Question

Table
x y
0 8
4 9
8 10
12 11
16 12
Find the line of least squares regression for the data. What is the equation for this line

Answer 1

To find the line of least squares regression for the given data, we can use the formula for the linear regression line:

\[
y = mx + b
\]

where \( m \) is the slope and \( b \) is the y-intercept. The formulas we will use to calculate these values are:

\[
m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}
\]

\[
b = \frac{\sum y - m(\sum x)}{n}
\]

Where:
- \( n \) is the number of data points,
- \( \sum x \) is the sum of x-values,
- \( \sum y \) is the sum of y-values,
- \( \sum xy \) is the sum of the product of x and y,
- \( \sum x^2 \) is the sum of the squares of x-values.

Let's calculate the needed sums using the provided data:

| \( x \) | \( y \) | \( xy \) | \( x^2 \) |
|---------|---------|----------|-----------|
| 0 | 8 | 0 | 0 |
| 4 | 9 | 36 | 16 |
| 8 | 10 | 80 | 64 |
| 12 | 11 | 132 | 144 |
| 16 | 12 | 192 | 256 |

Now, we can calculate the sums:

\[
\sum x = 0 + 4 + 8 + 12 + 16 = 40
\]

\[
\sum y = 8 + 9 + 10 + 11 + 12 = 50
\]

\[
\sum xy = 0 + 36 + 80 + 132 + 192 = 440
\]

\[
\sum x^2 = 0 + 16 + 64 + 144 + 256 = 480
\]

Now, plug these values into the formulas.

1. **Calculate \( n \)**:
\[
n = 5
\]

2. **Calculate \( m \)**:
\[
m = \frac{5(440) - (40)(50)}{5(480) - (40)^2}
\]
\[
m = \frac{2200 - 2000}{2400 - 1600}
\]
\[
m = \frac{200}{800} = 0.25
\]

3. **Calculate \( b \)**:
\[
b = \frac{50 - 0.25(40)}{5}
\]
\[
b = \frac{50 - 10}{5}
\]
\[
b = \frac{40}{5} = 8
\]

Therefore, the equation of the line of least squares regression is:

\[
y = 0.25x + 8
\]