Answer :
To find the least squares regression line, we use the formula
[tex]$$
\hat{y} = bx + a,
$$[/tex]
where the slope [tex]$b$[/tex] and the intercept [tex]$a$[/tex] are found using the formulas
[tex]$$
b = \frac{\sum (x-\bar{x})(y-\bar{y})}{\sum (x-\bar{x})^2}
$$[/tex]
and
[tex]$$
a = \bar{y} - b\bar{x}.
$$[/tex]
Given the data set:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
2 & 100 \\
4 & 83 \\
6 & 50 \\
8 & 35 \\
10 & 23 \\
\hline
\end{array}
\][/tex]
we first calculate the mean of the [tex]$x$[/tex]-values and [tex]$y$[/tex]-values.
The mean of [tex]$x$[/tex] is:
[tex]$$
\bar{x} = \frac{2 + 4 + 6 + 8 + 10}{5} = \frac{30}{5} = 6.
$$[/tex]
The mean of [tex]$y$[/tex] is:
[tex]$$
\bar{y} = \frac{100 + 83 + 50 + 35 + 23}{5} = \frac{291}{5} = 58.2.
$$[/tex]
Next, we calculate the numerator for [tex]$b$[/tex], which is [tex]$\sum (x-\bar{x})(y-\bar{y})$[/tex]:
[tex]\[
\begin{array}{c|c|c|c}
x & y & (x - \bar{x}) & (y - \bar{y}) \\
\hline
2 & 100 & 2 - 6 = -4 & 100 - 58.2 = 41.8 \\
4 & 83 & 4 - 6 = -2 & 83 - 58.2 = 24.8 \\
6 & 50 & 6 - 6 = 0 & 50 - 58.2 = -8.2 \\
8 & 35 & 8 - 6 = 2 & 35 - 58.2 = -23.2 \\
10 & 23 & 10 - 6 = 4 & 23 - 58.2 = -35.2 \\
\end{array}
\][/tex]
Now, compute each term [tex]$(x-\bar{x})(y-\bar{y})$[/tex]:
- For [tex]$x=2$[/tex]: [tex]$(-4)(41.8) = -167.2$[/tex]
- For [tex]$x=4$[/tex]: [tex]$(-2)(24.8) = -49.6$[/tex]
- For [tex]$x=6$[/tex]: [tex]$(0)(-8.2) = 0$[/tex]
- For [tex]$x=8$[/tex]: [tex]$(2)(-23.2) = -46.4$[/tex]
- For [tex]$x=10$[/tex]: [tex]$(4)(-35.2) = -140.8$[/tex]
Summing these products gives:
[tex]$$
\sum (x-\bar{x})(y-\bar{y}) = -167.2 - 49.6 + 0 - 46.4 - 140.8 = -404.0.
$$[/tex]
Next, we compute the denominator [tex]$\sum (x-\bar{x})^2$[/tex]:
[tex]\[
\begin{array}{c|c}
x & (x-\bar{x})^2 \\
\hline
2 & (-4)^2 = 16 \\
4 & (-2)^2 = 4 \\
6 & (0)^2 = 0 \\
8 & (2)^2 = 4 \\
10 & (4)^2 = 16 \\
\end{array}
\][/tex]
Summing these squares gives:
[tex]$$
\sum (x-\bar{x})^2 = 16 + 4 + 0 + 4 + 16 = 40.0.
$$[/tex]
Now, the slope is calculated as:
[tex]$$
b = \frac{-404.0}{40.0} = -10.1.
$$[/tex]
Using the formula for the intercept:
[tex]$$
a = \bar{y} - b\bar{x} = 58.2 - (-10.1)(6).
$$[/tex]
This simplifies to:
[tex]$$
a = 58.2 + 60.6 = 118.8.
$$[/tex]
Thus, the equation of the least squares regression line is:
[tex]$$
\hat{y} = -10.1x + 118.8.
