Question

question #8
use the given data set to determine the correlation coefficient r.

.878
-.907
.854
-.932

Explanation:

Step1: Calculate the means

Let \(x\) be weight and \(y\) be MPG.
\(\bar{x}=\frac{1924 + 2317+2640+3108+2051}{5}=\frac{12040}{5} = 2408\)
\(\bar{y}=\frac{45+30+31+19+36}{5}=\frac{161}{5}=32.2\)

Step2: Calculate numerator and denominator components

For each pair \((x_i,y_i)\):

• \((x_1 - \bar{x})(y_1-\bar{y})=(1924 - 2408)(45 - 32.2)=(- 484)\times12.8=-6203.2\)
• \((x_2 - \bar{x})(y_2-\bar{y})=(2317 - 2408)(30 - 32.2)=(-91)\times(-2.2) = 200.2\)
• \((x_3 - \bar{x})(y_3-\bar{y})=(2640 - 2408)(31 - 32.2)=232\times(-1.2)=-278.4\)
• \((x_4 - \bar{x})(y_4-\bar{y})=(3108 - 2408)(19 - 32.2)=700\times(-13.2)=-9240\)
• \((x_5 - \bar{x})(y_5-\bar{y})=(2051 - 2408)(36 - 32.2)=(-357)\times3.8=-1356.6\)

Sum of \((x_i - \bar{x})(y_i-\bar{y})\): \(S_{xy}=-6203.2 + 200.2-278.4-9240-1356.6=-16878\)

For \(S_{xx}\):

• \((x_1 - \bar{x})^2=(1924 - 2408)^2=(-484)^2 = 234256\)
• \((x_2 - \bar{x})^2=(2317 - 2408)^2=(-91)^2=8281\)
• \((x_3 - \bar{x})^2=(2640 - 2408)^2=(232)^2 = 53824\)
• \((x_4 - \bar{x})^2=(3108 - 2408)^2=(700)^2=490000\)
• \((x_5 - \bar{x})^2=(2051 - 2408)^2=(-357)^2 = 127449\)

\(S_{xx}=234256+8281 + 53824+490000+127449=913810\)

For \(S_{yy}\):

• \((y_1 - \bar{y})^2=(45 - 32.2)^2=(12.8)^2 = 163.84\)
• \((y_2 - \bar{y})^2=(30 - 32.2)^2=(-2.2)^2=4.84\)
• \((y_3 - \bar{y})^2=(31 - 32.2)^2=(-1.2)^2 = 1.44\)
• \((y_4 - \bar{y})^2=(19 - 32.2)^2=(-13.2)^2=174.24\)
• \((y_5 - \bar{y})^2=(36 - 32.2)^2=(3.8)^2 = 14.44\)

\(S_{yy}=163.84+4.84 + 1.44+174.24+14.44=358.8\)

Step3: Calculate the correlation coefficient \(r\)

The formula for \(r\) is \(r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}\)
\(r=\frac{-16878}{\sqrt{913810\times358.8}}\)
\(r=\frac{-16878}{\sqrt{327975028}}\)
\(r=\frac{-16878}{18110.08}\approx - 0.932\)

Answer:

• 0.932