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**Assignment Solutions on Epidemiology for Health and Medical Sciences**

**Question 1**

- Relationship between treatment and improvement after 8 weeks

**Stata Code: tab treatment improve_8wk, chi2**

The primary outcome variable improve_8wk is a ordinal variable (0 = No and 1 = Yes) represents whether patient experienced an improvement in PSD at 8 weeks. On the other hand, the treatment variable is also a qualitative variable that represents the treatment undertaken by the patient (0 = Sham and 1 = tacs). Thus, the most appropriate test to test the association between treatment and primary outcome is chi square test for independence

** 1. Null Hypothesis: H0:**

That is, there is no association between treatment and patient experienced an improvement in PSD at 8 weeks

** 2. Alternative Hypothesis: Ha:**

That is, there is an association between treatment and patient experienced an improvement in PSD at 8 weeks

** 3. Expected frequencies**

Under the assumption that the null hypothesis is true, the expected frequencies are calculated by using the formula given below

** 4. Expected Frequency Table**

- The workings of chi square test statistic is given below

S. No | Observed Frequency, Oi | Expected Frequency, Ei | (Oi – Ei) | (Oi – Ei)^2/Ei |

1 | 25 | 19.5 | 5.5 | 1.551282 |

2 | 25 | 30.5 | -5.5 | 0.991803 |

3 | 14 | 19.5 | -5.5 | 1.551282 |

4 | 36 | 30.5 | 5.5 | 0.991803 |

Total | 100 | 100 | 5.086 |

The chi square test statistic was calculated by adding the square of the difference between observed and expected frequency divided by the corresponding expected frequency. It is known fact that the greater the difference between the observed and expected frequency for each cell, the chi square test statistic becomes larger. Thus, the cell with greater difference is considered as the most contributing factor for the relationship between two qualitative variables

Degrees of freedom: df = (r – 1) * (c – 1) = (2 – 1) * (2 – 1) = 1

** 2. Stata Code: tab treatment improve_8wk, chi2**

The calculated value matches exactly with the chi square test statistic calculated using stata

Statistical Decision: Reject the Null Hypothesis

- The p–value falls below 0.05 indicating that there is an association between treatment and patient experienced an improvement in PSD at 8 weeks
- Thus, the odds ratio of improvement in the tACS group, relative to the Sham group is 2.571

This indicate that, when the patient is taking tACS treatment, then there is 2.571 times for him to experience an improvement in PSD at 8 weeks when compared to that of Sham Group patients

**Question 2**

** 1. Stata Command: by treatment, sort : summarize vas**

The mean VAS score in sham group is 3.014 with a standard deviation of 2.81 and the mean VAS score in tACS group is 4.558 with a standard deviation of 2.425. On comparing the mean values, we see that the mean VAS score in tACS group is high when compared to that of sham group.

- VAS contains numeric data (scores) from a Visual Analogue Scale and it is measured under ratio scale
- Null Hypothesis: H0: µSham = µtACS

That is, there is no difference in the average pain score between Sham and tACS treatment groups

** 2. Alternative Hypothesis: Ha: µSham ≠ µtACS**

That is, there is a difference in the average pain score between Sham and tACS treatment groups

The difference in mean pain score between Sham and tACS group is 3.014 – 4.558 = – 1.544

** 3. The expected difference mean pain score between Sham and tACS group is 0**

Here, the null hypothesis states that there is no difference in the average pain score between Sham and tACS treatment groups

- The p – value of t test statistic corresponding to 98 degrees of freedom is 0.0041
- Stata Command: ttest vas, by(treatment)

Here, the results obtained from the manual calculation and through stata output remains the same. Thus, the value of chi square test statistic is – 2.9397 and the p – value of t test statistic corresponding to 98 degrees of freedom is 0.0041

** 4. Statistical Decision: Reject the null hypothesis**

The 95% confidence interval of the difference in mean is (-2.587, -0.502), indicating that when repeated samples are taken, then 95 out of 100 times the true mean difference will fall within this interval

- Yes, based on the 95% confidence interval of the difference in mean (since the mean value do not include zero) we can say that the blinding of participants would be possible in this study

**Question 3**

1. The major reason for the dropout of the participants is:

**Adverse reactions caused by the treatment****Lack of treatment effectiveness****Health status of the patient**

2. Dropout in the study may lead to attrition bias and it may occur when the trial losses its patients considering to their health conditions and they may not able to continue the entire experiment. This certainly indicates that the patients who show significant improvement only continue in the trial. This can be solved by accounting all dropouts and the reason for their dropouts. If the reason seems to be a result of adverse event, then it can be included in the safety analysis.