y=_x+(). x 3,4,5,6,8 and y 4,5,6,10,15
$$\small{\text{
Determine the least squares regression line for (n=5):
$
\qquad
\begin{array}{rrr}
& x & y \\
1 & 3 & 4 \\
2 & 4 & 5 \\
3 & 5 & 6 \\
4 & 6 & 10 \\
5 & 8 & 15 \\
\end{array}
$
}}\\
\sum{X} = 3 + 4 + 5 + 6 + 8 = 26 \qquad \overline{X} = \frac{26}{5}=5.2\\
\sum{Y} = 4 + 5 + 6 + 10 + 15 = 40 \qquad \overline{Y} = \frac{40}{5}=8\\ \\
\small{\text{
$
\sum{ (X-\overline{X})^2 } = (3-5.2)^2 + (4-5.2)^2 + (5-5.2)^2 + (6-5.2)^2 + (8-5.2)^2 = 14.8
$
}}\\
\small{\text{
$
\sum{(Y-\overline{Y})^2} = (4-8)^2 + (5-8)^2 + (6-8)^2 + (10-8)^2 + (15-8)^2 =82
$
}}\\
\small{\text{
$
\sum{(X-\overline{X})(Y-\overline{Y})} =
(3-5.2)(4-8) + (4-5.2)(5-8) + (5-5.2)(6-8) + (6-5.2)(10-8) + (8-5.2)
(15-8) = 34
$
}}\\\\
\small{\text{
Standard Deviation in x und y:
}}\\
\small{\text{
$
\quad SD_x=\sqrt{
\dfrac{
\sum{ (X-\overline{X})^2 }
} {n-1}
} = \sqrt{ \dfrac{ 14.8 }{ 5-1 } } = 1.92353840617
$
}}\\
\small{\text{
$
\quad SD_y=\sqrt{
\dfrac{
\sum{ (Y-\overline{Y})^2 }
} {n-1}
} == \sqrt{ \dfrac{ 82 }{ 5-1 } } = 4.52769256907
$
}}
}}\\\\
\small{\text{
Correlation Coefficient, r :
}}\\
\small{\text{
$
r = \dfrac{
\sum (X-\overline{X})
(Y-\overline{Y})
}
{
\sqrt{ \sum (X-\overline{X})^2 }
\sqrt{ \sum (Y-\overline{Y})^2 }
} =
\dfrac{ 34 }
{
\sqrt{ 14.8 }
\sqrt{ 82 }
} = 0.97598048329
$
}}\\\\$$
$$$\\\\$
\small{\text{
PREDICTED $Y = a + b X $,
}}\\
\small{\text{
where the slope $b$ and intercept $a$ are calculated in the following order:
}}\\
\small{\text{
$
b = r \dfrac{ \mbox{SD}_Y}{\mbox{SD}_X} =0.97598048329
*\dfrac{4.52769256907
}{1.92353840617}=2.29729729730
$
}}\\ \\
\small{\text{
$
a = \overline{Y} - b \overline{X} = 8 - 2.29729729730* 5.2 = -3.94594594597
$
}}\\\\
\small{\text{
The regression line is
}}
\small{\text
$
\boxed{ \text{Predicted }\quad
Y = a + bX = -3.94594594597+ (2.29729729730
)X
$
}}$$
.
