Model Answer
0 min readIntroduction
In medical research, establishing the efficacy of a new drug requires rigorous testing and statistical analysis. The placebo effect, where a patient experiences a benefit from a treatment with no active medicinal properties, is a well-documented phenomenon. To determine if a drug’s effect is genuine and not simply due to the placebo effect, researchers employ statistical methods to compare the outcomes of a treatment group with a control group receiving a placebo. This question presents data from an experiment designed to assess the effectiveness of a drug against the common cold, and asks us to determine if the observed difference in outcomes between the drug and sugar pill groups is statistically significant.
Hypothesis Formulation
Before conducting the statistical test, we need to formulate the null and alternative hypotheses:
- Null Hypothesis (H0): There is no significant difference in the effect of the drug and sugar pills. In other words, the drug is no more effective than the placebo.
- Alternative Hypothesis (H1): There is a significant difference in the effect of the drug and sugar pills. The drug is more effective than the placebo.
Contingency Table and Expected Frequencies
The given data can be represented in a contingency table:
| Helped | Harmed | No Effect | Total | |
|---|---|---|---|---|
| Drug | 150 | 30 | 70 | 250 |
| Sugar Pills | 130 | 40 | 80 | 250 |
| Total | 280 | 70 | 150 | 500 |
To perform the Chi-Square test, we need to calculate the expected frequencies for each cell under the assumption that the null hypothesis is true. The expected frequency for each cell is calculated as:
Expected Frequency = (Row Total * Column Total) / Grand Total
Here's a table showing the calculated expected frequencies:
| Helped | Harmed | No Effect | |
|---|---|---|---|
| Drug | 140 | 35 | 75 |
| Sugar Pills | 140 | 35 | 75 |
Chi-Square Statistic Calculation
The Chi-Square statistic (χ2) is calculated using the following formula:
χ2 = Σ [(Observed Frequency - Expected Frequency)2 / Expected Frequency]
Calculating the Chi-Square statistic for each cell and summing them up:
- For Drug - Helped: (150 - 140)2 / 140 = 100 / 140 = 0.714
- For Drug - Harmed: (30 - 35)2 / 35 = 25 / 35 = 0.714
- For Drug - No Effect: (70 - 75)2 / 75 = 25 / 75 = 0.333
- For Sugar Pills - Helped: (130 - 140)2 / 140 = 100 / 140 = 0.714
- For Sugar Pills - Harmed: (40 - 35)2 / 35 = 25 / 35 = 0.714
- For Sugar Pills - No Effect: (80 - 75)2 / 75 = 25 / 75 = 0.333
χ2 = 0.714 + 0.714 + 0.333 + 0.714 + 0.714 + 0.333 = 3.522
Degrees of Freedom and P-value
The degrees of freedom (df) for a contingency table are calculated as:
df = (Number of Rows - 1) * (Number of Columns - 1)
In this case, df = (2 - 1) * (3 - 1) = 1 * 2 = 2
Using a Chi-Square distribution table or a statistical calculator, with a Chi-Square statistic of 3.522 and 2 degrees of freedom, the p-value is approximately 0.171.
Conclusion
We typically use a significance level (α) of 0.05. Since the p-value (0.171) is greater than the significance level (0.05), we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that there is a statistically significant difference in the effect of the drug and sugar pills. The observed difference could be due to random chance.
Conclusion
Based on the statistical analysis of the provided data, we cannot conclude that the drug is significantly more effective than the sugar pills in curing the common cold. The Chi-Square test yielded a p-value greater than the conventional significance level of 0.05, indicating that the observed differences in treatment outcomes could be attributed to random variation. Further research with a larger sample size or a more potent drug formulation might be necessary to establish a statistically significant effect.
Answer Length
This is a comprehensive model answer for learning purposes and may exceed the word limit. In the exam, always adhere to the prescribed word count.