Testing of difference – T test using R

Hi,

I have written about a hypothesis testing of independence in my previous post Testing of Independence – Chi Square test – Manual, LibreOffice, R. This post talks about testing of mean difference.

Lets take the same salary data set used in my previous post.

[code language=”r”]
> 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
[/code]

Let’s find out the difference of mean salary between male and female.

[code language=”r”]
> aggregate(Salary~gender, mean, data=sal)
gender Salary
1 female 16040
2 male 27000
[/code]

Mean salary of female μ₀ = 16040
Mean salary of female μ₁ = 27000
The symbol ~ differentiates between dependent and independent variables

Obviously there is a difference. Let’s see what a t-test in R shows us.

[code language=”r”]
> t.test(Salary~gender, data = sal)

Welch Two Sample t-test

data: Salary by gender
t = -1.4494, df = 25.039, p-value = 0.1596
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-26532.252 4612.252
sample estimates:
mean in group female mean in group male
16040 27000
[/code]

How to interpret this result?

The p-value is compared with the desired significance level of our test and, if it is smaller, the result is significant. That is, if the null hypothesis were to be rejected at the 5% significance level, this would be reported as “p < 0.05″. Small p-values suggest that the null hypothesis is unlikely to be true. But our p-value 0.15 > 0.05. Hence null hypothesis is rejected and alternate hypothesis is accepted. There is a difference in salary between both genders.

Higher the t value (ignore the sign), higher the difference.