Here you will find the top order batsmen performance comparison in different conditions like home and away matches by three special statistics.

1. PROJECTWORK 2020
TOPIC: COMPARISON OF TOP ORDER
BATSMAN IN HOME AND AWAY
CONDITION
Presented by
Pankaj Chowdhury
Bsc Sem-6
DEPARTMENTOF STATISTICS
SIKSHABHAVANA
VISVA BHARATI
SHANTINIKETAN

2. Born
5 November
1988 (age 31)
Delhi
Chikoo
1.75 m (5 ft 9 in)
Right-handed
Right-arm
medium
Nickname
Height
Batting
Bowling
Role top
(order
batsman)

3. Born
Full name Rohit Gurunath Sharma
30 April 1987 (age 33)
Nagpur
Nickname Shaana, Hitman, Ro
Batting Right-handed
Bowling Right-arm off break

4. Born
5 December 1985 (age 34)
Delhi
Nickname Gabbar
Height
Batting
1.80 m (5 ft 11 in)
Left-handed
Bowling Right-arm off spin
Role Opening batsman

5. ACKNOWLEDGEME
NTThe satisfaction and euphoria that accompany
the successful completion of any task would be
incomplete without mentioning the people who
made it possible whose consistent guidance
and encouragement crowned the effort with
success.
Fist of all , I am thankful to our project
supervisor
– Mr Tirthankar Ghosh ,under whose guidance I
am able to complete our project . I am
wholeheartedly thankful to him for giving me his
valuable to time and attention and providing me
a systematic way for completing our project in
time.
Also I will like to thank my friends,seniors,and
lab maintenance staff and everyone who has
helped me in completing this project.
Thank You.

6. INTRODUCTIO
N
AIM
To understand and compare the different
parameters in the field of cricket with the help
of statistical software like Microsoft Excel and
R .OBJECTIV
E
For carrying out the project the following objectives
have been formulated :
To learn the use of software for doing statistical
analysis.
To prepare hypothesis in order to compare the
home
and away match performance of different
cricketers of India .

7. Literature
Indian cricket team isoneof the mostsuccessfulteam
among all International cricket playing nations.
Traditionally India ismuchstronger at home than abroad,
the Indian team hasimproved its overseas from
,especially in limited-overs cricket, sincethe start of the
21stcentury ,winning test matches in Australia, England
and SouthAfrica.
Asof 10February 2020, India ranked 1stin test, 2nd in ODI,
4th in T20 byICC.
In this content I will discussthe batting performances of
Indian top three batsman in home and overseas matches
in recent two years .

8. Opener 1 :
Rohit Sharma
Opener 2:
Shikhar Dhawan
First down :
Virat Kohli

9. Source and description of the
data
Source of data
The data iscollected from
https://stats.espncricinfo.com,updated on 20th
February,2020.
Description of data
The data isgiven with individual away,home
match scoresof Virat Kohli , RohitSharma
,Shikhar Sharma.
We describe the whole data one by one and
graphical representation.

10. The data used in the analysis is given
below :
Name of the player
:
Virat kohli
Home match Away match
46
132
105
27
3
17
89
24
84
16
66
119
13
71
19
63
114
11
36
108
51
120
89
107
67
6
87
107
76
146
110
135
2
69
23
46
22
59
74
171

11. Name of the player
:
Shikhar Dhawan
Home match Away match
36
64
34
99
7
23
26
38
47
45
12
17
95
35
47
0
47
19
3
23
89
70
26
61
12
12
94
48
33
55
88
46
39
17
16
63
1
46
88
44

12. Name of the player
:
Rohit Sharma
Home match Away match
192
42
116
99
150
65
98
65
45
88
17
75
55
31
24
63
129
99
41
138
6
10
28
94
77
39
93
38
85
90
60
8
80
60
50
83
3
138
37
38

13. Objective of the data
:Here I want to test a player is equally efficient in home match and away match or not.
So,here we can use statistical tools to compare their mean performance in home and away
matches.
A well known test to compare two means is paired t-test when they have equal number of
sample size.
Statistical Analysis :
Let,
H0:the mean of two samples are
equal. vs
H1:the means are not equal.
Before we conduct the t-test we should test whether the two data have same variances by F-
test and for normality checking we can use Shapiro-wilk test.
Suppose there is n samples for home and away match each .
Let ,x(i) be the i th observation of home
match y(i) be the i th observation of away
match .
𝑚 𝑥 =mean of scores in home matches
𝑚 𝑦 = mean of scores in away matches
t=(𝑚 𝑥 -𝑚 𝑦 ) /√𝑆2(1/𝑛 𝑦 + 1/𝑛 𝑦 )
𝑆2=(∑(𝑥 − 𝑚 𝑥 )2 + ∑(𝑦 − 𝑚 𝑦 )2)/( 𝑛𝑥+ 𝑛𝑦-2)
The df is = 𝑛𝑥+ 𝑛𝑦-2
If the absolute value of t test statistics |t| is greater than the critical value ,then the
difference is significant . otherwise itisn’t .

