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:** $$d_i = \text{With CP}_i - \text{Without CP}_i$$ Differences: $1, -2, -10, -1, 5, -3, 8, 3, -4, -7$ 4. **Calculate sample mean and standard deviation of differences:** $$\bar{d} = \frac{1 - 2 - 10 - 1 + 5 - 3 + 8 + 3 - 4 - 7}{10} = \frac{-10}{10} = -1$$ Calculate variance: $$s_d^2 = \frac{1}{n-1} \sum (d_i - \bar{d})^2$$ Calculate each squared deviation: $(1+1)^2=4$, $(-2+1)^2=1$, $(-10+1)^2=81$, $(-1+1)^2=0$, $(5+1)^2=36$, $(-3+1)^2=4$, $(8+1)^2=81$, $(3+1)^2=16$, $(-4+1)^2=9$, $(-7+1)^2=36$ Sum: $4+1+81+0+36+4+81+16+9+36=268$ $$s_d^2 = \frac{268}{9} \approx 29.78$$ $$s_d = \sqrt{29.78} \approx 5.46$$ 5. **Calculate t-statistic:** $$t = \frac{\bar{d} - 0}{s_d / \sqrt{n}} = \frac{-1}{5.46 / \sqrt{10}} = \frac{-1}{1.726} \approx -0.58$$ 6. **Degrees of freedom:** $df = n-1 = 9$ 7. **Critical value for two-tailed test at $\alpha=0.05$:** From t-tables, $t^*_{0.025,9} \approx 2.262$ 8. **Conclusion:** Since $|t| = 0.58 < 2.262$, we fail to reject $H_0$. There is not enough 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.