1. **Problem Statement:** Find the least squares regression equation for yield $y$ as a linear function of temperature $x$ using the data: $$x: 0, 25, 50, 75, 100$$ $$y: 14, 38, 54, 76, 95$$ 2. **Formula for Least Squares Regression Line:** The regression line is given by: $$y = b_0 + b_1 x$$ where $$b_1 = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$$ $$b_0 = \bar{y} - b_1 \bar{x}$$ 3. **Calculate sums and means:** $$n=5$$ $$\sum x = 0 + 25 + 50 + 75 + 100 = 250$$ $$\sum y = 14 + 38 + 54 + 76 + 95 = 277$$ $$\sum x^2 = 0^2 + 25^2 + 50^2 + 75^2 + 100^2 = 0 + 625 + 2500 + 5625 + 10000 = 18750$$ $$\sum xy = 0\times14 + 25\times38 + 50\times54 + 75\times76 + 100\times95 = 0 + 950 + 2700 + 5700 + 9500 = 18850$$ $$\bar{x} = \frac{250}{5} = 50$$ $$\bar{y} = \frac{277}{5} = 55.4$$ 4. **Calculate slope $b_1$:** $$b_1 = \frac{5 \times 18850 - 250 \times 277}{5 \times 18750 - 250^2} = \frac{94250 - 69250}{93750 - 62500} = \frac{25000}{31250} = 0.8$$ 5. **Calculate intercept $b_0$:** $$b_0 = 55.4 - 0.8 \times 50 = 55.4 - 40 = 15.4$$ 6. **Regression equation:** $$y = 15.4 + 0.8x$$ 7. **Interpretation:** The yield increases by 0.8 grams for each degree Celsius increase in temperature. 8. **Significance test for relationship at 1% level:** Calculate correlation coefficient $r$ and test $H_0: \rho=0$ vs $H_a: \rho \neq 0$. 9. **Calculate $r$:** $$r = \frac{n\sum xy - \sum x \sum y}{\sqrt{(n\sum x^2 - (\sum x)^2)(n\sum y^2 - (\sum y)^2)}}$$ Calculate $\sum y^2$: $$14^2 + 38^2 + 54^2 + 76^2 + 95^2 = 196 + 1444 + 2916 + 5776 + 9025 = 19357$$ Calculate denominator: $$\sqrt{(5 \times 18750 - 250^2)(5 \times 19357 - 277^2)} = \sqrt{(93750 - 62500)(96785 - 76729)} = \sqrt{31250 \times 20056} \approx \sqrt{627000000} \approx 25039.99$$ Calculate numerator: $$5 \times 18850 - 250 \times 277 = 94250 - 69250 = 25000$$ So, $$r = \frac{25000}{25039.99} \approx 0.9984$$ 10. **Test statistic:** $$t = r \sqrt{\frac{n-2}{1-r^2}} = 0.9984 \sqrt{\frac{3}{1 - 0.9968}} = 0.9984 \sqrt{\frac{3}{0.0032}} = 0.9984 \times 30.62 = 30.57$$ 11. **Critical value:** For $\alpha=0.01$ and $df=3$, two-tailed $t$ critical value is approximately 5.841. 12. **Decision:** Since $30.57 > 5.841$, reject $H_0$. There is a significant relationship between yield and temperature at 1% significance level. **Final answers:** - Estimated regression equation: $$y = 15.4 + 0.8x$$ - Significant relationship exists between yield and temperature at 1% level.