14. The Shapiro-Wilk test is a way to tell if a random sample comes from a normal distribution. The
test gives you a W value; small values indicate your sample is not normally distributed (you can
reject the null hypothesis that your population is normally distributed if your values are under a
certain threshold). The formula for the W value is:
where:
xi are the ordered random sample values
ai are constants generated from the covariances, variances and means of the sample (size n) from
a normally distributed sample.
The test has limitations, most importantly that the test has a bias by sample size. The larger the
sample, the more likely you’ll get a statistically significat result

15. Let X1, ..., Xn and Y1, ..., Ym be samples from two populations which each has normal distribution.
The expected values.for the two populations can be different, and the hypothesis to be tested is
that the variances are equal. Let
𝑋 = ∑ 𝑛 𝑋𝑖/n; 𝑌= ∑ 𝑛 𝑌𝑖/n
𝑖=1 𝑖=1
be the sample means Let
2
y
2‫﷩‬‫﷩‬‫﷩‬S2
= 1
∑n (X − X) ,S=
x i=1 i(n−1) (n−1)
1
(Y − Y)2∑n
i=1 i
be the sample variences Then the test statistic
S2
xS2F= y
;
has an F distribution with n − 1 and m − 1 degrees of freedom if the null hypothesis of equality
of variances is true. Otherwise it follows an F-distribution scaled by the ratio of true variances.
The null hypothesis is rejected if F is either too large or too small based on the desired alpha
level

16. Test for this experiment:(Virat Kohli)
Shapiro-Wilk normality test
𝐻𝑜: the data follows normality .
𝐻1: the data is not following normal distribution.
Data: Virat Kohli HOME
W = 0.91069 p-value = 0.06571
Data: Virat Kohli AWAY
W = 0.98051, p-value = 0.9406
As p values in both the cases (>0.05), so data follows normal
distribution .
F testℎ 𝑎𝑆2,𝑆2are the sample variances of home and away
matches.
ℎ 𝑎𝐻𝑜: 𝑆2=𝑆2
ℎ 𝑎𝐻1: 𝑆2≠ 𝑆2
F = 0.83414, num df = 19, denom df = 19, p-value = 0.6967 alternative hypothesis: true
ratio of variances is not equal to 1
95 percent confidence interval:
0.3301624 2.1074118
sample estimates:
ratio of variances
0.834139
As p values in the case(>0.05), so data those two samples have
equal variances .
t-test
𝐻 ,𝐴are the mean of home and away match of Virat Kohli
𝐻𝑜: 𝐻 =𝐴;
𝐻1: 𝐻 ≠ 𝐴;
t = -1.4421,
df = 38,
p-value =0.1575
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-48.555803 8.155803

17. sample estimates:
mean of H:58.15
mean of A:78.35
As p values in the case(>0.05), so the mean of the home and away match
can not differ to each other .

18. barplot of virat's home match
020406080100120
barplot of virat's away match
050100150

19. home away
virat's avarage run comparison
run
0204060

20. Test for this experiment:(Shikhar
Dhawan)
Shapiro-Wilk normality test
𝐻𝑜: the data follows normality .
𝐻1: the data is not following normal distribution.
Data: Shikhar Dhawan HOME
W = 0.90533, p-value = 0.05195
Data: Shikhar Dhawan AWAY
W = 0.94757, p-value = 0.3317
As p values in both the cases (>0.05), so data follows normal
distribution .
F test
𝑆2,𝑆2are the sample variances of home and away matches.
ℎ 𝑎
𝐻𝑜 : 𝑆2 =𝑆2
ℎ 𝑎
𝐻1 : 𝑆2 ≠ 𝑆2
ℎ 𝑎
F = 0.87496, num df = 19, denomdf
= 19, p-value = 0.7739
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.3463191 2.2105387
sample estimates:
ratio of variances
0.8749581
As p values in the case(>0.05), so data those two samples have
equal variances .
t-test
𝐻 ,𝐴are the mean of home and away match of Shikhar Dhawan
𝐻𝑜: 𝐻 =𝐴;

21. 𝐻1: 𝐻 ≠ 𝐴;
t = -1.32, df = 38, p-value = 0.1947
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-29.263327 6.163327
sample estimates:
mean of H:35.85
mean of A:47.70
As p values in the case(>0.05), so the mean of the home and away match
can not differ to each other .

