1. **State the problem:** We want to test if there is evidence that the students improved, i.e., the mean difference of errors (before - after) is greater than 0. 2. **Given data:** - Sample size $n = 25$ - Mean difference $\bar{d} = 2.4$ (decrease means positive improvement) - Sum of squares of differences $\sum d_i^2 = 900$ - Significance level $\alpha = 0.05$ 3. **Calculate sample variance of differences:** $$ s_d^2 = \frac{\sum d_i^2 - n\bar{d}^2}{n-1} = \frac{900 - 25 \times 2.4^2}{24} $$ $$ = \frac{900 - 25 \times 5.76}{24} = \frac{900 - 144}{24} = \frac{756}{24} = 31.5 $$ 4. **Calculate sample standard deviation:** $$ s_d = \sqrt{31.5} \approx 5.61 $$ 5. **Set up hypotheses:** - Null hypothesis: $H_0: \mu_d \leq 0$ (no improvement or worse) - Alternative hypothesis: $H_a: \mu_d > 0$ (improvement) 6. **Calculate the test statistic (t-score):** $$ t = \frac{\bar{d} - 0}{s_d / \sqrt{n}} = \frac{2.4}{5.61 / \sqrt{25}} = \frac{2.4}{5.61/5} = \frac{2.4}{1.122} \approx 2.14 $$ 7. **Degrees of freedom:** $df = n - 1 = 24$ 8. **Find critical t-value for one-tailed test at 5%:** $$ t_{0.05, 24} \approx 1.711 $$ 9. **Decision:** Since $t = 2.14 > 1.711$, we reject the null hypothesis. 10. **Conclusion:** There is sufficient evidence at the 5% significance level to conclude that students' arithmetic proficiency improved (errors decreased).