Answer:
The sample correlation coefficient is, r = 0.8722.
The equation of the least-squares line is:
[tex]y= -0.161+0.154x[/tex]
Step-by-step explanation:
(a)
The scatter diagram displaying the data for X : total number of jobs in a given neighborhood and Y : number of entry-level jobs in the same neighborhood is shown below.
(b)
The table attached below verifies the values of [tex]\sum X,\ \sum Y,\ \sum X^{2},\ \sum Y^{2}\ \text{and}\ \sum XY[/tex].
The sample correlation coefficient is:
[tex]\begin{aligned}r~&=~\frac{n\cdot\sum{XY} - \sum{X}\cdot\sum{Y}} {\sqrt{\left[n \sum{X^2}-\left(\sum{X}\right)^2\right] \cdot \left[n \sum{Y^2}-\left(\sum{Y}\right)^2\right]}} \\r~&=~\frac{ 6 \cdot 1163 - 201 \cdot 30 } {\sqrt{\left[ 6 \cdot 7759 - 201^2 \right] \cdot \left[ 6 \cdot 182 - 30^2 \right] }} \approx 0.8722\end{aligned}[/tex]
Thus, the sample correlation coefficient is, r = 0.8722.
(c)
The slope and intercept are:
[tex]\begin{aligned} a &= \frac{\sum{Y} \cdot \sum{X^2} - \sum{X} \cdot \sum{XY} }{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} = \frac{ 30 \cdot 7759 - 201 \cdot 1163}{ 6 \cdot 7759 - 201^2} \approx -0.161 \\ \\b &= \frac{ n \cdot \sum{XY} - \sum{X} \cdot \sum{Y}}{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} = \frac{ 6 \cdot 1163 - 201 \cdot 30 }{ 6 \cdot 7759 - \left( 201 \right)^2} \approx 0.154\end{aligned}[/tex]
The equation of the least-squares line is:
[tex]y= -0.161+0.154x[/tex]
(d)
The least-squares line is graphed in the diagram below.