$$[/tex]
Therefore, the equation of the least squares regression line that most closely matches the data set is
[tex]$$
\boxed{\hat{y} = -10.1x + 118.8.}
$$[/tex]
[tex]$$
\hat{y} = bx + a,
$$[/tex]
where the slope [tex]$b$[/tex] and the intercept [tex]$a$[/tex] are found using the formulas
[tex]$$
b = \frac{\sum (x-\bar{x})(y-\bar{y})}{\sum (x-\bar{x})^2}
$$[/tex]
and
[tex]$$
a = \bar{y} - b\bar{x}.
$$[/tex]
Given the data set:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
2 & 100 \\
4 & 83 \\
6 & 50 \\
8 & 35 \\
10 & 23 \\
\hline
\end{array}
\][/tex]
we first calculate the mean of the [tex]$x$[/tex]-values and [tex]$y$[/tex]-values.
The mean of [tex]$x$[/tex] is:
[tex]$$
\bar{x} = \frac{2 + 4 + 6 + 8 + 10}{5} = \frac{30}{5} = 6.
$$[/tex]
The mean of [tex]$y$[/tex] is:
[tex]$$
\bar{y} = \frac{100 + 83 + 50 + 35 + 23}{5} = \frac{291}{5} = 58.2.
$$[/tex]
Next, we calculate the numerator for [tex]$b$[/tex], which is [tex]$\sum (x-\bar{x})(y-\bar{y})$[/tex]:
[tex]\[
\begin{array}{c|c|c|c}
x & y & (x - \bar{x}) & (y - \bar{y}) \\
\hline
2 & 100 & 2 - 6 = -4 & 100 - 58.2 = 41.8 \\
4 & 83 & 4 - 6 = -2 & 83 - 58.2 = 24.8 \\
6 & 50 & 6 - 6 = 0 & 50 - 58.2 = -8.2 \\
8 & 35 & 8 - 6 = 2 & 35 - 58.2 = -23.2 \\
10 & 23 & 10 - 6 = 4 & 23 - 58.2 = -35.2 \\
\end{array}
\][/tex]
Now, compute each term [tex]$(x-\bar{x})(y-\bar{y})$[/tex]:
- For [tex]$x=2$[/tex]: [tex]$(-4)(41.8) = -167.2$[/tex]
- For [tex]$x=4$[/tex]: [tex]$(-2)(24.8) = -49.6$[/tex]
- For [tex]$x=6$[/tex]: [tex]$(0)(-8.2) = 0$[/tex]
- For [tex]$x=8$[/tex]: [tex]$(2)(-23.2) = -46.4$[/tex]
- For [tex]$x=10$[/tex]: [tex]$(4)(-35.2) = -140.8$[/tex]
Summing these products gives:
[tex]$$
\sum (x-\bar{x})(y-\bar{y}) = -167.2 - 49.6 + 0 - 46.4 - 140.8 = -404.0.
$$[/tex]
Next, we compute the denominator [tex]$\sum (x-\bar{x})^2$[/tex]:
[tex]\[
\begin{array}{c|c}
x & (x-\bar{x})^2 \\
\hline
2 & (-4)^2 = 16 \\
4 & (-2)^2 = 4 \\
6 & (0)^2 = 0 \\
8 & (2)^2 = 4 \\
10 & (4)^2 = 16 \\
\end{array}
\][/tex]
Summing these squares gives:
[tex]$$
\sum (x-\bar{x})^2 = 16 + 4 + 0 + 4 + 16 = 40.0.
$$[/tex]
Now, the slope is calculated as:
[tex]$$
b = \frac{-404.0}{40.0} = -10.1.
$$[/tex]
Using the formula for the intercept:
[tex]$$
a = \bar{y} - b\bar{x} = 58.2 - (-10.1)(6).
$$[/tex]
This simplifies to:
[tex]$$
a = 58.2 + 60.6 = 118.8.
$$[/tex]
Thus, the equation of the least squares regression line is:
[tex]$$
\hat{y} = -10.1x + 118.8.
$$[/tex]
Therefore, the equation of the least squares regression line that most closely matches the data set is
[tex]$$
\boxed{\hat{y} = -10.1x + 118.8.}
$$[/tex]
