I have written about how to run the ANOVA test in my previous post Analysis of Variance ANOVA using R. We analyzed the salary difference between different level of education.
For ease of (my!) understanding, I would take the same data set in this post as well. So here is the same data set.
> sal id gender educ Designation Level Salary Last.drawn.salary Pre..Exp Ratings.by.interviewer 1 1 female UG Jr Engineer JLM 10000 1000 3 4 2 2 male DOCTORATE Chairman TLM 100000 100000 20 4 3 3 male DIPLOMA Jr HR JLM 6000 6000 1 3 4 4 male PG Engineer MLM 15000 15000 7 2 5 5 female PG Sr Engineer MLM 25000 25000 12 4 6 6 male DIPLOMA Jr Engineer JLM 6000 8000 1 1 7 7 male DIPLOMA Jr Associate JLM 8000 8000 2 4 8 8 female PG Engineer MLM 13000 13000 7 3 9 9 female PG Engineer MLM 14000 14000 7 2 10 10 female PG Engineer MLM 16000 16000 8 4 11 11 female UG Jr Engineer JLM 10000 1000 3 4 12 12 male DOCTORATE Chairman TLM 100000 100000 20 4 13 13 male DIPLOMA Jr HR JLM 6000 6000 1 3 14 14 male PG Engineer MLM 15000 15000 7 2 15 15 female PG Sr Engineer MLM 25000 25000 12 4 16 16 male DIPLOMA Jr Engineer JLM 6000 8000 1 1 17 17 male DIPLOMA Jr Associate JLM 8000 8000 2 4 18 18 female PG Engineer MLM 13000 13000 7 3 19 19 female PG Engineer MLM 14000 14000 7 2 20 20 female PG Engineer MLM 16000 16000 8 4 21 21 female PG Sr Engineer MLM 25000 25000 12 4 22 22 male DIPLOMA Jr Engineer JLM 6000 8000 1 1 23 23 male DIPLOMA Jr Associate JLM 8000 8000 2 4 24 24 female PG Engineer MLM 13000 13000 7 3 25 25 female PG Engineer MLM 14000 14000 7 2 26 26 female PG Engineer MLM 16000 16000 8 4 27 27 female UG Jr Engineer JLM 10000 1000 3 4 28 28 male DOCTORATE Chairman TLM 100000 100000 20 4 29 29 male DIPLOMA Jr HR JLM 6000 6000 1 3 30 30 male PG Engineer MLM 15000 15000 7 2 31 31 female PG Sr Engineer MLM 25000 25000 12 4 32 32 female PG Sr Engineer MLM 25000 25000 12 4 33 33 male DIPLOMA Jr Engineer JLM 6000 8000 1 1 34 34 male DIPLOMA Jr Associate JLM 8000 8000 2 4 35 35 female PG Engineer MLM 13000 13000 7 3 36 36 female PG Engineer MLM 14000 14000 7 2 37 37 female PG Engineer MLM 16000 16000 8 4 38 38 female UG Jr Engineer JLM 10000 1000 3 4 39 39 male DOCTORATE Chairman TLM 100000 100000 20 4 40 40 male DIPLOMA Jr HR JLM 6000 6000 1 3 41 41 male PG Engineer MLM 15000 15000 7 2 42 42 female PG Sr Engineer MLM 25000 25000 12 4 43 43 male DIPLOMA Jr Engineer JLM 6000 8000 1 1 44 44 male DIPLOMA Jr Associate JLM 8000 8000 2 4 45 45 female PG Engineer MLM 13000 13000 7 3 46 46 female PG Engineer MLM 16000 16000 8 4 47 47 female UG Jr Engineer JLM 10000 1000 3 4 48 48 male DOCTORATE Chairman TLM 100000 100000 20 4 49 49 male DIPLOMA Jr HR JLM 6000 6000 1 3 50 50 male PG Engineer MLM 15000 15000 7 2
We have already executed ANOVA test. Following is the output.
