1. **Problem Statement:** Determine whether the new treatment is comparatively superior to the conventional treatment based on the given data: | Treatment | Favourable | Non Favourable | |--------------|------------|---------------| | New | 60 | 30 | | Conventional | 40 | 70 | 2. **Method:** Use the Chi-square test for independence to check if the treatment type and outcome are related. 3. **Formula:** $$\chi^2 = \sum \frac{(O - E)^2}{E}$$ where $O$ is observed frequency and $E$ is expected frequency. 4. **Calculate totals:** - Total patients = $60 + 30 + 40 + 70 = 200$ - Total favourable = $60 + 40 = 100$ - Total non favourable = $30 + 70 = 100$ - Total new treatment = $60 + 30 = 90$ - Total conventional = $40 + 70 = 110$ 5. **Calculate expected frequencies:** - $E_{New,Favourable} = \frac{90 \times 100}{200} = 45$ - $E_{New,NonFavourable} = \frac{90 \times 100}{200} = 45$ - $E_{Conv,Favourable} = \frac{110 \times 100}{200} = 55$ - $E_{Conv,NonFavourable} = \frac{110 \times 100}{200} = 55$ 6. **Calculate Chi-square statistic:** $$\chi^2 = \frac{(60-45)^2}{45} + \frac{(30-45)^2}{45} + \frac{(40-55)^2}{55} + \frac{(70-55)^2}{55}$$ $$= \frac{15^2}{45} + \frac{(-15)^2}{45} + \frac{(-15)^2}{55} + \frac{15^2}{55}$$ $$= \frac{225}{45} + \frac{225}{45} + \frac{225}{55} + \frac{225}{55}$$ $$= 5 + 5 + 4.09 + 4.09 = 18.18$$ 7. **Degrees of freedom:** $df = (rows - 1)(columns - 1) = (2-1)(2-1) = 1$ 8. **Decision:** At $df=1$, the critical value of $\chi^2$ at 5% significance is 3.841. Since $18.18 > 3.841$, we reject the null hypothesis. 9. **Conclusion:** There is a significant association between treatment type and outcome. The new treatment is comparatively superior to the conventional treatment because it has more favourable outcomes. **Final answer:** The new treatment is statistically superior to the conventional treatment based on the Chi-square test.