1. The problem is to calculate the Pearson product-moment correlation coefficient $r$ given the data: sums for $X$, $Y$, $X^2$, $Y^2$, and $XY$ with degrees of freedom $Df = 103$. 2. The formula for the Pearson correlation coefficient is: $$r = \frac{n\sum XY - (\sum X)(\sum Y)}{\sqrt{[n\sum X^{2} - (\sum X)^{2}][n\sum Y^{2} - (\sum Y)^{2}]}}$$ 3. From the data we identify or infer sums: Let $n = Df + 1 = 103 + 1 = 104$ Given sums approximately: $\sum X \approx 84.54 + 241 = 325.54$ $\sum Y \approx 8877 + 7236 = 16113$ $\sum X^{2} \approx 7147.0116 + 58081 = 65228.0116$ $\sum Y^{2} \approx 78801129 + 52359696 = 131160825$ $\sum XY \approx 750461.6 + 1743876 = 2494337.6$ 4. Substitute into formula: $$r = \frac{104 \times 2494337.6 - 325.54 \times 16113}{\sqrt{[104 \times 65228.0116 - 325.54^{2}][104 \times 131160825 - 16113^{2}]}}$$ 5. Calculate numerator: $$104 \times 2494337.6 = 259410710.4$$ $$(\sum X)(\sum Y) = 325.54 \times 16113 = 5245096.02$$ Numerator = $259410710.4 - 5245096.02 = 254165614.38$ 6. Calculate denominator components: $$104 \times 65228.0116 = 6783713.2064$$ $$(\sum X)^{2} = 325.54^{2} = 105964.2916$$ First bracket = $6783713.2064 - 105964.2916 = 6677748.9148$ $$104 \times 131160825 = 13640649300$$ $$(\sum Y)^{2} = 16113^{2} = 259622769$$ Second bracket = $13640649300 - 259622769 = 13381026531$ 7. Calculate denominator: $$\sqrt{6677748.9148 \times 13381026531} = \sqrt{89330665944754187.988} \approx 9456712686.32$$ 8. Calculate $r$: $$r = \frac{254165614.38}{9456712686.32} \approx 0.0269$$ 9. Interpretation: The Pearson correlation coefficient $r \approx 0.027$ shows a very weak positive linear relationship between $X$ and $Y$ in the data provided. Final answer: $r \approx 0.027$