1. **State the problem:** We want to construct a 99% confidence interval for the true mean change in typing speeds before and after using the program. 2. **Identify the data:** We have paired samples (before and after) for 7 people: Before: $[51, 33, 47, 53, 41, 33, 37]$ After: $[32, 34, 40, 46, 51, 33, 30]$ 3. **Calculate the differences:** $d_i = \text{Before}_i - \text{After}_i$ $$d = [51-32, 33-34, 47-40, 53-46, 41-51, 33-33, 37-30] = [19, -1, 7, 7, -10, 0, 7]$$ 4. **Calculate the sample mean of differences:** $$\bar{d} = \frac{19 + (-1) + 7 + 7 + (-10) + 0 + 7}{7} = \frac{29}{7} \approx 4.14$$ 5. **Calculate the sample standard deviation of differences:** First find squared deviations: $$(19 - 4.14)^2 = 221.0, (-1 - 4.14)^2 = 26.5, (7 - 4.14)^2 = 8.2, (7 - 4.14)^2 = 8.2, (-10 - 4.14)^2 = 200.0, (0 - 4.14)^2 = 17.1, (7 - 4.14)^2 = 8.2$$ Sum of squared deviations: $221.0 + 26.5 + 8.2 + 8.2 + 200.0 + 17.1 + 8.2 = 489.2$ Sample variance: $$s_d^2 = \frac{489.2}{7 - 1} = \frac{489.2}{6} \approx 81.53$$ Sample standard deviation: $$s_d = \sqrt{81.53} \approx 9.03$$ 6. **Find the critical t-value for 99% confidence and df = 6:** $$t^* \approx 3.707$$ 7. **Calculate the margin of error:** $$ME = t^* \times \frac{s_d}{\sqrt{n}} = 3.707 \times \frac{9.03}{\sqrt{7}} \approx 3.707 \times 3.41 = 12.64$$ 8. **Construct the confidence interval:** $$\bar{d} \pm ME = 4.14 \pm 12.64 = (-8.5, 16.8)$$ **Interpretation:** We are 99% confident that the true mean change in typing speed (before minus after) lies between -8.5 and 16.8 words per minute. This interval includes zero, so there is not strong evidence that the program changes typing speed on average.