Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Mike Wilde is president of the teachers\' union for Otsego School District. In p

ID: 3219776 • Letter: M

Question

Mike Wilde is president of the teachers' union for Otsego School District. In preparing for upcoming negotiations, he would like to investigate the salary structure of classroom teachers in the district. He believes there are three factors that affect a teacher's salary: years of experience, a rating of teaching effectiveness given by the principal, and whether the teacher has a master's degree. A random sample of 20 teachers resulted in the following data:

A. Develop a correlation matrix. Which independent variable has the strongest correlation with the dependent variable? Does it appear there will be any problems with multicollinearity?

B. Determine the regression equation. What salary would you estimate for a teacher with five years' experience, a rating by the principal of 60, and no master's degree?

C. Conduct a global test of hypothesis to determine whether any of the net regression coefficients differ from zero. Use the 0.05 significance level.

D. Conduct a test of hypothesis for the individual regression coefficients. Would you consider deleting any of the independent variables? Use the 0.05 significance level.

E. If your conclusion in part (d) was to delete one or more independent variables, run the regression equation again without those variables.

Salary Years of Principal's Master's ($000), Experience. Rating, Degree." $55.1 35 57.6 43 51 60 14 80 54 55 55.7 49.7 30 546 17 62.4 68 57.6 14 25 54.7 62 80 1 = yes, 0 = no. S* ex:1 0 0 1 1 0 1 0 1 1 1 0 0 1 0 1 0 1 0 1 0 sg ae ND ag pn 253103064500 46850202 ctX-34567875593487629687 na e-| 8 5 2 5 1 4 97231537 48428 16306008677687468788 aOY 372960259450275466 es 56666 754 776 56 ye S$

Explanation / Answer

(A)

Experiance and Rating seems to have a high correlation with the dependent variable Salary as they have a corelation value >0.5 i.e. 0.86 and 0.54 respectievely. There does not appear to be a problem of multicollinearity as there does not appear to be a strong correlation among any of the independent variables.

(B)

The regression equation is given by,

Salary = 43.91519 + 0.89938*Experience + 0.15392*Rating - 0.66731*Masters Degree

Therefore Salary estimate of a teacher with 5 years experience, a principals rating of 60 and no masters degree is given by

Salary = 43.91519 + 0.89938*5 + 0.15392*60 - 0.66731*0

= 57.64729

(C)

We test the hypothesis, H0 = it does not differ from zero

H1 = it differ from zero

the table value of t for n-2 = 18 d.f with significance level of 0.05 is 1.734

now t is calculated as

t = (estimated coeff - hypothesied coefficient)/standard error

If the calculated tvalue is greater that the table value, we reject H0.

(i) Experience:

t = (0.89938 - 0)/0.08768 = 10.25753

(ii) Principals rating

t = (0.15392 - 0)/0.03144 = 4.895674

(iii) Masters Degree

t = (-0.66731 - 0)/1.21393 = -0.5497104

Thus except for the Masters degree rest all the coefficients differ from zero.

(D)

On fitting the regression model, it is observed that the p value of Experience and Principlas rationg is very small (approximately equal to zero) and hence are significant. But the p-value of Masters degree is 0.590111. Since the p-value is >0.05, we reject our Null Hypothesis that the variable is significant and conclude that the Masters Degree is not a significant variable.

(E)

Fitting the regression equation by removing the Masters Degree shows that there is not much change in the R2 values, in fact the adjusted R2 values have only increased. The fitted model is given by,

Salary = 44.11570 + 0.89265*Experience + 0.14638*Rating

as you can see, there is also not much difference in the coefficients of the fitted model even after removing the Masters Degree.

The rcode for the same is given below:

# reading data
data1 = read.csv("C:\Users\Admin\Desktop\data1.csv")
summary(data1)
cor(data1)

#model 1
model1 = lm(Salary ~ ., data=data1)
summary(model1)

model2 = lm(Salary ~ Exp + Rating, data=data1)
summary(model2)