Answer:
Solved below.
Step-by-step explanation:
The data is provided for the starting salary (in $1,000) at some company and years of prior working experience in the same field for randomly selected 10 employees.
(a)
The formula to compute the correlation coefficient is:
[tex]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]}} \\[/tex]
The required values are computed in the Excel sheet below.
[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{ 10 \cdot 7252 - 97 \cdot 661 } {\sqrt{\left[ 10 \cdot 1335 - 97^2 \right] \cdot \left[ 10 \cdot 45537 - 661^2 \right] }} \approx 0.9855\end{aligned}[/tex]
Thus, the sample correlation coefficient r is 0.9855.
(b)
The slope of the regression line is:
[tex]b_{1} &= \frac{ n \cdot \sum{XY} - \sum{X} \cdot \sum{Y}}{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} \\\\= \frac{ 10 \cdot 7252 - 97 \cdot 661 }{ 10 \cdot 1335 - \left( 97 \right)^2} \\\\\approx 2.132[/tex]
Thus, the slope of the regression line is 2.132.
(c)
The y-intercept of the line is:
[tex]b_{0} &= \frac{\sum{Y} \cdot \sum{X^2} - \sum{X} \cdot \sum{XY} }{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} \\\\= \frac{ 661 \cdot 1335 - 97 \cdot 7252}{ 10 \cdot 1335 - 97^2} \\\\\approx 45.418[/tex]
Thus, the y-intercept of the line is 45.418.
(d)
The equation of the sample regression line is:
[tex]y=45.418+2.132x[/tex]
(e)
Compute the predicted starting salary for a person who spent 15 years working in the same field as follows:
[tex]y=45.418+2.132x\\\\=45.418+(2.132\times15)\\\\=45.418+31.98\\\\=77.398\\\\\approx 77.4[/tex]
Thus, the predicted starting salary for a person who spent 15 years working in the same field is $77.4 K.
Answer:
Yes correct
Step-by-step explanation:
I think this is correct becase: 2 50
5 55
7 62
etc
these are all correct
