Answer to Question #215519 in Statistics and Probability for Ahmad

Question

Conduct a study taking cricket players. Captain is interested to select the players on the basis of past 10 matches. Use one Way Anova, justify that there are performance difference among the players. Sample size should be 3 to 4 players. Use tables

Answer 1

Let the three players be Player1 , Player2 and Player3, we would take records of past 10 matches of those players.

The study is shown in below table(in terms of runs scored)

STEP-1: Calculation of correction term (C)

$$C=\dfrac{G^2}{N}$$

where, $G=\sum X_1+\sum X_2+\sum X_3=310+337+298=945$

and $N=10+10+10=30$

So,

$C=\dfrac{G^2}{N}=\dfrac{(945)^2}{30}=29767.5$

STEP-2: Calculation of total sum of squares(SST)

$SST=\sum X_1^2+\sum X_2^2+\sum X_3^2-C\\SST=(14348+16297+11982)-29767.5=12859.5$

STEP-3: Calculation of sum of squares between groups (SSB)

$SSB=[\dfrac{(\sum X_1)^2+(\sum X_2)^2+(\sum X_3)^2}{10}]-C\\SSB= \dfrac{(310)^2+(337)^2+(298)^2}{10}-29767.5\\SSB=29847.3-29767.5=79.8$

STEP-4: Sum of squares between groups(SSW)

$SSW=SST-SSB=12859.5-79.8=12779.7$

Number of players (k) = 3

Now we have to form one way ANOVA table:

Hence, F-static df = (2,27) is $F_{(2,27)}$

$F_{(2,27)}$ $= \dfrac{MSSB}{MSSW}=\dfrac{39.9}{473.32}=0.08$

Now Hypothesis for testing:

$H_0:$ There is no performance difference among the players.

$H_A:$ There is performance difference among the players.

For above testing we would be using F-static.

Let our level of significance ($\alpha$) = 0.05

critical value for F-stat at df = (2,27) at $\alpha$ = 0.05 is ; $F_{critical }=3.354$

As $F_{calculated}=0.08$ which is less than $F_{critical }=3.354$

We cannot reject our null hypothesis.

CONCLUSION: From the given study we cannot tell that there exist performance difference among players.