Determine the least squares regression line.

y=_x+().   x 3,4,5,6,8 and y 4,5,6,10,15

$$\small{\text{ Determine the least squares regression line for (n=5): $ \qquad \begin{array}{rrr} & x & y \\ 1 & 3 & 4 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \\ 4 & 6 & 10 \\ 5 & 8 & 15 \\ \end{array} $ }}\\ \sum{X} = 3 + 4 + 5 + 6 + 8 = 26 \qquad \overline{X} = \frac{26}{5}=5.2\\ \sum{Y} = 4 + 5 + 6 + 10 + 15 = 40 \qquad \overline{Y} = \frac{40}{5}=8\\ \\ \small{\text{ $ \sum{ (X-\overline{X})^2 } = (3-5.2)^2 + (4-5.2)^2 + (5-5.2)^2 + (6-5.2)^2 + (8-5.2)^2 = 14.8 $ }}\\ \small{\text{ $ \sum{(Y-\overline{Y})^2} = (4-8)^2 + (5-8)^2 + (6-8)^2 + (10-8)^2 + (15-8)^2 =82 $ }}\\ \small{\text{ $ \sum{(X-\overline{X})(Y-\overline{Y})} = (3-5.2)(4-8) + (4-5.2)(5-8) + (5-5.2)(6-8) + (6-5.2)(10-8) + (8-5.2) (15-8) = 34 $ }}\\\\ \small{\text{ Standard Deviation in x und y: }}\\ \small{\text{ $ \quad SD_x=\sqrt{ \dfrac{ \sum{ (X-\overline{X})^2 } } {n-1} } = \sqrt{ \dfrac{ 14.8 }{ 5-1 } } = 1.92353840617 $ }}\\ \small{\text{ $ \quad SD_y=\sqrt{ \dfrac{ \sum{ (Y-\overline{Y})^2 } } {n-1} } == \sqrt{ \dfrac{ 82 }{ 5-1 } } = 4.52769256907 $ }} }}\\\\ \small{\text{ Correlation Coefficient, r : }}\\ \small{\text{ $ r = \dfrac{ \sum (X-\overline{X}) (Y-\overline{Y}) } { \sqrt{ \sum (X-\overline{X})^2 } \sqrt{ \sum (Y-\overline{Y})^2 } } = \dfrac{ 34 } { \sqrt{ 14.8 } \sqrt{ 82 } } = 0.97598048329 $ }}\\\\$$

$$$\\\\$ \small{\text{ PREDICTED $Y = a + b X $, }}\\ \small{\text{ where the slope $b$ and intercept $a$ are calculated in the following order: }}\\ \small{\text{ $ b = r \dfrac{ \mbox{SD}_Y}{\mbox{SD}_X} =0.97598048329 *\dfrac{4.52769256907 }{1.92353840617}=2.29729729730 $ }}\\ \\ \small{\text{ $ a = \overline{Y} - b \overline{X} = 8 - 2.29729729730* 5.2 = -3.94594594597 $ }}\\\\ \small{\text{ The regression line is }} \small{\text $ \boxed{ \text{Predicted }\quad Y = a + bX = -3.94594594597+ (2.29729729730 )X $ }}$$