Statistics project on comparison of two batsmen
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The document presents a project comparing the performance of top-order Indian batsmen Rohit Sharma, Shikhar Dhawan, and Virat Kohli in home and away conditions using statistical analysis techniques such as t-tests and Shapiro-Wilk tests. It concludes that there are no significant differences in their performances between home and away matches, with graphical representations demonstrating average scores. The analysis is based on data collected from ESPN Cricinfo, and acknowledges the support from various individuals and resources throughout the project.
Statistics project on comparison of two batsmen
- 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 2S2 = 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.