1. **State the problem:** We want to test if there is a difference in taste ratings between fruit juice treated with Cold Plasma (CP) and without CP using matched pairs data from 10 experts. 2. **Hypotheses:** - Null hypothesis $H_0$: There is no difference in mean taste ratings, i.e., $\mu_d = 0$ where $d$ is the difference (With CP - Without CP). - Alternative hypothesis $H_a$: There is a difference, i.e., $\mu_d \neq 0$. 3. **Calculate differences:** For each expert, $d_i = \text{With CP}_i - \text{Without CP}_i$: $$d = [93-92, 65-67, 37-47, 78-79, 89-84, 49-52, 88-80, 55-52, 63-67, 62-69] = [1, -2, -10, -1, 5, -3, 8, 3, -4, -7]$$ 4. **Calculate sample mean and standard deviation of differences:** - Mean difference $\bar{d} = \frac{1 + (-2) + (-10) + (-1) + 5 + (-3) + 8 + 3 + (-4) + (-7)}{10} = \frac{-10}{10} = -1$ - Calculate squared deviations: $$\sum (d_i - \bar{d})^2 = (1+1)^2 + (-2+1)^2 + (-10+1)^2 + (-1+1)^2 + (5+1)^2 + (-3+1)^2 + (8+1)^2 + (3+1)^2 + (-4+1)^2 + (-7+1)^2$$ $$= 2^2 + (-1)^2 + (-9)^2 + 0^2 + 6^2 + (-2)^2 + 9^2 + 4^2 + (-3)^2 + (-6)^2$$ $$= 4 + 1 + 81 + 0 + 36 + 4 + 81 + 16 + 9 + 36 = 268$$ - Sample variance $s_d^2 = \frac{268}{10-1} = \frac{268}{9} \approx 29.78$ - Sample standard deviation $s_d = \sqrt{29.78} \approx 5.46$ 5. **Calculate test statistic:** $$t = \frac{\bar{d} - 0}{s_d / \sqrt{n}} = \frac{-1}{5.46 / \sqrt{10}} = \frac{-1}{5.46 / 3.162} = \frac{-1}{1.726} \approx -0.58$$ 6. **Degrees of freedom:** $df = n - 1 = 9$ 7. **Find critical value $t^*$ for two-tailed test at $\alpha=0.05$:** From $t$-distribution table, $t^*_{0.025,9} \approx 2.262$ 8. **Conclusion:** Since $|t| = 0.58 < 2.262$, we fail to reject $H_0$. There is insufficient evidence to conclude a difference in taste ratings between juices with and without CP treatment. **Final answer:** No significant difference in taste ratings at 5% significance level.