1. **State the problem:** We want to test whether social media usage ($x$) affects reading performance ($y$) of students. 2. **Define the null hypothesis ($H_0$):** There is no association between social media usage and reading performance, i.e., the population correlation coefficient $\rho = 0$. 3. **Data given:** Social Media Usage ($x$): 5, 9, 4, 10, 14, 15, 6, 12, 6, 9, 5, 4 Reading Performance ($y$): 85, 125, 80, 225, 115, 175, 135, 145, 120, 130, 87, 90 4. **Calculate necessary statistics:** Number of observations, $n = 12$. Calculate means: $$\bar{x} = \frac{5+9+4+10+14+15+6+12+6+9+5+4}{12} = \frac{99}{12} = 8.25$$ $$\bar{y} = \frac{85+125+80+225+115+175+135+145+120+130+87+90}{12} = \frac{1512}{12} = 126$$ Calculate sample covariance $S_{xy}$ and standard deviations $S_x$, $S_y$: Compute $\sum (x_i - \bar{x})(y_i - \bar{y})$, $\sum (x_i - \bar{x})^2$, and $\sum (y_i - \bar{y})^2$: | $x_i$ | $y_i$ | $x_i - \bar{x}$ | $y_i - \bar{y}$ | $(x_i - \bar{x})(y_i - \bar{y})$ | $(x_i - \bar{x})^2$ | $(y_i - \bar{y})^2$ | |-------|-------|-----------------|-----------------|----------------------------------|---------------------|---------------------| | 5 | 85 | -3.25 | -41 | 133.25 | 10.5625 | 1681 | | 9 | 125 | 0.75 | -1 | -0.75 | 0.5625 | 1 | | 4 | 80 | -4.25 | -46 | 195.5 | 18.0625 | 2116 | | 10 | 225 | 1.75 | 99 | 173.25 | 3.0625 | 9801 | | 14 | 115 | 5.75 | -11 | -63.25 | 33.0625 | 121 | | 15 | 175 | 6.75 | 49 | 330.75 | 45.5625 | 2401 | | 6 | 135 | -2.25 | 9 | -20.25 | 5.0625 | 81 | | 12 | 145 | 3.75 | 19 | 71.25 | 14.0625 | 361 | | 6 | 120 | -2.25 | -6 | 13.5 | 5.0625 | 36 | | 9 | 130 | 0.75 | 4 | 3 | 0.5625 | 16 | | 5 | 87 | -3.25 | -39 | 126.75 | 10.5625 | 1521 | | 4 | 90 | -4.25 | -36 | 153 | 18.0625 | 1296 | Summations: $$\sum (x_i - \bar{x})(y_i - \bar{y}) = 1127.75$$ $$\sum (x_i - \bar{x})^2 = 164.125$$ $$\sum (y_i - \bar{y})^2 = 18711$$ Calculate sample covariance and standard deviations: $$S_{xy} = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{n-1} = \frac{1127.75}{11} = 102.52$$ $$S_x = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} = \sqrt{\frac{164.125}{11}} = 3.86$$ $$S_y = \sqrt{\frac{\sum (y_i - \bar{y})^2}{n-1}} = \sqrt{\frac{18711}{11}} = 41.20$$ 5. **Calculate correlation coefficient $r$:** $$r = \frac{S_{xy}}{S_x S_y} = \frac{102.52}{3.86 \times 41.20} = \frac{102.52}{159.03} = 0.64$$ 6. **Test statistic for correlation:** $$t = \frac{r \sqrt{n-2}}{\sqrt{1-r^2}} = \frac{0.64 \times \sqrt{10}}{\sqrt{1-0.64^2}} = \frac{0.64 \times 3.162}{\sqrt{1-0.4096}} = \frac{2.023}{\sqrt{0.5904}} = \frac{2.023}{0.768} = 2.635$$ 7. **Determine degrees of freedom:** $$df = n-2 = 10$$ 8. **Level of significance:** Commonly $\alpha = 0.05$. 9. **Critical value:** Two-tailed test with $df=10$ and $\alpha=0.05$, critical $t$ approximately $2.228$. 10. **Decision:** Since $t_{calculated} = 2.635 > t_{critical} = 2.228$, reject the null hypothesis. 11. **Interpretation:** There is significant evidence at the 5% level to conclude that social media usage is associated with reading performance in this group of students.