In: Math
Employees |
Age |
Salary |
Mary |
23 |
28.6 |
Frieda |
31 |
53.3 |
Alicia |
44 |
73.8 |
Tom |
22 |
26.0 |
Gillian |
25 |
34.3 |
Bob |
54 |
63.5 |
Vivian |
51 |
96.4 |
Cacil |
60 |
122.9 |
Barry |
40 |
63.8 |
Jaime |
64 |
111.1 |
Wanda |
34 |
82.5 |
Sam |
63 |
80.4 |
Saundra |
40 |
69.3 |
Pete |
31 |
52.8 |
steve |
28 |
54.0 |
Juan |
36 |
58.7 |
Dave |
58 |
72.3 |
Lee |
52 |
88.6 |
Judd |
43 |
60.2 |
Sunil |
28 |
61.0 |
Marcia |
54 |
75.8 |
Ellen |
44 |
79.8 |
Iggy |
36 |
70.2 |
The regression output is:
Find the sample regression equation and interpret the coefficients. Remember your interpretations should be in terms of the problem. (4 points)
The regression equation is:
Salary = 8.1906 + 1.4474*Age
The constant is meaningless since age cannot be zero
The coefficient age interprets that when age increases by 1, the salary will increase by 1.4474.
Find the coefficient of determination, and interpret its value. (3 points)
The coefficient of determination value is 0.663. It means that 66.3% of the variation in Salary is explained by Age.
Use residual analysis to check the validity of the model and fully explain your findings and conclusions. (6 points)
The residual output is:
Observation | Salary | Predicted | Residual |
1 | 28.60 | 41.48 | -12.88 |
2 | 53.30 | 53.06 | 0.24 |
3 | 73.80 | 71.87 | 1.93 |
4 | 26.00 | 40.03 | -14.03 |
5 | 34.30 | 44.37 | -10.07 |
6 | 63.50 | 86.35 | -22.85 |
7 | 96.40 | 82.01 | 14.39 |
8 | 122.90 | 95.03 | 27.87 |
9 | 63.80 | 66.09 | -2.29 |
10 | 111.10 | 100.82 | 10.28 |
11 | 82.50 | 57.40 | 25.10 |
12 | 80.40 | 99.37 | -18.97 |
13 | 69.30 | 66.09 | 3.21 |
14 | 52.80 | 53.06 | -0.26 |
15 | 54.00 | 48.72 | 5.28 |
16 | 58.70 | 60.30 | -1.60 |
17 | 72.30 | 92.14 | -19.84 |
18 | 88.60 | 83.45 | 5.15 |
19 | 60.20 | 70.43 | -10.23 |
20 | 61.00 | 48.72 | 12.28 |
21 | 75.80 | 86.35 | -10.55 |
22 | 79.80 | 71.87 | 7.93 |
23 | 70.20 | 60.30 | 9.90 |
The residual plot is:
From the residual plot, it is clear that the data is linearly related and there is a linear relationship between Age and Salary.
Estimate with 95% confidence the average employee salary for all employees that are 35 years old. Predict with 95% confidence the estimated salary for an individual employee that is 35 years old. Write at least one sentence using your confidence interval and at least one sentence using your prediction interval. (6 points)
Predicted values for: Salary | ||||||
95% Confidence Interval | 95% Prediction Interval | |||||
Age | Predicted | lower | upper | lower | upper | Leverage |
35 | 58.8483 | 52.0302 | 65.6665 | 29.1281 | 88.5685 | 0.056 |
The 95% confidence interval of the average employee salary for all employees that are 35 years old is between 52.0302 and 65.6665. It includes the predicted value of the average employee salary for all employees that are 35 years old.
The 95% prediction interval of the average employee salary for all employees that are 35 years old is between 29.1281 and 88.5685. It includes the predicted value of the average employee salary for all employees that are 35 years old.
Verify that the p-value for the F is the same as the slope’s t statistic’s p-value, and show that t2 = F. (3 points)
The P-value for the slope is the same for the F and t statistic which is 0.000.
t = 6.42
t2 = 6.422 = 41.25 which is equal to F.
All Minitab outputs are attached.