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3. The table below shows the regression output of a multiple regression model relating the beginning salaries of employees in

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Answer #1

The regression model that is to be estimated is

salary 30 + β| Sex + β:Education + β3Experience + β4Months E

where \epsilon\stackrel{iid}\sim \mathcal{N}(0,\sigma^2) is a random disturbance

the estimated regression equation is

salary = 3572.6+724·75ex+86.29Education+1.269Experience+23.537Months

a) Using this

Df 4 23,763,297 5,940,824 23.35 8822,387,82 254,407 92 46.151.118 Source MS F-value Regression Residual Total

We know

k=4 is the number of independent variables

Sum of square Error (residuals), SSE = 22,387,821

degrees of freedom residuals = n-k-1 = 88

Sum of square Total (residuals), SST = 46,151,118

degrees of freedom total is n-1 = 92

The adjusted R-square is

SS7.1 SSEn-1 22.387.821 92 46, 151.118 88 × 0.4929

ans: The adjusted R-square is 0.4929

b) We want to test the following hypotheses

Ho : β1-Ba = β3-За 0 ← null hypothesis: No independent variable can explain variation in salary. Ha : at least one of ß1 2 0The test statistics has F distribution. The test statistics is given below

Source Regressiorn Residual Total Df 4 23,763,297 5,940,82423.35 88 22,387,821 254,407 92 46,151,118 MS F-value

The test statistics is F=23.35 with numerator df=4 and denominator df=88

Using F table for alpha=0.05 and numerator df=4 and denominator df=120 (The closest we can get to 88) we get the critical value of F = 2.45

We will reject the null hypothesis if the test statistics is greater than the critical value.

Here the test statistics is 23.35 and it is greater than the critical value, 2.45. Hence we reject the null hypothesis.

We conclude that the overall model is significant

c) There would be positive linear relationship between salary and experience if the slope coefficient \begin{align*} \beta_3 \end{align*} of Experience in the regression model is greater than 0.

That is we want to test the following hypotheses

Ho : β3 < 0 ← null hypothesis: Salary and Experience do not have a positive liner relationship Ha : > 0 ← alternative hypothe

This is a right tailed test (the alternative hypothesis has ">").

The hypothesized value of \begin{align*} \beta_3 \end{align*} (from the null hypothesis) is 33Ho

Using the following

Coefficient Standard error Variable Constant Sex Education Experience Months 3572.6 724.7 86.29 1.269 23.537 326.5 120.3 26.8

the test statistics is

\begin{align*}t=\frac{\hat{\beta}_3-\beta_{3H_0}}{s.e(\hat{\beta}_{3})} =\frac{1.269-0}{0.5877}=2.16\end{align*}

The degrees of freedom for t statistics is n-k-1=88

The p-value given in the output is 0.034 is for a 2 tailed test. For one tailed test the p-value is half of that, that is p-value=0.034/2=0.017

We will reject the null hypothesis if the p-value is less than level of significance

Here the p-value is 0.017 and it is less than 0.05 level of significance. Hence we reject the null hypothesis.

We conclude that there is a positive linear relationship between Salary and Experience, after accounting for the effect of the variables, Sex, Education, and Months

d) the predicted salary for Sex=1 (man), Education = 15, Experience = 20 , months=10 is

salary = 3572.6+724.7 × 1 86.29 × 15+ 1.269 × 20+23.537 × 10 = 5852.4

ans: The salary for a man with 15 years of education, 20 months of experience and 10 months with in the company is $5,852.4

(Important: In the question pasted, the unit of the salary is not given. Please express the figure 5,852.4 accordingly)

e) To know if there is an interaction between Sex and Experience we will modify the model as below

salary, Bo + β1 Sex + Education β3Experience Зам。nths + β5 (Sex × Experience) + e

If the interaction is significant it means that the coefficient 35メ0

The salary model for a man is (by setting Sex=1)

\begin{align*} \widehat{\text{salary}}_{\text{man}}&= \hat{\beta}_0+\hat{\beta}_1\times 1+\hat{\beta}_2\text{Education}+\hat{\beta}_3\text{Experience}+\beta_4\text{Months}+\hat{\beta}_5(1\times \text{Experience})\\ &=(\hat{\beta}_0+\hat{\beta}_1)+\hat{\beta}_2\text{Education}+(\hat{\beta}_3+\hat{\beta}_5)\text{Experience}+\beta_4\text{Months} \end{align*}

The salary model for a woman (by setting sex=0) is

\begin{align*} \widehat{\text{salary}}_{\text{woman}}&= \hat{\beta}_0+\hat{\beta}_1\times 0+\hat{\beta}_2\text{Education}+\hat{\beta}_3\text{Experience}+\beta_4\text{Months}+\hat{\beta}_5(0\times \text{Experience})\\ &=\hat{\beta}_0+\hat{\beta}_2\text{Education}+\hat{\beta}_3\text{Experience}+\beta_4\text{Months} \end{align*}

It means that the predicted salary changes by \begin{align*} (\hat{\beta}_3+\hat{\beta}_5) \end{align*} for one month increase in the experience for a man compared to  \begin{align*} \hat{\beta}_3 \end{align*} for a woman (while keeping other variables the same)

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