> aov1 <-aov(Salary~educ, data=sal)
> aov1
Call:
aov(formula = Salary ~ educ, data = sal)
Terms:
educ Residuals
Sum of Squares 35270186667 538293333
Deg. of Freedom 3 46
Residual standard error: 3420.823
Estimated effects may be unbalanced
> summary(aov1)
Df Sum Sq Mean Sq F value Pr(>F)
educ 3 3.527e+10 1.176e+10 1005 <2e-16 ***
Residuals 46 5.383e+08 1.170e+07
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Variance between groups
What we get above is the overall significant difference between DV (salary) and IV (Education)
To compare the difference between the group, we use Post hoc test. We shall use TukeyHSD() for this. We need to go for this approach, only if the anova is significant. If anova is not significant, there is no need for posthoc.
We’d see how to run get the variances across the groups in this post.
> tukey <- TukeyHSD(aov1)
> tukey
Tukey multiple comparisons of means
95% family-wise confidence level
Fit: aov(formula = Salary ~ educ, data = sal)
$educ
diff lwr upr p adj
DOCTORATE-DIPLOMA 93333.333 88624.720 98041.947 0.0000000
PG-DIPLOMA 10373.333 7395.345 13351.322 0.0000000
UG-DIPLOMA 3333.333 -1375.280 8041.947 0.2477298
PG-DOCTORATE -82960.000 -87426.983 -78493.017 0.0000000
UG-DOCTORATE -90000.000 -95766.850 -84233.150 0.0000000
UG-PG -7040.000 -11506.983 -2573.017 0.0006777
- diff – mean difference between education level
- lwr – lower mean
- upr – upper mean
If signs between lwr and upr are same, irrelevant of + or -, that denotes significant difference.
When you compare diploma (lower degree) with doctorate (higher degree), the difference would be +ve and vice versa. If you just want to see the difference, + or – is not significant.
Let’s plot this in a graph.
> plot(tukey)
0 is the mid point. So, anything near 0 do not have significant difference.
From the top, first plot is for the comparison between DOCTORATE-DIPLOMA. You would see a high positive difference. If you see the plot for UG-DOCTORATE, is it second highest difference, but this is negative difference. Anything near 0 like UG-DIPLOMA, does not have significant difference.
ANOVA between multiple variables
We received a new data set from company now, which has a new column Loan.deducation. Last.drawn.salary changes with respect to his loans.
> sal <- read.csv("sal.csv")
> head(sal)
id gender educ Designation Level Salary Loan.deduction Last.drawn.salary Pre..Exp Ratings.by.interviewer
1 1 female UG Jr Engineer JLM 10000 5901.74 4098.26 3 4
2 2 male DOCTORATE Chairman TLM 100000 4247.31 95752.69 20 4
3 3 male DIPLOMA Jr HR JLM 6000 3895.76 2104.24 1 3
4 4 male PG Engineer MLM 15000 9108.36 5891.64 7 2
5 5 female PG Sr Engineer MLM 25000 4269.39 20730.61 12 4
6 6 male DIPLOMA Jr Engineer JLM 6000 4137.31 1862.69 1 1
Company wants to see the differences among Salary (column 6), Loan.deduction (column 7) and Last.drawn.salary (column 8). We combine apply and anova as given below.
> aovset <- apply(sal[,6:8], 2, function(x)aov(x~educ, data = sal))
- sal[,6:8] takes all rows of columns 6, 7 and 8
- aov is our function
Following is the variance between education and Last.drawn.salary.
> summary(aovset$Last.drawn.salary)
Df Sum Sq Mean Sq F value Pr(>F)
educ 3 3.342e+10 1.114e+10 674.2 <2e-16 ***
Residuals 46 7.602e+08 1.653e+07
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
F value is 674, which means, the change is significant. Following would be more interesting.
> summary(aovset$Loan.deduction)
Df Sum Sq Mean Sq F value Pr(>F)
educ 3 25577616 8525872 1.14 0.343
Residuals 46 343898395 7476052
F value for Loan.deduction is lesser than 4. So, there is no change in the deductions between different education level.
See you in another interesting post.
JavaShine 
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