying poor reliability of age predicting BMD.
Figure 8: Regression Output
(Source: Self-calculated in Stata)
d) Assumptions of Simple Linear Regression
- Residuals vs. Fitted Values Plot
The plot chosen for the diagnostic check of homoscedasticity is the plot of residuals against fitted values.
Figure 9: Residuals vs. Fitted Values Plot
(Source: Self-created in Stata)
- Histogram of Residuals
This is the histogram of the residuals. Residuals are the differences between the observed and predicted values of the dependent variable.”
Figure 10: Histogram of Residuals
(Source: Self-created in Stata)
- Shapiro-Wilk Test for Normality
From the Shapiro-Wilk test, the p-values are as follows: p = 0. On the same note, the calculated significance value of 90890 is way higher than the acceptable level of significance of 0.05. This, in a way, is a good indication that the residuals are normally distributed (Ueno et al. 2021).
Figure 11: Shapiro-Wilk Test
(Source: Self-calculated in Stata)
Conclusion: Among the most commonly made assumptions, the normality of residuals as well as homoscedasticity are met in the course of the analysis when necessary variables are selected, and non-linear relationships are dismissed.
e) Summary of Regression in brief Conclusion
The regression analysis of the impact of bone mineral density and age also revealed that it is very weak and statistically insignificant. By having an R-squared of 0, 19 this means that age has a minimal impact on the fluctuation of BMD (Nurhayati et al. 2020). The residuals are almost normally distributed and, thus, one cannot notice such problems as homoscedasticity in the given plots.
Question 4: Randomised Controlled Trial on Sleep and Mindfulness App
a) Sample Size Calculation of the Desired Reduction
To analyse the successful identification of the required reduction in the share of university students who sleep less than 6 hours. Five hours per night from 34 per cent to 20 per cent using the “9 per cent power and a two-sided level of significance,” the sample size that is required will be 420 participants (Jahrami et al. 2020). The sample size recruited for this study is split half and half between the intervention and control groups; 210 participants in total [Referred to Appendix 1].
b) Sample Size Calculation for a Population of a Higher Proportion
Thus, if the percentage of university students with insufficient sleep is assumed to be 50%, and the target level is to reduce it to 20%, with 90% power, and two-tailed α = 0.05, the sample size required is 104 participants. This is subdivided into 52 participants per group [Referred to Appendix 2].
c) Recommended Scenario
The recommended scenario would be, according to part (a) more conservative estimate of the 34% to 20% based on the required sample size of 420 participants for the initial estimate. This helps to increase the overall confidence of the results achieved and decrease the chances of making Type II errors with regard to the identified effect (Yakhdani et al. 2021). It means that the higher proportion of the main type of cancer to the other types, that is 50% to 20%, would need a smaller sample size, but may not be very effective in achieving the intended power for the effect, given the large difference.
d) Attribute to Completion Rate
However, only 75% of the students undertake research studies; therefore, the total sample size arrived at by the study in part (a) should be adjusted to account for dropouts. To achieve the estimated “sample size of 420 participants” while accounting for a 75% completion rate, the total number of participants to enrol should be calculated as follows:
Required Sample Size = Desired Sample Size * Completion Rate
Required Sample Size = Population * Proportion = 420 * 0. 75 = 560
Therefore, 560 participants should be recruited to get the staggered sample size of 420 to be considered after the exercise of dropping out [Referred to Appendix 3].
e) Calculation of the sample size when the groups are unequal
If students are in the ratio 3:1, where “the number of students in the intervention group is three times the number of students in the control group, assuming the” same power and significance level as that of part a above is to be used, the sample size for unequal groups can be determined from the formula (Aziz et al. 2021). Here, 3:1 is the allocated ratio concerning the intervention and control groups.