22. barplot of shikhars's home match
020406080
barplot of shikhars's away match
020406080

23. home away
shikhar dhawan's avarage run comparison
run
010203040

24. Test for this experiment:(Rohit
Sharma)Shapiro-Wilk normality test
𝐻𝑜: the data follows normality .
𝐻1: the data is not following normal distribution.
Data: Rohit Sharma HOME
W = 0.95311, p-value = 0.4168
Data: Rohit Sharma AWAY
W = 0.94808, p-value = 0.3389
As p values in both the cases (>0.05), so data follows normal
distribution .
F test
𝑆2,𝑆2are the sample variances of home and away matches.
ℎ 𝑎
𝐻𝑜 : 𝑆2 =𝑆2
ℎ 𝑎
𝐻1 : 𝑆2 ≠ 𝑆2
ℎ 𝑎
F = 1.6157, num df = 19, denom df = 19, p-value = 0.3043 alternative hypothesis: true
ratio of variances is not equal to 1
95 percent confidence interval:
0.6395158 4.0819999
sample estimates:
ratio of variances
1.615705
As p values in the case(>0.05), so data those two samples have
equal variances .
t-test
𝐻 ,𝐴are the mean of home and away match of Rohit Sharma
𝐻𝑜: 𝐻 =𝐴;
𝐻1: 𝐻 ≠ 𝐴;
t = 1.9567
df = 38
p-value = 0.05776
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.8903839 52.3903839
sample estimates:
mean of x : 81.60
mean of y: 55.85

25. barplot of rohit's home match
050100150
barplot of rohit's away match
020406080100120

26. home away
rohit's avarage run comparison
run
020406080

27. Rcode
Virat Kohli
vi=read.csv("C:/Users/hp/Desktop/project
work/virat.csv",header=T) vi
vix=vi$home
viy=vi$awa
y vix
viy
p=data.frame(vix,vi
y) p
shapiro.test(vi
x)
shapiro.test(vi
y)
var.test(vix,viy,alternative="two.sided")
t.test(vix,viy,alternative = "two.sided",var.equal =
TRUE) summary(vix)
summary(viy)
par(mfrow=c(2,1
))
barplot(vix,main="barplot of virat's home match",horiz
=F ) barplot(viy,main="barplot of virat's away
match",horiz = F) mvix=mean(vix)
mviy=mean(viy)
mvi=c(mvix,mvi
y)
barplot(mvi,horiz=F,col="red",ylab="run",names.arg=c("home","aw
ay"))

28. Rohit Sharma
rh=read.csv("C:/Users/hp/Desktop/project
work/rohit.csv",header=T) rh
rhx=rh$hom
e
rhy=rh$awa
y rhx
rhy
p=data.frame(rhx,rh
y) p
shapiro.test(rhx
)
shapiro.test(rhy
)
var.test(rhx,rhy,alternative="two.sided")
t.test(rhx,rhy,alternative = "two.sided",var.equal =
TRUE)
summary(rhx)
summary(rhy)
par(mfrow=c(2,1
))
barplot(rhx,main="barplot of virat's home match",horiz
=F ) barplot(rhy,main="barplot of virat's away
match",horiz = F) mrhx=mean(rhx)
mrhy=mean(rhy)
mrh=c(mrhx,mrh

29. Shikhar Dhawan
sh=read.csv("C:/Users/hp/Desktop/project work/shikhar.csv",header
= T) sh
shx=sh$home
shy=sh$aw
ay shx
shy
p=data.frame(shx,sh
y) p
shapiro.test(sh
x)
shapiro.test(sh
y)
var.test(shx,shy,alternative="two.sided")
t.test(shx,shy,alternative = "two.sided",var.equal =
TRUE) summary(shx)
summary(shy)
par(mfrow=c(2,1
))
barplot(shx,main="barplot of shikhars's home match",horiz =F )
barplot(shy,main="barplot of shikhars's away match",horiz
= F) mshx=mean(shx)
mshy=mean(shy)
msh=c(mshx,msh
y)
barplot(msh,horiz=F,col="red",ylab="run",names.arg=c("home","away"),main="shikhar
dhawan's avarage run comparison")

30. Conclusio
nThere is no difference in performance due to home match away match of Virat Kohli,
Shikhar Dhawan,Rohit Sharama.
By graphical representation we can see the average score of Rohit Sharma in home matches
is higher tham away match and for Virat Kohli and Shikhar Dhawan away match average
score is higher than home match.