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 \(y = 8-2x.\)