Question

The following data set is obtained by a randomly selected sample of 93 employees working at a bank.

SALARY	EDUC	EXPER	TIME
39000	12	0	1
40200	10	44	7
42900	12	5	30
43800	8	6	7
43800	8	8	6
43800	12	0	7
43800	12	0	10
43800	12	5	6
44400	15	75	2
45000	8	52	3
45000	12	8	19
46200	12	52	3
48000	8	70	20
48000	12	6	23
48000	12	11	12
48000	12	11	17
48000	12	63	22
48000	12	144	24
48000	12	163	12
48000	12	228	26
48000	12	381	1
48000	16	214	15
49800	8	318	25
51000	8	96	33
51000	12	36	15
51000	12	59	14
51000	15	115	1
51000	15	165	4
51000	16	123	12
51600	12	18	12
52200	8	102	29
52200	12	127	29
52800	8	90	11
52800	8	190	1
52800	12	107	11
54000	8	173	34
54000	8	228	33
54000	12	26	11
54000	12	36	33
54000	12	38	22
54000	12	82	29
54000	12	169	27
54000	12	244	1
54000	15	24	13
54000	15	49	27
54000	15	51	21
54000	15	122	33
55200	12	97	17
55200	12	196	32
55800	12	133	30
56400	12	55	9
57000	12	90	23
57000	12	117	25
57000	15	51	17
57000	15	61	11
57000	15	241	34
60000	12	121	30
60000	15	79	13
61200	12	209	21
63000	12	87	33
63000	15	231	15
46200	12	12	22
50400	15	14	3
51000	12	180	15
51000	12	315	2
52200	12	29	14
54000	12	7	21
54000	12	38	11
54000	12	113	3
54000	15	18	8
54000	15	359	11
57000	15	36	5
60000	8	320	21
60000	12	24	2
60000	12	32	17
60000	12	49	8
60000	12	56	33
60000	12	252	11
60000	12	272	19
60000	15	25	13
60000	15	36	32
60000	15	56	12
60000	15	64	33
60000	15	108	16
60000	16	46	3
63000	15	72	17
66000	15	64	16
66000	15	84	33
66000	15	216	16
68400	15	42	7
69000	12	175	10
69000	15	132	24
81000	16	55	33

This data set was obtained by collecting information on a randomly selected sample of 93 employees working at a bank.

SALARY-  starting annual salary at the time of hire

EDUC  -  number of years of schooling at the time of the hire

EXPER -  number of months of previous work experience at the time of hire

TIME   -  number of months that the employee has been working at the bank until now

2. Use the least squares method to fit a simple linear model that relates the salary (dependent variable) toeducation (independent variable).

a)  What is your model? State the hypothesis that is to be tested, the decision rule, the test statistic, and your decision, usinga level of significance of 5%.

b)  What percentage of the variation in salary has been explained by the regression?

c) Provide a 95% confidence interval estimate for the true slope value.

d) Based on your model, what is the expected salary of a new hire with 12 years of education

e ) What is the 95% prediction interval for the salary of a new hire with 12 years of education? Use the fact that the distance value = 0.011286

Please explain clearly.

Answer 1

Sol:

Perform in R studio

use lm function in R to fit a linear model of salary on educ

Use sumamry function to get the coeffcient and p value

Predict function to get the confidence and predicttion interval for newdata=12 educ

Rcode:

