1. **Problem statement:** Find the least squares linear regression line $y = ax + b$ for the given data and estimate $y$ when $x=30$. 2. **Given data:** $x$: 5, 6, 8, 10, 12, 13, 15, 16, 17 $y$: 16, 19, 23, 28, 36, 41, 44, 45, 50 3. **Formula for least squares regression line:** $$a = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$$ $$b = \frac{\sum y - a \sum x}{n}$$ where $n$ is the number of data points. 4. **Calculate sums:** $n = 9$ $$\sum x = 5 + 6 + 8 + 10 + 12 + 13 + 15 + 16 + 17 = 102$$ $$\sum y = 16 + 19 + 23 + 28 + 36 + 41 + 44 + 45 + 50 = 302$$ $$\sum x^2 = 5^2 + 6^2 + 8^2 + 10^2 + 12^2 + 13^2 + 15^2 + 16^2 + 17^2 = 5^2 + 6^2 + 8^2 + 10^2 + 12^2 + 13^2 + 15^2 + 16^2 + 17^2$$ Calculate each: $25 + 36 + 64 + 100 + 144 + 169 + 225 + 256 + 289 = 1308$ $$\sum xy = (5)(16) + (6)(19) + (8)(23) + (10)(28) + (12)(36) + (13)(41) + (15)(44) + (16)(45) + (17)(50)$$ Calculate each: $80 + 114 + 184 + 280 + 432 + 533 + 660 + 720 + 850 = 3853$ 5. **Calculate slope $a$:** $$a = \frac{9 \times 3853 - 102 \times 302}{9 \times 1308 - 102^2} = \frac{34677 - 30804}{11772 - 10404} = \frac{3873}{1368}$$ Simplify fraction: $$\frac{\cancel{3873}}{\cancel{1368}} = \frac{3873/3}{1368/3} = \frac{1291}{456}$$ Approximate: $$a \approx 2.832$$ 6. **Calculate intercept $b$:** $$b = \frac{302 - 2.832 \times 102}{9} = \frac{302 - 288.864}{9} = \frac{13.136}{9} \approx 1.459$$ 7. **Regression line:** $$y = 2.832x + 1.459$$ 8. **Estimate $y$ when $x=30$:** $$y = 2.832 \times 30 + 1.459 = 84.96 + 1.459 = 86.419$$ **Final answer:** The least squares regression line is $$y = 2.832x + 1.459$$ and the estimated value of $y$ at $x=30$ is approximately $$86.419$$.