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The data below show the hourly wages earned by six employees. The second column shows the number of years of work experience for each employee. The third column is a binary variable indicating whether or not the employee has completed a training program. For the following questions, you may check your work using a calculator but must show how you set up the expression you are using (i.e. show your work).
wage experience training
8 1 0
20 10 0
10 2 0
12 4 0
16 4 1
14 6 1
a) What are the mean, variance, and standard deviation of wages?
b) What is the mean, variance, and standard deviation of experience?
c) Draw a scatter plot with experience on the horizontal axis and wages on the vertical axis. Draw a straight line that best fits the data points. Based on this line, what is the approximate wage of someone with 0 years of experience?
d) What is the covariance of wages and experience? Is this consistent with the line you drew in part c)? Explain.
e) What is the correlation of wages and experience?

Respuesta :

MrRoyal

The approximate wage of someone with 0 years of experience is -4.64

a) What are the mean, variance, and standard deviation of wages?

The dataset of wages is

Wages = 8, 20, 10, 12, 16, 14

The mean is calculated as:

Mean = Sum/Count

This gives

Mean = (8+ 20+ 10+ 12+ 16+ 14)/6

Mean = 13.33

The variance is

Variance = ∑(x- mean)²/n-1

This gives

Variance = [(8 - 13.33)^2 + (20 - 13.33)^2 + (10 - 13.33)^2 + (12 - 13.33)^2 + (16 - 13.33)^2 + (14 - 13.33)^2]/5

Evaluate

Variance = 18.67

The standard deviation is

Standard deviation = √Variance

This gives

Standard deviation = √18.67

Evaluate

Standard deviation = 4.32

b) What is the mean, variance, and standard deviation of experience?

The dataset of experience is

Wages = 1, 10, 2, 4, 4, 6

The mean is calculated as:

Mean = Sum/Count

This gives

Mean = (1+ 10+ 2+ 4+ 4+ 6)/6

Mean = 4.5

The variance is

Variance = ∑(x- mean)²/n-1

This gives

Variance = [(1 - 4.5)^2 + (10 - 4.5)^2 + (2 - 4.5)^2 + (4 - 4.5)^2 + (4 - 4.5)^2 + (6 - 4.5)^2]/5

Evaluate

Variance = 10.3

The standard deviation is

Standard deviation = √Variance

This gives

Standard deviation = √10.3

Evaluate

Standard deviation = 3.21

c) Draw a scatter plot with experience on the horizontal axis and wages on the vertical axis.

See attachment for the scatter plot and the straight line that best fits the data points.

The equation of the straight line is

y = 0.69x - 4.64

The approximate wage of someone with 0 years of experience is

y = 0.69 * 0 - 4.64

Evaluate

y = -4.64

Hence, the approximate wage of someone with 0 years of experience is -4.64

d) What is the covariance of wages and experience?

This is calculated as:

Cov(x,y) = ∑(x - x mean)(y - y mean)/N - 1

So, we have:

Cov(x,y) = [(8 - 13.33) * (1 - 4.5) + (20 - 13.33)*(10 - 4.5) + (10 - 13.33)*(2 - 4.5) + (12 - 13.33)*(4 - 4.5) + (16 - 13.33)*(4 - 4.5) + (14 - 13.33)*(6 - 4.5)]/5

Evaluate

Cov(x,y) = 12.80

Hence, the covariance of wages and experience is 12.80

Is this consistent with the line you drew in part c?

Yes it is

e) What is the correlation of wages and experience?

Here, we use a graphing calculator.

From the graphing calculator, we have:

r = 0.9231

Hence, the correlation of wages and experience is 0.9231


Read more about linear regression at:

https://brainly.com/question/17844286

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