Answer:
Solved below.
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:
![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]}} \\](https://img.qammunity.org/2021/formulas/mathematics/college/hrpu3hx4fqrqlrinwcj7fqol6sgqq4rdr3.png)
The required values are computed in the Excel sheet below.
![\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 } {√(\left[ 10 \cdot 1335 - 97^2 \right] \cdot \left[ 10 \cdot 45537 - 661^2 \right] )} \approx 0.9855\end{aligned}](https://img.qammunity.org/2021/formulas/mathematics/college/b2ko3f6zvdvqszp1fve474sg009i8flex4.png)
Thus, the sample correlation coefficient r is 0.9855.
(b)
The slope of the regression line is:

Thus, the slope of the regression line is 2.132.
(c)
The y-intercept of the line is:

Thus, the y-intercept of the line is 45.418.
(d)
The equation of the sample regression line is:

(e)
Compute the predicted starting salary for a person who spent 15 years working in the same field as follows:

Thus, the predicted starting salary for a person who spent 15 years working in the same field is $77.4 K.