M.R.L.N. Panchanana
(M.Sc Statistics from University of Hyderabad, India)

M.Tech (Compueter Science & Technology
from Jawaharlal Nehru University, Delhi)

**Abstract**
Fitting a parametric distribution to a data set is a key
step in analyzing the data. In many statistical procedures like ANOVA, several
hypothesis tests are performed assuming the data as normal. In this presentation, it is discussed how to
perform statistical techniques using R Software, to fit a statistical
distribution to oneday cricket Score made by Sachin Ramess Tendulkar. DNB (Does
Not Batted) cases are not considered.
**Introduction**
Few
functions used in R^{®} is useful in deciding, which Univariate
distribution is suitable to the specified variable from the data set. It provides
Univariate discrete and continuous distributions. Before going to perform, goodness of fit test,
it is advisable to know the characteristics of the data by descriptive
statistics methods like **summary**
and **stem** and
Density estimation methods like **hist**,
**boxplot**, **ecdf**, **density, fitdistr**,**ks**.**test**. Graphical illustrative methods are **qqplot** and **qqline**.

**Oneday cricket Runs**
The data for illustrative examples is collected from **http://www.howstat.com.au/ . **Only the oneday runs made by few famous
batsmen are used for the analysis. The data is saved in playername.ssa7bdat
files. Primarly “Sachin Tendulkar’s oneday match runs are analyzed.
Open R Software and load the
package sas7bdat to import sas data files.
> library(sas7bdat)

Read the data file as follows:
SACHIN <-
read.sas7bdat("D:/Modeling/sachin.sas7bdat")
The following commands invoke the
function and computes mean, median, and
different percentiles.
> summary(SACHIN)
RUNS
Min. :
0.00
1st Qu .
: 8.00

Median : 28.50

Mean : 40.77

3rd Qu .
: 63.00

Max. :200.00

Sachin's Avarage is around 40 runs. Here, we considered every match whether he is out or notout.

With quantile function, distribution of runs at each percentile can be observed.

> quantile(sachin,seq(0, 1, by=.1))
0% 10% 20% 30% 40% 50%
60% 70% 80%
90% 100%
0.0 2.0 5.0 11.0
18.4 28.5 39.0 53.0 72.8 100.0 200.0
To have a look at the shape of the distribution of the Runs,
Stem and leaf plot can be generated with the following statements.

> sachin <-
as.numeric(SACHIN$RUNS)
> stem(sachin)
0 |
00000000000000000000111111111111111111112222222222222222223333333333+46

1 |
00000011111111112222233344444445555555666677777788888889999

2 |
00011111111222233344445555667777778888888999

3 |
00000011112222223444555556666666777788889999999

4 | 0000011112334444555556777888899

5 | 0012222333334444557777

6 | 0112222233334555555677788999

7 | 0012347789

8 | 011122223455678899

9 | 011333334556778999

10 | 000000112455

11 | 001234457788

12 | 002234778

13 | 4789

14 | 0111366

15 | 2

16 | 3

17 | 5

18 | 6

19 |

20 | 0

The Graph shows the runs made by Sachin in nice formatt. For Example, Sachin made 200 runs is shown in the lowest part and he did not make any runs in 190s.

Histograms and Nonparametric density
estimation methods
Histogram is a density estimation
method for continuous variable, where each bin contains the observations. Bin
is an interval, which partitions range of observations into intervals. The
following is the input.

> hist(sachin, col="darkblue",
border = " yellow ",
+ xlab="Oneday match Runs",
+ ylab="count",
+ main="Oneday Runs made by Sachin Ramesh Tendulkar")
The output is:

Since the variable RUNS is greater than or equal to 0 and the height of the first bar is high and you can observe the gradual decline in height of the bars from second bar onwards. But, it is not forgettable that the histogram depends on the width of the bar and number of bars. If number of bars is increased, histogram is plotted with small intervals by decreasing the smoothness. At this point it is unable to decide what type of function it is. If we decrease the number of intervals to 1, then the function is uniform. R Software gives an option breaks= to the user to change the numbers of bars.
> bins=c(seq(0,150,by=10),200)

>
hist(sachin,breaks=bins,col="darkblue", border = " yellow
", xlab="Oneday match
Runs", ylab="count", main="Oneday Runs made by
Sachin Ramesh Tendulkar")

> lines(density(sachin), col=
"red")

> rug(sachin)

The Output is as follows:
You can observe that there is difference in the shape of the
histogram by giving the bins as 0-10, 11-20, …,150-200. With rugs() function
you can observe the data points on X-axis. You can clearly identify the runs
mad in each interval, particularly 100+ runs.

Before deciding the shape of the density function generate
box plot. These are useful in deciding the symmetric and asymmetric shapes of
the function of the existing data.

> boxplot(sachin, horizontal =
TRUE, col = "orange", xlab="Oneday match
Runs",main="Oneday Runs made by Sachin Ramesh Tendulkar")

<Bookmark(10You can observe the Outliers in box plot after 149+ runs.

