Question

linear regression and correlation.
compute the equation of the linear regression line in the form y = mx + b, where m is the slope and b is the intercept.
use at least 3 decimal places. (round if necessary)
y = x +
compute the correlation coefficient for this data set. use at least 3 decimal places. (round if necessary)
r =
question help: message instructor post to forum
submit question

Explanation:

Step1: Calculate the necessary sums

Let \(n = 5\).
\(\sum_{i = 1}^{n}x_i=2 + 3+4 + 5+6=20\)
\(\sum_{i = 1}^{n}y_i=0.7 + 5.2+11.1+9.3+16.4 = 42.7\)
\(\sum_{i = 1}^{n}x_i^2=2^2+3^2+4^2+5^2+6^2=4 + 9+16+25+36 = 90\)
\(\sum_{i = 1}^{n}x_iy_i=2\times0.7+3\times5.2 + 4\times11.1+5\times9.3+6\times16.4=1.4 + 15.6+44.4+46.5+98.4 = 206.3\)

Step2: Calculate the slope \(m\)

The formula for the slope \(m\) of the regression - line is \(m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^2-(\sum_{i = 1}^{n}x_i)^2}\)
Substitute the values:

$$
LATEXBLOCK0
$$

Step3: Calculate the y - intercept \(b\)

The formula for the y - intercept \(b\) is \(b=\bar{y}-m\bar{x}\), where \(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\) and \(\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}\)
\(\bar{x}=\frac{20}{5}=4\), \(\bar{y}=\frac{42.7}{5}=8.54\)
\(b = 8.54-3.550\times4=8.54 - 14.2=- 5.660\)

Step4: Calculate the correlation coefficient \(r\)

First, calculate \(S_{xx}=\sum_{i = 1}^{n}x_i^2-\frac{1}{n}(\sum_{i = 1}^{n}x_i)^2=90-\frac{20^2}{5}=90 - 80 = 10\)
\(S_{yy}=\sum_{i = 1}^{n}y_i^2-\frac{1}{n}(\sum_{i = 1}^{n}y_i)^2\)
\(\sum_{i = 1}^{n}y_i^2=0.7^2+5.2^2+11.1^2+9.3^2+16.4^2=0.49+27.04+123.21+86.49+268.96 = 506.19\)
\(S_{yy}=506.19-\frac{42.7^2}{5}=506.19-\frac{1823.29}{5}=506.19 - 364.658=141.532\)
\(S_{xy}=\sum_{i = 1}^{n}x_iy_i-\frac{1}{n}\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i=206.3-\frac{20\times42.7}{5}=206.3 - 170.8 = 35.5\)
The correlation coefficient \(r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{35.5}{\sqrt{10\times141.532}}=\frac{35.5}{\sqrt{1415.32}}\approx\frac{35.5}{37.621}\approx0.944\)

Answer:

The equation of the linear regression line is \(y = 3.550x-5.660\)
The correlation coefficient \(r\approx0.944\)