Question

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? what is the strength of the model? miles driven: 27, 65, 83, 109, 142, 175. gallons in tank: 13, 12, 11, 10, 9, 8.

Explanation:

Step1: Denote miles driven as \(x\) and gallons in tank as \(y\). We have data points \((x_1,y_1),(x_2,y_2),\cdots,(x_6,y_6)\) where \(x = [27,65,83,109,142,175]\) and \(y=[13,12,11,10,9,8]\).

Step2: Calculate the means of \(x\) and \(y\).

The mean of \(x\), \(\bar{x}=\frac{27 + 65+83+109+142+175}{6}=\frac{501}{6}=83.5\)
The mean of \(y\), \(\bar{y}=\frac{13 + 12+11+10+9+8}{6}=\frac{63}{6}=10.5\)

Step3: Calculate the numerator and denominator for the correlation - coefficient \(r\).

The numerator \(S_{xy}=\sum_{i = 1}^{6}(x_i-\bar{x})(y_i - \bar{y})\)
\((27-83.5)(13 - 10.5)=(- 56.5)\times2.5=-141.25\)
\((65 - 83.5)(12-10.5)=(-18.5)\times1.5=-27.75\)
\((83 - 83.5)(11 - 10.5)=(-0.5)\times0.5=-0.25\)
\((109 - 83.5)(10 - 10.5)=(25.5)\times(-0.5)=-12.75\)
\((142 - 83.5)(9 - 10.5)=(58.5)\times(-1.5)=-87.75\)
\((175 - 83.5)(8 - 10.5)=(91.5)\times(-2.5)=-228.75\)
\(S_{xy}=-141.25-27.75 - 0.25-12.75-87.75-228.75=-498.5\)

The denominator:
\(S_{xx}=\sum_{i = 1}^{6}(x_i-\bar{x})^2\)
\((27-83.5)^2=(-56.5)^2 = 3192.25\)
\((65 - 83.5)^2=(-18.5)^2=342.25\)
\((83 - 83.5)^2=(-0.5)^2 = 0.25\)
\((109 - 83.5)^2=(25.5)^2=650.25\)
\((142 - 83.5)^2=(58.5)^2=3422.25\)
\((175 - 83.5)^2=(91.5)^2=8372.25\)
\(S_{xx}=3192.25+342.25 + 0.25+650.25+3422.25+8372.25=15979.5\)

\(S_{yy}=\sum_{i = 1}^{6}(y_i-\bar{y})^2\)
\((13 - 10.5)^2=(2.5)^2 = 6.25\)
\((12 - 10.5)^2=(1.5)^2=2.25\)
\((11 - 10.5)^2=(0.5)^2 = 0.25\)
\((10 - 10.5)^2=(-0.5)^2=0.25\)
\((9 - 10.5)^2=(-1.5)^2=2.25\)
\((8 - 10.5)^2=(-2.5)^2=6.25\)
\(S_{yy}=6.25+2.25 + 0.25+0.25+2.25+6.25=17.5\)

\(r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{-498.5}{\sqrt{15979.5\times17.5}}=\frac{-498.5}{\sqrt{279641.25}}\approx\frac{-498.5}{528.81}\approx - 0.94\)
Since \(|r|\approx0.94\) which is close to \(1\), the strength of the model is strong.

For the linear - regression model \(y=mx + b\), \(m=\frac{S_{xy}}{S_{xx}}=\frac{-498.5}{15979.5}\approx - 0.0312\)
\(b=\bar{y}-m\bar{x}=10.5-(-0.0312)\times83.5=10.5 + 2.6152=13.1152\)
The linear - regression model is \(y=-0.0312x + 13.1152\)

Answer:

The correlation coefficient is approximately \(-0.94\). The strength of the model is strong. The linear - regression model is \(y=-0.0312x + 13.1152\)