QUESTION IMAGE
Question
find the equation for the least squares regression line of the data described below.
mr. ballard is trying to convince his middle school typing students that it is better to slow down when they are learning to type. he argues that students who type too quickly make too many errors.
to prove his point, mr. ballard has students type a sample paragraph. he records their average typing speed (in words per minute), x, and the number of errors they made, y.
| typing speed | errors |
|---|---|
| 49 | 0 |
| 64 | 14 |
| 77 | 8 |
| 77 | 14 |
round your answers to the nearest thousandth.
$y = \square x + \square$
Step1: Calculate necessary sums
First, we list the data points: \((44, 4)\), \((49, 0)\), \((64, 14)\), \((77, 8)\), \((77, 14)\).
Let \(n = 5\) (number of data points).
Calculate \(\sum x\), \(\sum y\), \(\sum xy\), and \(\sum x^2\):
- \(\sum x = 44 + 49 + 64 + 77 + 77 = 311\)
- \(\sum y = 4 + 0 + 14 + 8 + 14 = 40\)
- \(\sum xy = (44\times4)+(49\times0)+(64\times14)+(77\times8)+(77\times14)\)
\(= 176 + 0 + 896 + 616 + 1078 = 2766\)
- \(\sum x^2 = 44^2 + 49^2 + 64^2 + 77^2 + 77^2\)
\(= 1936 + 2401 + 4096 + 5929 + 5929 = 20291\)
Step2: Calculate the slope \(m\)
The formula for the slope \(m\) of the least - squares regression line is:
\(m=\frac{n\sum xy-\sum x\sum y}{n\sum x^{2}-(\sum x)^{2}}\)
Substitute the values:
\(n = 5\), \(\sum xy = 2766\), \(\sum x = 311\), \(\sum y = 40\), \(\sum x^{2}=20291\)
\(m=\frac{5\times2766 - 311\times40}{5\times20291-(311)^{2}}\)
First, calculate the numerator: \(5\times2766=13830\), \(311\times40 = 12440\), so the numerator is \(13830 - 12440=1390\)
Then, calculate the denominator: \(5\times20291 = 101455\), \((311)^{2}=96721\), so the denominator is \(101455 - 96721 = 4734\)
\(m=\frac{1390}{4734}\approx0.2936\approx0.294\) (rounded to the nearest thousandth)
Step3: Calculate the y - intercept \(b\)
The formula for the y - intercept \(b\) is:
\(b=\frac{\sum y - m\sum x}{n}\)
Substitute the values: \(\sum y = 40\), \(m\approx0.2936\), \(\sum x = 311\), \(n = 5\)
\(b=\frac{40-0.2936\times311}{5}\)
First, calculate \(0.2936\times311\approx0.2936\times300+0.2936\times11 = 88.08+3.2296 = 91.3096\)
Then, \(40 - 91.3096=- 51.3096\)
\(b=\frac{-51.3096}{5}\approx - 10.2619\approx - 10.262\) (rounded to the nearest thousandth)
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\(y = 0.294x-10.262\)