1. **State the Problem:** We want to test if there is an association between social media usage (x) and reading performance (RA, y). 2. **Null Hypothesis ($H_0$):** There is no association between social media usage and reading performance. That is, the correlation coefficient $\rho = 0$. 3. **Alternative Hypothesis ($H_a$):** There is an association between social media usage and reading performance ($\rho \neq 0$). 4. **Data:** 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 5. **Calculate sample size:** $$ n = 12 $$ 6. **Calculate means:** $$ \bar{x} = \frac{5+9+4+10+14+15+6+12+6+9+5+4}{12} = \frac{99}{12} = 8.25 $$ $$ \bar{y} = \frac{85+125+80+225+115+175+135+145+120+130+87+90}{12} = \frac{1512}{12} = 126 $$ 7. **Calculate sample covariance and sample variances:** $$ S_{xy} = \sum (x_i - \bar{x})(y_i - \bar{y}) $$ $$ S_{xx} = \sum (x_i - \bar{x})^2 $$ $$ S_{yy} = \sum (y_i - \bar{y})^2 $$ Calculating each term: | $x_i$ | $y_i$ | $x_i - \bar{x}$ | $y_i - \bar{y}$ | $(x_i - \bar{x})(y_i - \bar{y})$ | $(x_i - \bar{x})^2$ | $(y_i - \bar{y})^2$ | |---|---|---|---|---|---|---| |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| Sum each column: $$ S_{xy} = 133.25 -0.75 +195.5 +173.25 -63.25 +330.75 -20.25 +71.25 +13.5 +3 +126.75 +153 = 1,112.75 $$ $$ S_{xx} = 10.5625 +0.5625 +18.0625 +3.0625 +33.0625 +45.5625 +5.0625 +14.0625 +5.0625 +0.5625 +10.5625 +18.0625 = 164.674 $$ $$ S_{yy} = 1681 +1 +2116 +9801 +121 +2401 +81 +361 +36 +16 +1521 +1296 = 19,432 $$ 8. **Calculate correlation coefficient:** $$ r = \frac{S_{xy}}{\sqrt{S_{xx} S_{yy}}} = \frac{1112.75}{\sqrt{164.674 \times 19432}} = \frac{1112.75}{\sqrt{3,200,008.768}} = \frac{1112.75}{1788.85} = 0.622 \approx 0.62 $$ 9. **Test statistic:** Use t-test for correlation: $$ t = r \sqrt{\frac{n-2}{1-r^2}} = 0.622 \sqrt{\frac{10}{1-0.622^2}} = 0.622 \sqrt{\frac{10}{1-0.387}} = 0.622 \sqrt{16.26} = 0.622 \times 4.033 = 2.51 $$ 10. **Degrees of freedom:** $$ df = n - 2 = 10 $$ 11. **Significance level:** Commonly $\alpha = 0.05$ 12. **Critical value:** From t-distribution table for two-tailed test and $df=10$ at 0.05 significance: $$ t_{critical} = 2.228 $$ 13. **Decision:** Since calculated $t = 2.51 > 2.228$, we reject the null hypothesis. 14. **Interpretation:** We have sufficient evidence at the 5% level of significance to conclude that there is an association between social media usage and reading performance among the students. **Summary Table:** | Parameter | Value | |---|---| | Null Hypothesis($H_0$) | $\rho = 0$ (No association) | | Computed value (t) | 2.51 | | Level of significance ($\alpha$) | 0.05 | | Degrees of freedom (df) | 10 | | Critical value | 2.228 | | Decision | Reject $H_0$ | | Interpretation | There is a significant association between social media usage and reading performance. |