Question

Stat 3309 - Statistical Analysis for Business Applications I

Consider the following data representing the starting salary (in $1,000) at some company and years of prior working experience in the same ï¬eld. The sample of 10 employees was taken and the following data is reported.

Years of experience

Starting Salary (in $1,000)
0
45

2 50
5 55
7 62
8 63
10 70
12 68
15 75
18 81
20 92
Part 1: Use the formulas provided on the 3rd formula sheet to compute the following quantities. Open an Excel spreadsheet and write the table with data given above. Add columns for x2, y2, and xy, as well as the last row for Σ. For each of the following quantities, write the formula for it in a cell and evaluate it.

(a) Find the sample correlation coeï¬cient r.

(b) Find the slope b1 of the sample regression line.

(c) Find the y-intercept b0 of the sample regression line.

(d) What is the equation of the sample regression line?

(e) Find the predicted starting salary for a person who spent 15 years working in the same ï¬eld.

(f) Find the observed starting salary for a person who spent 15 years working in the same ï¬eld.

(g) What is the diï¬erence between the observed and the predicted starting salary for a person who spent 15 years working in the same ï¬eld?

(h) Find the total sum of squares SST.

(i) Find the sum of squares error SSE.

(j) Find the sum of squares regression SSR.

(k) Use the answers from (h)-(j) to conï¬rm that SST = SSR + SSE. (l) Find the coeï¬cient of determination R2.

(m) Use your answers for (a), (b) and (l), to conï¬rm that r = ±âR2.

(n) What proportion of variation is explained using the regression model?

(o) Find the standard error of the estimate se.

(p) Find the standard error of the regression slope sb.

(q) Does the number of years of prior working experience in the same ï¬eld aï¬ect the starting salary at this company ? Use the sample provided above and the signiï¬cance level of 0.05.

(hint: perform the hypothesis test for H0 : β1 = 0 vs. H1 : β1 6= 0.)

Part 2: Find and use Excel built-in-functions to check your answers for r, b1, and b0. Next to each cell from Part 1, calculate these three quantities using Excel built-in-functions and conï¬rm your answers from Part 1.

(hint: for example, for r the Excel built-in function is "CORREL")

Part 3: Bellow your answers from Parts 1 and 2, perform the regression analysis using Excel built-in-module which can be found under "DATA" â "Data Analysis" â "Regression" and double check your answers from Part 1. Draw the scatter plot of the data and, by visually observing the graph, determine if there is a linear relationship between the number of years of prior working experience in the same ï¬eld and the starting salary at this company.

Answer 1

Yes correct

Explanation:

I think this is correct becase: 2 50

5 55

7 62

etc

these are all correct

Answer 2

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]}} \\`

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}`

Thus, the sample correlation coefficient r is 0.9855.

(b)

The slope of the regression line is:

`b_(1) &= \frac{ n \cdot \sum{XY} - \sum{X} \cdot \sum{Y}}{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} \\\\= ( 10 \cdot 7252 - 97 \cdot 661 )/( 10 \cdot 1335 - \left( 97 \right)^2) \\\\\approx 2.132`

Thus, the slope of the regression line is 2.132.

(c)

The y-intercept of the line is:

`b_(0) &= \frac{\sum{Y} \cdot \sum{X^2} - \sum{X} \cdot \sum{XY} }{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} \\\\= ( 661 \cdot 1335 - 97 \cdot 7252)/( 10 \cdot 1335 - 97^2) \\\\\approx 45.418`

Thus, the y-intercept of the line is 45.418.

(d)

The equation of the sample regression line is:

`y=45.418+2.132x`

(e)

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

`y=45.418+2.132x\\\\=45.418+(2.132*15)\\\\=45.418+31.98\\\\=77.398\\\\\approx 77.4`

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