# Find the regression line using the given points.

Find the regression line using the given points.

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Jayden-James Duffy

Given:
The given bivariate data is(1,6),(3,2), and (2,4).
Calculation:
First determine $$\displaystyle\sum{x}_{{i}}{y}_{{i}},\sum{x}_{{i}}{\quad\text{and}\quad}\sum{y}_{{i}}$$.
$$\displaystyle{x}_{{i}}\rightarrow$$ first coordinates of the ordered pairs.
$$\displaystyle{y}_{{i}}\rightarrow$$ second coordinates of the ordered pairs.
$$\displaystyle\sum{x}_{{i}}{y}_{{i}}={1}\cdot{6}+{3}\cdot{2}+{2}\cdot{4}={20}$$
$$\displaystyle\sum{x}_{{i}}={1}+{3}+{2}={6}$$
$$\displaystyle\sum{{x}_{{i}}^{{2}}}={1}^{{2}}+{3}^{{2}}+{2}^{{2}}={14}$$
$$\displaystyle\sum{y}_{{i}}={6}+{2}+{4}={12}$$
$$\displaystyle\sum{{y}_{{i}}^{{2}}}={6}^{{2}}+{2}^{{2}}+{4}^{{2}}={56}$$
Find the sample variance $$\displaystyle{s}^{{2}}$$ using the formula $$\displaystyle{s}^{{2}}=\frac{{\sum{{x}_{{i}}^{{2}}}-\frac{{\sum{{x}_{{i}}^{{2}}}}}{{{n}}}}}{{{n}-{1}}}$$
$$\displaystyle{{s}_{{x}}^{{2}}}=\frac{{{14}-\frac{{{6}^{{2}}}}{{3}}}}{{{3}-{1}}}=\frac{{{14}-{12}}}{{2}}={1}$$
$$\displaystyle{{s}_{{y}}^{{2}}}=\frac{{{56}-\frac{{{12}^{{2}}}}{{3}}}}{{{3}-{1}}}=\frac{{{56}-{48}}}{{2}}={4}$$
Find sample standard deviation.
$$\displaystyle{s}_{{x}}=\sqrt{{1}}={1}$$
$$\displaystyle{s}_{{y}}=\sqrt{{4}}={2}$$
For covariance $$\displaystyle{s}_{{{x}{y}}}$$ using formula $$\displaystyle{s}_{{{x}{y}}}=\frac{{\sum{x}_{{i}}{y}_{{i}}-\frac{{\sum{x}_{{i}}-\sum{y}_{{i}}}}{{{n}}}}}{{{n}-{1}}}$$.
Where $$n =$$ number of ordered pairs.
$$n=3$$
$$\displaystyle{s}_{{{x}{y}}}=\frac{{{20}-\frac{{{6}\cdot{12}}}{{3}}}}{{{3}-{1}}}$$
$$\displaystyle=\frac{{{20}-\frac{{{72}}}{{3}}}}{{{3}-{1}}}$$
$$\displaystyle=\frac{{{20}-{24}}}{{2}}$$
$$\displaystyle=-\frac{{4}}{{2}}=-{2}$$
Find the correlation coefficient r using the formula $$\displaystyle{r}=\frac{{{s}_{{{x}{y}}}}}{{{s}_{{x}}{s}_{{y}}}}$$.
$$\displaystyle{r}=-\frac{{2}}{{{1}\cdot{2}}}=-{1}$$
Find the slope b using the formula $$\displaystyle{b}={r}\cdot\frac{{s}_{{y}}}{{s}_{{x}}}$$
$$\displaystyle{b}=-{1}\cdot\frac{{2}}{{1}}=-{2}$$
Find the value of y-intercept.
$$\displaystyle{a}=\overline{{y}}-{b}\overline{{x}}=\frac{{\sum{y}_{{i}}}}{{n}}-{b}\frac{{\sum{x}_{{i}}}}{{n}}$$
$$\displaystyle=\frac{{12}}{{3}}-{\left(-{2}\right)}\frac{{6}}{{3}}={4}+{2}\cdot{2}={8}$$
Regression line is $$y = a+bx$$
$$y=8-2x$$
Hence the regression line is $$y = 8-2x.$$