Data was collected from 40 employees to develop a regression model to predict th
ID: 3295526 • Letter: D
Question
Data was collected from 40 employees to develop a regression model to predict the employee’s annual salary using their years with the company (Years), their starting salary (Starting), and their Gender (Male = 0, Female = 1). The results from Excel regression analysis are shown below:
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.718714957
R Square
0.516551189
Standard Error
10615.63461
Observations
40
ANOVA
df
SS
MS
F
Significance F
Regression
3
4334682510
1444894170
12.82165585
7.48476E-06
Residual
36
4056901131
112691698.1
Total
39
8391583641
Coefficients
Standard Error
t Stat
P-value
Intercept
27946.57894
4832.438706
5.783121245
1.35464E-06
Years
1665.251558
425.0829092
3.917474737
0.000383313
Starting
0.266374185
0.12610443
2.112330112
0.041661598
Gender
-3285.541043
5617.145392
-0.584912943
0.56225464
a. What is the regression equation?
b. In testing the null hypothesis that the regression equation is not significant at the 0.05 level, what is the appropriate conclusion?
c. In testing the significance of the partial regression coefficient associated with the Years variable at the 0.05 significance level, what is the appropriate conclusion?
d. In testing the significance of the partial regression coefficient associated with the Starting variable at the 0.05 significance level, what is the appropriate conclusion?
e. In testing the significance of the partial regression coefficient associated with the Gender variable at the 0.05 significance level, what is the appropriate conclusion?
f. For a male employee with 5 years of experience and a starting salary of $30,000, what is the approximate 95% confidence interval for his annual salary?
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.718714957
R Square
0.516551189
Standard Error
10615.63461
Observations
40
ANOVA
df
SS
MS
F
Significance F
Regression
3
4334682510
1444894170
12.82165585
7.48476E-06
Residual
36
4056901131
112691698.1
Total
39
8391583641
Coefficients
Standard Error
t Stat
P-value
Intercept
27946.57894
4832.438706
5.783121245
1.35464E-06
Years
1665.251558
425.0829092
3.917474737
0.000383313
Starting
0.266374185
0.12610443
2.112330112
0.041661598
Gender
-3285.541043
5617.145392
-0.584912943
0.56225464
Explanation / Answer
a) The regression equation is Y = 27946.57894 + 1665.251558Years + 0.266374185 Starting - 3285.541043 Gender
b) P-value = 7.48476E-06 < alpha 0.05, so we reject H0
Thus we conclude that the regression equation is significant at the 0.05 level
i.e. regression equation is best fit to the given data
c)
P-value =0.000383313 < alha 0.05, so we reject H0
Thus we conclude that the significance of the partial regression coefficient associated with the Years variable at the 0.05 significance level
d)
P-value = 0.041661598 < alpha 0.05,so we reject H0
Thus we conclude that the significance of the partial regression coefficient associated with the Starting variable at the 0.05 significance level
e) P-value = 0.56225464 > alpha 0.05, we accept H0
the partial regression coefficient no associated with the Gender variable at the 0.05 significance level
f) The predict value of salary is
salary = 27946.57894 + 1665.251558(5) + 0.266374185 (30,000) - 3285.541043 (0)
= 44264.06228
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