Question

In: Statistics and Probability

The HR Department of Vanderizing Bulb Company wanted to know if among its salaried employees there...

The HR Department of Vanderizing Bulb Company wanted to know if among its salaried employees there was a relationship between years of school and annual compensation. A random sample of employees yielded the following data:

Employee Years of School Compensation ($)
1 20 69,582
2 15 55,433
3 13 64,678
4 14 74,465
5 17 70,159
6 16 62,487
7 17 69,763
8 18 71,125
9 20 64,350
10 14 53,290

a. Find the correlation coefficient for the relationship between Years of School and Compensation.

b. Find the best equation which should be used to predict compensation on the basis of years of schooling. Justify by a test of a hypothesis.

c. Maggie Radford has 15 years of schooling. What is her predicted salary if she works at Vanderizing?

Solutions

Expert Solution

  1. Correlation coefficient for the relationship between Years of School and Compensation is calculated by the formula r(x,y)= cov(x,y) / (std(x) * std(y)
Cov(years , compensation)= 5643.356     standard deviation(years) = 2.458545
 
 std(compensation)= 6912.532
 
r(x,y)= cov(x,y) / [(std(x) * std(y)] = 5643.356 / (6912.532 *  2.458545) = 0.3320642
 
therefore the Correlation coefficient for the relationship between Years of School and Compensation is is 0.3320642.


    

  1. the best equation which should be used to predict compensation
on the basis of years of schooling is
    2. 



Y^= 50221.4 + 933.6 * Years.


H0:  B1=0


H1: B1=/ 0.


As p value is 0.349. P value > 0.05.Thus, the years of
schooling is not significant with compensation. We accept H0


    

  1.   The regression equation is Y^= 50221.4 + 933.6 *
Years. Need to calculate compensation for 15 years .It is done by
:
    2. 



Y^= 50221.4 + 933.6 * Years = 50221.4 + 933.6 * 15 =
64226.1.


Hence,


The compensation for 15 years of schooling is 64226.1.


The Values can be checked by R codes .


View(dat1)


> attach(dat1)


 


 


> cov(years,compensation)


[1] 5643.356


 


> sqrt(var(years))


[1] 2.458545


 


> sqrt(var(compensation))


[1] 6912.532


 


 


> cor(years,compensation)


[1] 0.3320642


 


> model <- lm(compensation~years)


> model


 


Call:


lm(formula = compensation ~ years)


 


Coefficients:


(Intercept)        years  


    50221.4        933.6  


 


> predict(model,data.frame(years=15))


      1 


64226.1

                    

                                        
                    


        

            


         

             
orchestra answered 1 year ago

         

    

        

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