Answer :
We begin with 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]
The least squares regression line is given by
[tex]$$
\hat{y} = bx + a,
$$[/tex]
where the slope [tex]$b$[/tex] and the [tex]$y$[/tex]-intercept [tex]$a$[/tex] are computed using the following formulas:
[tex]$$
b = \frac{n \sum xy - (\sum x)(\sum y)}{n \sum x^2 - (\sum x)^2}
$$[/tex]
and
[tex]$$
a = \frac{\sum y - b\sum x}{n}.
$$[/tex]
We have [tex]$n = 5$[/tex] data points.
1. First, calculate the sums:
- The sum of the [tex]$x$[/tex] values is
[tex]$$
\sum x = 2 + 4 + 6 + 8 + 10 = 30.
$$[/tex]
- The sum of the [tex]$y$[/tex] values is
[tex]$$
\sum y = 100 + 83 + 50 + 35 + 23 = 291.
$$[/tex]
- The sum of the products [tex]$xy$[/tex] is
[tex]$$
\sum xy = (2)(100) + (4)(83) + (6)(50) + (8)(35) + (10)(23) = 200 + 332 + 300 + 280 + 230 = 1342.
$$[/tex]
- The sum of the squares of [tex]$x$[/tex] is
[tex]$$
\sum x^2 = 2^2 + 4^2 + 6^2 + 8^2 + 10^2 = 4 + 16 + 36 + 64 + 100 = 220.
$$[/tex]
2. Substitute these sums into the formula for the slope [tex]$b$[/tex]:
[tex]$$
b = \frac{5 \cdot 1342 - 30 \cdot 291}{5 \cdot 220 - 30^2} = \frac{6710 - 8730}{1100 - 900} = \frac{-2020}{200} = -10.1.
$$[/tex]
3. Then, compute the intercept [tex]$a$[/tex]:
[tex]$$
a = \frac{291 - (-10.1)(30)}{5} = \frac{291 + 303}{5} = \frac{594}{5} = 118.8.
$$[/tex]
Thus, the equation of the regression line is:
[tex]$$
\hat{y} = -10.1x + 118.8.
$$[/tex]
Comparing with the provided options, we see that the equation that most closely matches the data set is
[tex]$$
\hat{y} = -10.1x + 118.8.
$$[/tex]
[tex]$$
\begin{array}{|c|c|}
\hline
x & y \\
\hline
2 & 100 \\
4 & 83 \\
6 & 50 \\
8 & 35 \\
10 & 23 \\
\hline
\end{array}
$$[/tex]
The least squares regression line is given by
[tex]$$
\hat{y} = bx + a,
$$[/tex]
where the slope [tex]$b$[/tex] and the [tex]$y$[/tex]-intercept [tex]$a$[/tex] are computed using the following formulas:
[tex]$$
b = \frac{n \sum xy - (\sum x)(\sum y)}{n \sum x^2 - (\sum x)^2}
$$[/tex]
and
[tex]$$
a = \frac{\sum y - b\sum x}{n}.
$$[/tex]
We have [tex]$n = 5$[/tex] data points.
1. First, calculate the sums:
- The sum of the [tex]$x$[/tex] values is
[tex]$$
\sum x = 2 + 4 + 6 + 8 + 10 = 30.
$$[/tex]
- The sum of the [tex]$y$[/tex] values is
[tex]$$
\sum y = 100 + 83 + 50 + 35 + 23 = 291.
$$[/tex]
- The sum of the products [tex]$xy$[/tex] is
[tex]$$
\sum xy = (2)(100) + (4)(83) + (6)(50) + (8)(35) + (10)(23) = 200 + 332 + 300 + 280 + 230 = 1342.
$$[/tex]
- The sum of the squares of [tex]$x$[/tex] is
[tex]$$
\sum x^2 = 2^2 + 4^2 + 6^2 + 8^2 + 10^2 = 4 + 16 + 36 + 64 + 100 = 220.
$$[/tex]
2. Substitute these sums into the formula for the slope [tex]$b$[/tex]:
[tex]$$
b = \frac{5 \cdot 1342 - 30 \cdot 291}{5 \cdot 220 - 30^2} = \frac{6710 - 8730}{1100 - 900} = \frac{-2020}{200} = -10.1.
$$[/tex]
3. Then, compute the intercept [tex]$a$[/tex]:
[tex]$$
a = \frac{291 - (-10.1)(30)}{5} = \frac{291 + 303}{5} = \frac{594}{5} = 118.8.
$$[/tex]
Thus, the equation of the regression line is:
[tex]$$
\hat{y} = -10.1x + 118.8.
$$[/tex]
Comparing with the provided options, we see that the equation that most closely matches the data set is
[tex]$$
\hat{y} = -10.1x + 118.8.
$$[/tex]
