1. **State the problem:** We want to find the Pearson correlation coefficient $r$ between daily advertising expenditure (in $100s) and daily units sold, test the significance of this correlation, and conclude if there is a significant positive correlation at the 0.05 significance level. 2. **Data:** Ad Expense $x$: 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 $n=5$ is the number of data points. 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:** $$n\sum xy - \sum x \sum y = 5 \times 2786 - 125 \times 108 = 13930 - 13500 = 430$$ 6. **Calculate denominator:** $$\sqrt{(n\sum x^2 - (\sum x)^2)(n\sum y^2 - (\sum y)^2)} = \sqrt{(5 \times 3193 - 125^2)(5 \times 2442 - 108^2)}$$ $$= \sqrt{(15965 - 15625)(12210 - 11664)} = \sqrt{340 \times 546} = \sqrt{185640} \approx 430.85$$ 7. **Calculate $r$:** $$r = \frac{430}{430.85} \approx 0.998$$ 8. **Test statistic for hypothesis testing:** The test statistic $t$ for testing $H_0: \rho=0$ is $$t = r \sqrt{\frac{n-2}{1-r^2}}$$ Calculate denominator inside square root: $$1 - r^2 = 1 - (0.998)^2 = 1 - 0.996 = 0.004$$ Calculate $t$: $$t = 0.998 \sqrt{\frac{5-2}{0.004}} = 0.998 \sqrt{\frac{3}{0.004}} = 0.998 \sqrt{750} = 0.998 \times 27.386 = 27.332$$ 9. **Conclusion:** Degrees of freedom $df = n-2 = 3$. The critical $t$ value for a one-tailed test at $\alpha=0.05$ and $df=3$ is approximately 2.353. Since $t=27.332 > 2.353$, we reject the null hypothesis. There is sufficient evidence at the 0.05 significance level to conclude a significant positive correlation between advertising expenditure and units sold. **Final answers:** - Pearson correlation coefficient $r = 0.998$ - Test statistic $t = 27.3320$ - Conclusion: Significant positive correlation exists at 0.05 level.