1. **State the problem:** Given data points $x = \{1,2,3,4,5,6\}$ and $y = \{5.6,4.6,4.5,3.7,3.2,2.7\}$, we need to find sums of squares and cross-products ($S_{xx}$, $S_{yy}$, $S_{xy}$), the least-squares regression line, predict $y$ at $x=3.5$, and construct the ANOVA table. 2. **Calculate sums:** Calculate $\sum x$, $\sum y$, $\sum x^2$, $\sum y^2$, and $\sum xy$: $$\sum x = 1+2+3+4+5+6 = 21$$ $$\sum y = 5.6+4.6+4.5+3.7+3.2+2.7 = 24.3$$ $$\sum x^2 = 1^2+2^2+3^2+4^2+5^2+6^2 = 91$$ $$\sum y^2 = 5.6^2+4.6^2+4.5^2+3.7^2+3.2^2+2.7^2 = 108.03$$ $$\sum xy = (1)(5.6)+(2)(4.6)+(3)(4.5)+(4)(3.7)+(5)(3.2)+(6)(2.7) = 75.1$$ 3. **Calculate sums of squares:** $$S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 91 - \frac{21^2}{6} = 91 - 73.5 = 17.5$$ $$S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n} = 108.03 - \frac{24.3^2}{6} = 108.03 - 98.48 = 9.55$$ $$S_{xy} = \sum xy - \frac{(\sum x)(\sum y)}{n} = 75.1 - \frac{21 \times 24.3}{6} = 75.1 - 85.05 = -9.95$$ 4. **Calculate regression coefficients:** $$b = \frac{S_{xy}}{S_{xx}} = \frac{-9.95}{17.5} = -0.5686$$ $$\bar{x} = \frac{\sum x}{n} = \frac{21}{6} = 3.5$$ $$\bar{y} = \frac{\sum y}{n} = \frac{24.3}{6} = 4.05$$ $$a = \bar{y} - b \bar{x} = 4.05 - (-0.5686)(3.5) = 4.05 + 1.99 = 6.04$$ So the least-squares line is: $$\hat{y} = 6.04 - 0.5686x$$ 5. **Calculate SSR, SSE, and Total SS:** $$SSR = \frac{S_{xy}^2}{S_{xx}} = \frac{(-9.95)^2}{17.5} = \frac{99.00}{17.5} = 5.66$$ $$SSE = S_{yy} - SSR = 9.55 - 5.66 = 3.89$$ $$\text{Total SS} = S_{yy} = 9.55$$ 6. **Predict $y$ at $x=3.5$:** $$\hat{y} = 6.04 - 0.5686 \times 3.5 = 6.04 - 1.99 = 4.05$$ 7. **Construct ANOVA table:** | Source | df | SS | MS | F | |-------------|-----|------|----------|------------| | Regression | 1 | 5.66 | 5.66 | $\frac{5.66}{\frac{3.89}{4}} = 5.82$ | | Error | 4 | 3.89 | 0.9725 | | | Total | 5 | 9.55 | | | **Summary:** - $S_{xx} = 17.5$ - $S_{yy} = 9.55$ - $S_{xy} = -9.95$ - Regression line: $\hat{y} = 6.04 - 0.5686x$ - Predicted $y$ at $x=3.5$ is $4.05$ - ANOVA F-statistic is approximately $5.82$ indicating the regression is significant at typical levels.