Question

The CIO of an IT company would like to investigate how a software developer’s work experience (in number of years), professional certifications (number of certificates), and knowledge of various computer languages (number of programming languages) contribute to his/her work performance. The work performance is measured on the scale of 1 to 1000 so that the higher one’s score, the better his/her work performance. He collects data from 20 software developers in his company Years of experience Num of certificates Num of progLang Performance

Years of expiernce , number of certificates , number of prolong, performance

1 27 3 8 880

2 11 1 2 686

3 6 2 1 681

4 21 3 4 877

5 26 3 4 975

6 21 3 4 902

7 18 3 3 900

8 15 1 5 882

9 14 2 3 200

10 5 1 3 650

11 17 2 5 908

12 14 2 4 687

13 15 3 4 891

14 7 2 3 566

15 5 3 1 959

16 13 2 5 793

17 11 1 4 778

18 31 3 7 975

19 16 2 5 846

20 12 2 6 779.

Using the pivot table on this data what is the standard deviation for the performance of software developers who have. obtained 2 to certificates and know 3 programming languages?

a 117.4

b 67.3

c 56.9

d 258.80

we would like to create a regression model that would describe how performance depends on work experience, professional , certifications, and knowledge of varies computer languages.

what is the respond (or dependent) variable in this regression model

a performance

b employee id

c knowledge of computer languages

d work experience

e professional certifications

What is the value of r-squared of this regression model?

a .28

b .53

c .45

d .76

e .37

Answer 1

Ans : a. 0.28

> exp<-c(27,11,6,21,26,21,18,15,14,5,17,14,15,7,5,13,11,31,16,12)
> cert<-c(3,1,2,3,3,3,3,1,2,1,2,2,3,2,3,2,1,3,2,2)
> prolan<-c(8,2,1,4,4,4,3,5,3,3,5,4,4,3,1,5,4,7,5,6)
> perfo<-c(880,686,681,877,975,902,900,882,200,650,908,687,891,566,959,793,778,975,846,779)

> dmat1<-cbind(exp,cert,prolan)
> mod1<-lm(perfo~dmat1)
> mod1

Call:
lm(formula = perfo ~ dmat1)

Coefficients:
(Intercept) dmat1exp dmat1cert dmat1prolan
490.170 3.862 78.903 16.815

> summary(mod1)

Call:
lm(formula = perfo ~ dmat1)

Residuals:
Min 1Q Median 3Q Max
-552.49 -9.27 18.97 59.99 196.00

Coefficients:
Estimate Std. Error t value Pr(>|t|)   
(Intercept) 490.170 143.128 3.425 0.00347 **
dmat1exp 3.862 9.811 0.394 0.69904   
dmat1cert 78.903 66.548 1.186 0.25307   
dmat1prolan 16.815 33.554 0.501 0.62311   
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 167.9 on 16 degrees of freedom
Multiple R-squared: 0.2778,   Adjusted R-squared: 0.1424
F-statistic: 2.052 on 3 and 16 DF, p-value: 0.1471

So, we get r-squared value = 0.28