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In: Statistics and Probability

Question 3 Using R for calculations In a competitive company, the income levels of its employees...

Question 3 Using R for calculations

In a competitive company, the income levels of its employees are not standardised and are awarded on a case-by-case basis after negotiations between individual employees and the company directors. An industry body wants an analysis of employee income with respect to their relative rank in the company. A random sample of 184 individuals in the company were recruited and their relative Rank and Income values were anonymously recorded.

The data is available below. The variables are defined below.

Income - The yearly income of the employee.

Rank - The relative rank of the employee’s position at the company (1 being highest rank, 9 being lowest)

a. Fit a simple linear regression to the data with Income as the response variable and worker Rank as the predictor.

b. Using your analysis in a. or otherwise, explain why simple linear regression is inadequate to explain the structure in this dataset.

c. Fit a polynomial regression model to the data and select the best order polynomial to explain the data using the significance testing techniques discussed in lectures.

d. Predict the Income for a person planning to apply for a position at Rank 5 at one of these competitive companies in the near future

Rank	Income
7	106790
6	70916
9	70495
3	191968
6	59373
6	106390
8	31339
3	235000
3	209008
5	115081
1	510684
1	557015
8	115096
2	311281
8	83348
6	118896
1	523692
3	230699
5	127867
6	103211
5	97534
8	68099
4	72454
6	129781
2	360613
6	73465
9	93146
6	104356
7	42327
5	145520
9	55853
7	77324
3	216965
1	532028
5	120256
6	37870
7	89948
1	511271
4	193372
2	281334
6	83604
8	53887
9	64738
9	72541
4	164709
9	56205
4	181247
5	92034
4	177882
1	483163
6	97319
1	484151
1	492368
6	120574
8	52470
7	46166
5	155870
6	76479
3	218382
8	91030
3	200678
2	364445
6	78075
7	77990
1	530666
6	136092
4	132705
7	120456
6	115115
2	296011
6	64033
1	512753
3	167713
7	60436
7	61206
3	266501
4	227492
1	514100
2	384562
2	271253
1	505753
4	148516
2	338896
9	70202
2	288968
7	116571
9	92788
4	166387
8	84762
6	92757
3	243974
8	44752
2	311745
4	165152
3	216874
4	224083
6	125820
4	196454
9	21565
2	340717
9	48784
5	105917
9	25375
8	103300
6	107669
7	93197
4	154516
8	59497
8	68733
1	540871
1	590015
3	134095
8	87005
7	45888
6	73332
4	217111
9	86037
1	463367
4	202798
4	213355
4	216602
9	35764
8	65762
2	352920
2	279612
2	349812
5	166996
6	107851
5	128139
6	166045
8	47305
9	60798
8	37471
8	10184
1	528574
4	164696
9	25789
5	140320
1	499333
2	336158
6	89999
8	104567
6	143554
5	163795
1	513261
4	165280
4	161781
7	81081
9	41830
9	22884
2	338717
6	89851
6	77929
9	29934
3	205850
5	84776
5	125247
6	80336
1	591938
9	74762
9	53977
5	107757
7	60626
5	111661
4	149466
2	346352
1	534712
6	147205
2	288935
7	96857
4	164486
6	65347
9	36389
9	102282
8	53647
3	263337
6	56293
6	78559
1	550526
9	79542
8	35019
8	133983
5	161509
5	127704

Expert Solution

Use below R code :

dim(Employee_Analysis)

184 2

184 rows with 2 columns

names(Employee_Analysis)

"Rank" "Income"
reg.mod <- lm(Employee_Analysis$Income~Employee_Analysis$Rank)
summary(reg.mod)

(Intercept) Employee_Analysis$Rank
438835.04 -50353.65

Regression equation is

income=438835.04 -50353.65*Rank

b. Using your analysis in a. or otherwise, explain why simple linear regression is inadequate to explain the structure in this dataset.

R sq=0.7629

76.29% variation in Income is explained by model.

F(1.182)=585.6

p=0.0000

P<0.05 Model is significant.We can use model for prediction.

Use R code to get residual plot as:

plot(fitted(reg.mod),residuals(reg.mod))

From Residual plot we observe polynomial equation is the best fit

c. Fit a polynomial regression model to the data and select the best order polynomial to explain the data using the significance testing techniques discussed in lectures.

using excel trend line:

Income = 10031*Rank^2 - 150945*Rank + 624270
R² = 0.9312

Use below R to get polynomial eq

poly.mod <- lm(Income ~ poly(Rank, 2, raw=TRUE),data=Employee_Analysis)
summary(poly.mod)

m(formula = Income ~ poly(Rank, 2, raw = TRUE), data = Employee_Analysis)

Residuals:

Min 1Q Median 3Q Max

-127622 -22179 -132 27474 108581

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 624270.1 10936.5 57.08 <2e-16 ***

poly(Rank, 2, raw = TRUE)1 -150944.8 4910.0 -30.74 <2e-16 ***

poly(Rank, 2, raw = TRUE)2 10031.2 476.6 21.05 <2e-16 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 39260 on 181 degrees of freedom

Multiple R-squared: 0.9312, Adjusted R-squared: 0.9305

F-statistic: 1225 on 2 and 181 DF, p-value: < 2.2e-16

d. Predict the Income for a person planning to apply for a position at Rank 5 at one of these competitive companies in the near future

x=5

using polymial eq

Income = 10031*Rank^2 - 150945*Rank + 624270

Income = 10031*5^2 - 150945*5 + 624270

Income=120320

predicted income uisng polynomial equation is 120320

using linear regression we get

income=438835.04 -50353.65*Rank

income=438835.04 -50353.65*5

=187066.8

Income=187067

predicted income uisng linear equation is 187067

ENTIRE R CODE IS

dim(Employee_Analysis)
names(Employee_Analysis)
reg.mod <- lm(Employee_Analysis$Income~Employee_Analysis$Rank)
summary(reg.mod)
coefficients(reg.mod)
plot(fitted(reg.mod),residuals(reg.mod))
poly.mod <- lm(Income ~ poly(Rank, 2, raw=TRUE),data=Employee_Analysis)
summary(poly.mod)

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