1. **Stating the problem:** We have data on motivation (X) and work enthusiasm (Y) and want to find the relationship between them using correlation and regression analysis. 2. **Calculate sums and sums of squares:** We first compute $\sum X$, $\sum Y$, $\sum X^2$, $\sum Y^2$, and $\sum XY$ from the data. Given data: $\sum X = 6+6+7+8+9+8+9+8+10+12 = 83$ $\sum Y = 7+6+7+8+9+10+10+8+10+12 = 87$ Calculate $\sum X^2$: $6^2+6^2+7^2+8^2+9^2+8^2+9^2+8^2+10^2+12^2 = 36+36+49+64+81+64+81+64+100+144 = 719$ Calculate $\sum Y^2$: $7^2+6^2+7^2+8^2+9^2+10^2+10^2+8^2+10^2+12^2 = 49+36+49+64+81+100+100+64+100+144 = 787$ Calculate $\sum XY$: $(6\times7)+(6\times6)+(7\times7)+(8\times8)+(9\times9)+(8\times10)+(9\times10)+(8\times8)+(10\times10)+(12\times12) = 42+36+49+64+81+80+90+64+100+144 = 750$ 3. **Calculate correlation coefficient $r$:** Formula: $$r = \frac{n\sum XY - \sum X \sum Y}{\sqrt{(n\sum X^2 - (\sum X)^2)(n\sum Y^2 - (\sum Y)^2)}}$$ Where $n=10$. Calculate numerator: $$10 \times 750 - 83 \times 87 = 7500 - 7221 = 279$$ Calculate denominator: $$\sqrt{(10 \times 719 - 83^2)(10 \times 787 - 87^2)} = \sqrt{(7190 - 6889)(7870 - 7569)} = \sqrt{301 \times 301} = 301$$ So, $$r = \frac{279}{301} \approx 0.927$$ 4. **Test significance of $r$ using t-test:** Formula: $$t = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$$ Calculate: $$t = \frac{0.927 \times \sqrt{8}}{\sqrt{1-0.927^2}} = \frac{0.927 \times 2.828}{\sqrt{1-0.859}} = \frac{2.622}{\sqrt{0.141}} = \frac{2.622}{0.375} \approx 6.99$$ Degrees of freedom $df = n-2 = 8$. For $\alpha=0.05$, critical $t_{0.05,8} \approx 2.306$. Since $6.99 > 2.306$, the correlation is significant. 5. **Find linear regression equation $Y = a + bX$:** Slope $b$: $$b = \frac{n\sum XY - \sum X \sum Y}{n\sum X^2 - (\sum X)^2} = \frac{279}{301} \approx 0.927$$ Intercept $a$: $$a = \bar{Y} - b \bar{X} = \frac{87}{10} - 0.927 \times \frac{83}{10} = 8.7 - 0.927 \times 8.3 = 8.7 - 7.7 = 1.0$$ Regression equation: $$Y = 1.0 + 0.927X$$ 6. **Calculate coefficient of determination $r^2$ to find influence:** $$r^2 = (0.927)^2 = 0.859$$ This means 85.9% of the variation in work enthusiasm is explained by motivation. **Final answers:** - Correlation coefficient $r \approx 0.927$ (strong positive correlation). - Correlation is significant ($t=6.99 > 2.306$). - Linear regression equation: $Y = 1.0 + 0.927X$. - Motivation explains about 85.9% of the variation in work enthusiasm.