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vova2212 [387]
3 years ago
7

An economist was interested in modeling the relation among annual income, level of education, and work experience. The following

data was obtained from a random sample of 12 individuals. Level of education is the number of years of education beyond eighth grade, so 1 represents completing 1 year of high school, 8 means completing 4 years of college, and so on. Work experience is the number of years employed in the current profession. Annual income is measured in thousands of dollars.
Work Experience Level of Education Annual Income ($ Thousands)
21 6 34.7
14 3 17.9
4 8 22.7
16 8 63.1
12 4 33.0
20 4 41.4
25 1 20.7
8 3 14.6
24 12 97.3
28 9 72.1
4 11 49.1
15 4 52.0

A) Construct a correlation matrix between work experience, level of education, and annual income. Is there any reason to be concerened with multicollinearity based on the correlation matrix?

Mathematics
1 answer:
professor190 [17]3 years ago
4 0

Answer:

Check the explanation

Step-by-step explanation:

A) We use Minitab to solve the question.

The correlation matrix is,

Correlation: Level of Education, Work Experience, Annual Income ($ ,000s)

Work Experience Level of Education

Level of Education -0.042 0.463

Annual Income ($ 0.463 0.756

From correlation matrix there is no any reason to concern multicoinearity.

B)

The Regression Analysis: Annual Income ($ Thousands) vs Work Experience and Level of Education

Analysis of Variance

Source D F Adj SS Adj MS F-Value P-Value

Regression 2 5577 2788.4 19.96 0.000

Work Experience 1 1675 1674.7 11.99 0.007

Level of Education 1 4114 4114.1 29.44 0.000

Error 9 1257 139.7

Total 11 6834

Model Summary

S R-sq R-sq(adj) R-sq(pred)

11.8204 81.60% 77.51% 69.81%

Coefficients

Term Coef SE Coef T-Value P-Value VIF

Constant -15.2 10.2 -1.49 0.171

Work Experience 1.545 0.446 3.46 0.007 1.00

Level of Education 5.57 1.03 5.43 0.000 1.00

E)

The value of R-sq is 81.60% & R-sq (adj) is 77.51% indicates that adequacy of the fitted model is good.

G & H )

Coefficients

Term Coef SE Coef T-Value P-Value

Constant -15.2 10.2 -1.49 0.171 > 0.05 (significant)

Work Experience 1.545 0.446 3.46 0.007 > 0.05 (significant) ______Reject H0 B1 = 0

Level of Education 5.57 1.03 5.43 0.000 < 0.05 (Not Significant) ______do not Reject H0 B2 = 0

H)

Regression Equation

Annual Income ($ Thousands) = -15.2 + 1.545 Work Experience + 5.57 Level of Education

= -15.2 + 1.545 * 12 + 5.57 * 4

= 25.62

I & J)

Regression Equation

Annual Income ($ Thousands) = -15.2 + 1.545 Work Experience + 5.57 Level of Education

Variable Setting

Work Experience 12

Level of Education 4

Predicted Income is 25.5649 thousands:

Kindly check the attached image below.

Fit SE Fit 95% CI 95% PI

25.5649 4.42545 (15.5538, 35.5759) (-2.98733, 54.1170)

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viva [34]

Answer:

79. [7, 12], [1, 6]

80. The total revenue in the month of May is $11,575

81. The total revenue in the month of November is $4,630

82. Values obtained from the model and actual values are almost the same, so the given model is precise.

Step-by-step explanation:

79.

The domain of a function is the set of x-values that we are allowed to plug into our function.

The graph below represents f(x),

the graph that the domain of the first part of the function is [7,12]. The domain of the second part of the function is [1,6]. The solution is:

f(x)=\left \{ {{-1.97x+26.3, 7 \leq x \leq 12} \atop {0.505x^{2} -1.47x+6.3, 1 \leq x \leq 6}} \right. \\\\

80.

1 ≤ 5 ≤ 6, so we'll plug the value 5 in the function f(x) = 0.505x² - 1.47x + 6.3

f(x) = y = (0.505)(5)² - 1.47(5) + 6.3

f(x) = y = (0.505)(25) - 7.35 + 6.3

f(x) = y = 12.625 - 1.05

f(x) = y = 11.575

For x=5 which represents 5th month May, we got y=11.575

Therefore, the total revenue in the month of May is 11.575 thousands dollars, ie 11,575 dollars.

81.

7 ≤ 11 ≤ 12, so we'll plug the value 11 in the function f(x) = -1.97x + 26.3

f(x) = − 1.97(11) + 26.3

     = − 21.67 + 26.3

     = 4.630.

For x=11 which represents 11th month November, we got y=4.630 Therefore, the total revenue in the month of November is 4.630 thousands dollars, 4,630 dollars.

82.

Values obtained from the model are y=11.575 (x=5) and y=4.630 (x=11). Actual data values are y=11.5 (x=5) and y=4.4 (x=11)

By comparing these values, we can conclude that the model is precise, because these values are almost the same.

Learn more about Piecewise Functions here: brainly.com/question/13882670

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