1. **State the problem:** We want to find the Pearson correlation coefficient $r$ between daily advertising expenditure (in $100s) and daily units sold for 5 days, then test if the correlation is significantly positive at the 0.05 significance level. 2. **Data:** Advertising expenditure $X$: 20, 22, 25, 28, 30 Units sold $Y$: 15, 18, 22, 25, 28 3. **Formula for Pearson correlation coefficient:** $$r = \frac{n\sum XY - \sum X \sum Y}{\sqrt{(n\sum X^2 - (\sum X)^2)(n\sum Y^2 - (\sum Y)^2)}}$$ where $n=5$. 4. **Calculate sums:** $$\sum X = 20 + 22 + 25 + 28 + 30 = 125$$ $$\sum Y = 15 + 18 + 22 + 25 + 28 = 108$$ $$\sum XY = 20\times15 + 22\times18 + 25\times22 + 28\times25 + 30\times28 = 300 + 396 + 550 + 700 + 840 = 2786$$ $$\sum X^2 = 20^2 + 22^2 + 25^2 + 28^2 + 30^2 = 400 + 484 + 625 + 784 + 900 = 3193$$ $$\sum Y^2 = 15^2 + 18^2 + 22^2 + 25^2 + 28^2 = 225 + 324 + 484 + 625 + 784 = 2442$$ 5. **Calculate numerator:** $$5 \times 2786 - 125 \times 108 = 13930 - 13500 = 430$$ 6. **Calculate denominator:** $$\sqrt{(5 \times 3193 - 125^2)(5 \times 2442 - 108^2)} = \sqrt{(15965 - 15625)(12210 - 11664)} = \sqrt{340 \times 546}$$ 7. **Simplify denominator:** $$\sqrt{340 \times 546} = \sqrt{185640} \approx 430.87$$ 8. **Calculate $r$:** $$r = \frac{430}{430.87} \approx 0.998$$ 9. **Test statistic for hypothesis testing:** $$t = r \sqrt{\frac{n-2}{1-r^2}} = 0.998 \sqrt{\frac{3}{1 - (0.998)^2}}$$ 10. **Calculate denominator inside square root:** $$1 - (0.998)^2 = 1 - 0.996004 = 0.003996$$ 11. **Calculate $t$:** $$t = 0.998 \sqrt{\frac{3}{0.003996}} = 0.998 \sqrt{750.75} = 0.998 \times 27.41 = 27.36$$ 12. **Conclusion:** Degrees of freedom $df = n-2 = 3$. The critical $t$ value for a one-tailed test at 0.05 significance and 3 df is approximately 2.353. Since $27.36 > 2.353$, we reject the null hypothesis and conclude there is sufficient evidence of a significant positive correlation. **Final answers:** - Pearson correlation coefficient $r = 0.998$ - Test statistic $t = 27.3600$ - There is sufficient evidence at the 0.05 significance level to conclude a significant positive correlation.