D. Is there a block effect? why? During a typical Professional Golf Association

Question

D. Is there a block effect? why?

During a typical Professional Golf Association (PGA) tournament, the competing golfers play four rounds of golf, where the hole locations are changed for each round. Here are the scores for the top five finishers at the 2009 U.S. Open. The following statistics were computed: SST = 272.95 SSB = 93.2 SSE = 127.6 a. Construct the ANOVA table. b. At the 5% significance level, can you conclude that the average scores produced by the four different rounds differ? c. At the 5% significance level, can you conclude that the average scores produced by the five different players differ?

Explanation / Answer

The r-code for the above problem is as given below:

data=c(69,69,67,67,70,64,70,70,65,78,70,69,70,70,79,73,70,71,76,72)
round=c(rep('1',5),rep('2',5),rep('3',5),rep('4',5))
Golfer=rep(c("Lucas","Phil","David","Ricky","Ross"),4)
fr=data.frame(data,round,Golfer)
fr
summary(fr)
str(fr)
fr.aov=aov(data~round+Golfer, data=fr)
fr.aov
summary(fr.aov)
fr1.aov=aov(data~round*Golfer, data=fr)
fr1.aov
summary(fr1.aov)
anova(fr.aov,fr1.aov)

Here,

H01: There is no significant difference among the rounds.

H02: There is no significant difference among the Golfers.

Vs

H11: There is significant difference among the rounds.

H1: There is significant difference among the rounds.

Output:

fr.aov=aov(data~round+Golfer, data=fr)
> fr.aov
Call:
aov(formula = data ~ round + Golfer, data = fr)

Terms:
round Golfer Residuals
Sum of Squares 52.15 93.20 127.60
Deg. of Freedom 3 4 12

Residual standard error: 3.260879
Estimated effects may be unbalanced
> summary(fr.aov)
Df Sum Sq Mean Sq F value Pr(>F)
round 3 52.15 17.38 1.635 0.233
Golfer 4 93.20 23.30 2.191 0.132
Residuals 12 127.60 10.63   
> fr1.aov=aov(data~round*Golfer, data=fr)
> fr1.aov
Call:
aov(formula = data ~ round * Golfer, data = fr)

Terms:
round Golfer round:Golfer
Sum of Squares 52.15 93.20 127.60
Deg. of Freedom 3 4 12

Estimated effects may be unbalanced
> summary(fr1.aov)
Df Sum Sq Mean Sq
round 3 52.15 17.38
Golfer 4 93.20 23.30
round:Golfer 12 127.60 10.63
> anova(fr.aov,fr1.aov)
Analysis of Variance Table

Model 1: data ~ round + Golfer
Model 2: data ~ round * Golfer
Res.Df RSS Df Sum of Sq F Pr(>F)
1 12 127.6
2 0 0.0 12 127.6   

Here, the null hypothesis is accepted.

Therefore,

b,The average scores produced by 4 rounds do not differ.

c,The average scores produced by 5 different players do not differ.

d, There is no block effect.Since even when we create ANOVA for interaction effect the null hypothesis is accepted.

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