Question

miles driven gallons in tank
27 13
65 12
83 11
109 10
142 9
175 8
janelle tracks the number of miles she drives and the
number of gallons of gas she has left.
what is the linear regression model for this scenario?
what is the correlation coefficient?
he strength of the model?
-0.996 egative
-0.035
0.035
0.996

Explanation:

Step1: Calculate the means

Let \(x\) be miles driven and \(y\) be gallons in tank.
\(\bar{x}=\frac{27 + 65+83+109+142+175}{6}=\frac{601}{6}\approx100.17\)
\(\bar{y}=\frac{13 + 12+11+10+9+8}{6}=\frac{63}{6} = 10.5\)

Step2: Calculate numerator and denominator for \(r\)

\(n = 6\)
\(\sum(x_i-\bar{x})(y_i - \bar{y})=(27-100.17)(13 - 10.5)+(65 - 100.17)(12-10.5)+(83-100.17)(11 - 10.5)+(109-100.17)(10 - 10.5)+(142-100.17)(9 - 10.5)+(175-100.17)(8 - 10.5)\)
\(=(-73.17)\times2.5+(-35.17)\times1.5+(-17.17)\times0.5+(8.83)\times(-0.5)+(41.83)\times(-1.5)+(74.83)\times(-2.5)\)
\(=-182.925-52.755 - 8.585-4.415-62.745 - 187.075=-500.5\)

\(\sum(x_i-\bar{x})^2=(27 - 100.17)^2+(65 - 100.17)^2+(83-100.17)^2+(109-100.17)^2+(142-100.17)^2+(175-100.17)^2\)
\(=(-73.17)^2+(-35.17)^2+(-17.17)^2+(8.83)^2+(41.83)^2+(74.83)^2\)
\(=5352.8489+1236.9289+294.8089+77.9689+1749.7489+5600.5289 = 14312.8334\)

\(\sum(y_i-\bar{y})^2=(13 - 10.5)^2+(12 - 10.5)^2+(11 - 10.5)^2+(10 - 10.5)^2+(9 - 10.5)^2+(8 - 10.5)^2\)
\(=(2.5)^2+(1.5)^2+(0.5)^2+(-0.5)^2+(-1.5)^2+(-2.5)^2\)
\(=6.25 + 2.25+0.25 + 0.25+2.25+6.25=17.5\)

\(r=\frac{\sum(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum(x_i-\bar{x})^2\sum(y_i-\bar{y})^2}}=\frac{- 500.5}{\sqrt{14312.8334\times17.5}}\)
\(=\frac{-500.5}{\sqrt{250474.5845}}\approx\frac{-500.5}{500.47}\approx - 0.996\)

For linear regression model \(y=mx + b\), \(m=\frac{\sum(x_i-\bar{x})(y_i - \bar{y})}{\sum(x_i-\bar{x})^2}=\frac{-500.5}{14312.8334}\approx - 0.035\)
\(b=\bar{y}-m\bar{x}=10.5-(-0.035)\times100.17\approx10.5 + 3.506 = 14.006\approx14\)
So \(y=-0.035x + 14\)

Answer:

The linear regression model is \(y=-0.035x + 14\). The correlation coefficient is \(-0.996\)