Find the equation for the least squares regression line of the data described below.

Jared is an editor reviewing short story submissions for a literary magazine. He collected data showing that he's an efficient reviewer and wants to leverage this to get a raise.
For a week, Jared recorded the number of words in each short story submission, x, and how long it took him to review that story (in minutes), y.
Words Minutes
1,586 30.65
3,444 31.43
5,009 44.52
5,572 37.20
7,069 57.41
Round your answers to the nearest thousandth.
y=
x+

1 answer

To find the equation for the least squares regression line \( y = mx + b \), we need to calculate the slope \( m \) and the y-intercept \( b \).

Given the data:

\[ \begin{array}{|c|c|} \hline \text{Words (x)} & \text{Minutes (y)} \ \hline 1586 & 30.65 \ 3444 & 31.43 \ 5009 & 44.52 \ 5572 & 37.20 \ 7069 & 57.41 \ \hline \end{array} \]

First, we calculate various necessary sums:

  1. \( n \) = number of data points = 5
  2. \( \sum x \) = \( 1586 + 3444 + 5009 + 5572 + 7069 = 22780 \)
  3. \( \sum y \) = \( 30.65 + 31.43 + 44.52 + 37.20 + 57.41 = 201.21 \)
  4. \( \sum xy \) = \( 1586 \times 30.65 + 3444 \times 31.43 + 5009 \times 44.52 + 5572 \times 37.20 + 7069 \times 57.41 \)

Calculating \( \sum xy \):

\[ \begin{align*} 1586 \times 30.65 & = 48575.90 \ 3444 \times 31.43 & = 108297.92 \ 5009 \times 44.52 & = 222492.68 \ 5572 \times 37.20 & = 207946.40 \ 7069 \times 57.41 & = 405851.29 \ \end{align*} \]

Adding these together: \[ \sum xy = 48575.90 + 108297.92 + 222492.68 + 207946.40 + 405851.29 = 1093074.19 \]

  1. \( \sum x^2 \) = \( 1586^2 + 3444^2 + 5009^2 + 5572^2 + 7069^2 \)

Calculating \( \sum x^2 \):

\[ \begin{align*} 1586^2 & = 2510596 \ 3444^2 & = 11871536 \ 5009^2 & = 25090181 \ 5572^2 & = 31018384 \ 7069^2 & = 49972461 \ \end{align*} \]

Adding these together: \[ \sum x^2 = 2510596 + 11871536 + 25090181 + 31018384 + 49972461 = 113394158 \]

Now we can calculate the slope \( m \) and intercept \( b \):

The formulas are:

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

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

Substituting the values into the formula for \( m \): \[ m = \frac{5 \times 1093074.19 - 22780 \times 201.21}{5 \times 113394158 - (22780)^2} \]

Calculating the numerator: \[ 5 \times 1093074.19 = 5465370.95 \] \[ 22780 \times 201.21 = 4589669.80 \] \[ \text{Numerator} = 5465370.95 - 4589669.80 = 875701.15 \]

Calculating the denominator: \[ 5 \times 113394158 = 566970790 \] \[ (22780)^2 = 518448484 \] \[ \text{Denominator} = 566970790 - 518448484 = 48422306 \]

Now calculating \( m \): \[ m = \frac{875701.15}{48422306} \approx 0.0181 \quad (\text{rounded to 4 decimal places}) \]

Now, calculating \( b \) with the value of \( m \): \[ b = \frac{201.21 - 0.0181 \times 22780}{5} \] Calculating the product: \[ 0.0181 \times 22780 \approx 412.858 \] Now calculating \( b \): \[ b = \frac{201.21 - 412.858}{5} = \frac{-211.648}{5} \approx -42.3296 \]

So, rounding \( m \) and \( b \) to the nearest thousandth gives: \[ m \approx 0.018 \quad \text{and} \quad b \approx -42.330 \]

Finally, the equation of the least squares regression line is: \[ y = 0.018x - 42.330 \]