With the above statistics and graphs, the distribution is not Symmetric. Once again, we can look at summary statistics.
Load the library "fBasics" and look at the Statistics:
> library(fBasics)
> basicStats(sachin)

sachin
nobs 452.000000
NAs 0.000000
Minimum 0.000000
Maximum 200.000000
1. Quartile 8.000000
3. Quartile 63.000000
Mean 40.765487
Median 28.500000
Sum 18426.000000
SE Mean 1.883299
LCL Mean 37.064357
UCL Mean 44.466617
Variance 1603.159959
Stdev 40.039480
Skewness 1.152907
Kurtosis 0.753945

With high variance, mean is almost equal to standard deviation. the distribution is heavily weighted far from the mean. Since skewness > 0, the distribution is right skewed as shown in the Histogram too, most values are concentrated on left side of the mean, with extreme values to the right. with all the above interpretation, Symmetric distributions will not be a fit to the data.
You can estimate the density function and Cumulative density
function as follows:

>
plot(density(sachin),main="Density estimate of sachin’s Oneday Runs")

>
plot(ecdf(sachin),main="Empirical cumulative distribution function of
Sachin’s Oneday Runs")

From all the above, it can be observed that the data is
asymmetric and skewed positively. So, You can look at distributions like
exponential, weibull, etc..,
###
Quantile-Quantile Plots

Before,
going to decide the distribution, examine with other graphical procedures,
Quantile-Quantile plots and Probability plots. The quantile of a sample is the
data point corresponding to a given fraction of the data. A one-sample quantile
plot looks like a cumulative sample distribution function. qqnorm()is used
to test the goodness of fit of a Normal distribution and qqplot() for any kind of other distribution. To plot
Q-Q plot for variable RUNS, the input is:

> x.teo<-rweibull(n=452,shape=0.83908325, scale=37.62877983)
> qqplot(sachin,x.teo,main="QQ-plot for Weibull distribution")
> abline(0,1)

The output is:

**fitdistr**
Coverted 0 RUNS to
0.1; From all the above methods, the variable with positive numbers greater
than zero may follow some asymmetric distributions, with more observations in
the first intervals and decreasing number of observations in the later
intervals.
> SACHIN1
<-read.sas7bdat("D:/Modeling/SYSTAT/Cluster/GRAPHS/Rdatastore/sachin1.sas7bdat")
> sachin1 <-
as.numeric(SACHIN1$RUNS)
We can see, whether any of the weibul or exponential
distributions fit to the data with the help of **fitdistr**.
Parameters of the distribution can be calculated by maximum likelihood
estimation methods. The following
statements will fit distributions to the variable RUNS.

> library(MASS)
> fitdistr(sachin1,"Exponential")

The Output is:
rate
0.024527892
(0.001153695)
> fitdistr(sachin1,"weibull")
The Output is:
shape scale
0.83908325 37.62877983
( 0.03253613) ( 2.20929924)

**Goodness of fit tests**

KS test can be conducted for Goodness of fit test.
>
ks.test(sachin1,"pexp",rate=0.024527892)
The output is as follows:

One-sample Kolmogorov-Smirnov test
data: sachin1
D = 0.0917, p-value = 0.0009926

alternative hypothesis:
two-sided
>ks.test(sachin1,"pweibull",shape=0.83908325
,scale=37.62877983)

The output is as follows:

One-sample
Kolmogorov-Smirnov test
data: sachin1
D = 0.0596, p-value = **0.08023**
alternative hypothesis: two-sided
We accept null
hypothesis that the data follow a Weibull distribution because the p-value is
enough higher than significance levels usually referred. Since shape
parameter of weibull is <1,** it is harder to get Sachin out as his score
increases.**

In an
unpublished work done by me and Prof Traimbakam Krishnan during 2004 to 2007,
analysis was done on censored data for several famous players. Not-outs are considered as right censored.
The results will be kept in separate Article.
**Conclusions**

By using function available in R software, we can analyze
univariate data very well.
Similar kind of Analysis can be done on all famous oneday match
players.

**References**
[1] Johnson, N. L., Kotz, S., and, Balakrishnan, N. (1994). *Univariate
continuous **distributions*.
Vol. 1, 2nd ed. :New York: John Wiley & Sons.

[2] Johnson, N. L., Kotz, S., and Balakrishnan, N. (1995). *Univariate
continuous **distributions*.Vol.
2, 2nd ed. :New York: John Wiley & Sons.

[3] Johnson, N. L., Kotz, S., and Kemp, A. W. (1993). *Univariate
discrete distributions*. 2nd ed.: New
York: John Wiley & Sons.

[4] A. Law and D. Kelton. *Simulation
Modelling and Analysis*. McGraw Hill, New York, 1990.

[5] http://cran.r-project.org/doc/manuals/R-intro.pdf