1. **State the problem:** Determine if there is an association between social media usage (x) and reading performance (RA, y) among students. 2. **Null hypothesis (H0):** There is no association between social media usage and reading performance. In other words, the correlation coefficient $\rho = 0$. 3. **Calculate the sums needed for correlation:** Given data: $x = [5,9,4,10,14,15,6,12,6,9,5,4]$ $y = [85,125,80,225,115,175,135,145,120,130,87,90]$ Compute sums: $$\sum x = 95$$ $$\sum y = 1412$$ $$\sum xy = 5\cdot85 + 9\cdot125 + \cdots + 4\cdot90 = 11453$$ $$\sum x^2 = 5^2 + 9^2 + \cdots + 4^2 = 913$$ $$\sum y^2 = 85^2 + 125^2 + \cdots + 90^2 = 175826$$ Number of pairs, $n = 12$ 4. **Calculate Pearson correlation coefficient $r$: ** $$r = \frac{n\sum xy - \sum x \sum y}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}}$$ Substitute values: $$r = \frac{12\times11453 - 95\times1412}{\sqrt{[12\times913 - 95^2][12\times175826 - 1412^2]}}$$ Calculate numerator: $$12 \times 11453 = 137436$$ $$95 \times 1412 = 134140$$ $$\Rightarrow \text{numerator} = 137436 - 134140 = 3296$$ Calculate denominator parts: $$12 \times 913 = 10956, \quad 95^2 = 9025 \, \Rightarrow \quad 10956 - 9025 = 2931$$ $$12 \times 175826 = 2109912, \quad 1412^2 = 1994944\quad \Rightarrow \quad 2109912 - 1994944 = 114968$$ Denominator: $$\sqrt{2931 \times 114968} = \sqrt{337119708} \approx 18358.67$$ Calculate $r$: $$r = \frac{3296}{18358.67} \approx 0.1795$$ 5. **Determine the test statistic $t$ for correlation:** $$t = r \sqrt{\frac{n-2}{1-r^2}} = 0.1795 \times \sqrt{\frac{10}{1 - 0.1795^2}}$$ Calculate denominator inside sqrt: $$1 - 0.1795^2 = 1 - 0.0322 = 0.9678$$ Calculate sqrt: $$\sqrt{\frac{10}{0.9678}} = \sqrt{10.34} = 3.215$$ Calculate $t$: $$t = 0.1795 \times 3.215 = 0.577$$ 6. **Set significance level and degrees of freedom:** - Level of significance $\alpha = 0.05$ - Degrees of freedom: $df = n - 2 = 10$ 7. **Find critical value:** From $t$-distribution table for $df=10$ and two-tailed $\alpha=0.05$, critical value $t_{0.025,10} \approx 2.228$ 8. **Decision:** Since computed $|t| = 0.577 < 2.228$, fail to reject the null hypothesis. 9. **Interpretation:** There is insufficient evidence to conclude a significant association between social media usage and reading performance among these students at the 5% significance level. --- **Summary Table:** | Null Hypothesis | Computed t-value | Level of Significance | Degrees of Freedom | Critical t-value | Decision | |-----------------|------------------|-----------------------|--------------------|------------------|----------| | $H_0: \rho=0$ | 0.577 | 0.05 | 10 | 2.228 | Fail to reject |