df1 =read.table(header = TRUE, text ="
SALARY   EDUC   EXPER   TIME
39000   12   0   1
40200   10   44   7
42900   12   5   30
43800   8   6   7
43800   8   8   6
43800   12   0   7
43800   12   0   10
43800   12   5   6
44400   15   75   2
45000   8   52   3
45000   12   8   19
46200   12   52   3
48000   8   70   20
48000   12   6   23
48000   12   11   12
48000   12   11   17
48000   12   63   22
48000   12   144   24
48000   12   163   12
48000   12   228   26
48000   12   381   1
48000   16   214   15
49800   8   318   25
51000   8   96   33
51000   12   36   15
51000   12   59   14
51000   15   115   1
51000   15   165   4
51000   16   123   12
51600   12   18   12
52200   8   102   29
52200   12   127   29
52800   8   90   11
52800   8   190   1
52800   12   107   11
54000   8   173   34
54000   8   228   33
54000   12   26   11
54000   12   36   33
54000   12   38   22
54000   12   82   29
54000   12   169   27
54000   12   244   1
54000   15   24   13
54000   15   49   27
54000   15   51   21
54000   15   122   33
55200   12   97   17
55200   12   196   32
55800   12   133   30
56400   12   55   9
57000   12   90   23
57000   12   117   25
57000   15   51   17
57000   15   61   11
57000   15   241   34
60000   12   121   30
60000   15   79   13
61200   12   209   21
63000   12   87   33
63000   15   231   15
46200   12   12   22
50400   15   14   3
51000   12   180   15
51000   12   315   2
52200   12   29   14
54000   12   7   21
54000   12   38   11
54000   12   113   3
54000   15   18   8
54000   15   359   11
57000   15   36   5
60000   8   320   21
60000   12   24   2
60000   12   32   17
60000   12   49   8
60000   12   56   33
60000   12   252   11
60000   12   272   19
60000   15   25   13
60000   15   36   32
60000   15   56   12
60000   15   64   33
60000   15   108   16
60000   16   46   3
63000   15   72   17
66000   15   64   16
66000   15   84   33
66000   15   216   16
68400   15   42   7
69000   12   175   10
69000   15   132   24
81000   16   55   33

"
)
df1
linmod=lm(SALARY ~ EDUC ,data=df1)
coefficients(linmod)
summary(linmod)
newdata=data.frame(EDUC=12)
attach(df1)
predict(linmod,newdata,level=0.95,interval="confidence")
predict(linmod,newdata,level=0.95,interval="predict")
Output:

> coefficients(linmod)
(Intercept) EDUC
38185.598 1280.859
> summary(linmod)

Call:
lm(formula = SALARY ~ EDUC, data = df1)

Residuals:
Min 1Q Median 3Q Max
-14555.9 -4632.5 444.1 3767.5 22320.7

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 38186 3774 10.117 < 2e-16 ***
EDUC 1281 297 4.313 4.08e-05 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 6501 on 91 degrees of freedom
Multiple R-squared: 0.1697,   Adjusted R-squared: 0.1606
F-statistic: 18.6 on 1 and 91 DF, p-value: 4.077e-05

> newdata=data.frame(EDUC=12)
> attach(df1)
> predict(linmod,newdata,level=0.95,interval="confidence")
fit lwr upr
1 53555.91 52184.04 54927.78
> predict(linmod,newdata,level=0.95,interval="predict")
fit lwr upr
1 53555.91 40569.57 66542.25

ANSWER:(2A)

linear regression model is

salary= 38185.598+1280.859 *Educ

slope=1280.859

y intercept=38185.598

Ho:

no linear relationship between salary and educ

Ha:

linear relationship between salary and educ

alpha=0.05

F statistic= 18.6 p-value: 4.077e-05

p<0.05

Reject Ho

Accept Ha

Conclusion:

There is suffcient statistcial evidence at 5% level of significance to conclude that there is a linear relationship between salary and educ

Model is significant

we can use this model to predict SALARY from EDUC

Solution-b:

R sq=0.1697

=0.1697*100

=16.97% variation in salary is explained by educ

Explained variance=16.97%

unexplained variance=100-16.97=83.03%

c) Provide a 95% confidence interval estimate for the true slope value.

confint(linmod)
2.5 % 97.5 %
(Intercept) 30688.2625 45682.933
EDUC 690.9706 1870.748

95% confidence interval estimate for the true slope value lies in between 690.9706 and 1870.748

d) Based on your model, what is the expected salary of a new hire with 12 years of education

salary= 38185.598+1280.859 *Educ

for Educ=12 substitute in regression eq

salary= 38185.598+1280.859 *12

=53555.91

e ) What is the 95% prediction interval for the salary of a new hire with 12 years of education? Use the fact that the distance value = 0.011286

95% prediction interval from output is

40569.57 and 66542.25