Dear R-helps,

I did an experiment with FAs ['High' and 'Zero'(no w-3) quality; n=24 for
each group]. Then I did AI to see their sperm competitiveness based on
their paternity performance. My data is as below where Fish ID- Blind ID
for each fish; Group ID- Dietary group ID; Diet quality - High=1, zero=0;
Babies for paternity- total no. of babies got from females; Success -
Babies shared/paterned by focal male; Failure - Babies shared/paterned by
competitor, Proportion - Success/(Success+Failure).

Fish ID

Group ID

Diet quality

Babies for paternity

Success

Failure

Proportion

1

High

1

9

5

4

0.556

12

High

1

7

5

2

0.714

15

High

1

7

4

3

0.571

20

High

1

6

5

1

0.833

32

High

1

7

2

5

0.286

37

High

1

3

1

2

0.333

48

High

1

4

1

3

0.25

53

High

1

10

0

10

0

65

High

1

3

3

0

1

70

High

1

4

4

0

1

77

High

1

7

2

5

0.286

82

High

1

6

6

0

1

96

High

1

8

2

6

0.25

104

High

1

12

10

2

0.833

111

High

1

4

3

1

0.75

123

High

1

6

5

1

0.833

128

High

1

8

8

0

1

133

High

1

6

5

1

0.833

144

High

1

12

6

6

0.5

152

High

1

13

11

2

0.846

159

High

1

8

1

7

0.125

164

High

1

4

1

3

0.25

169

High

1

6

2

4

0.333

5

Zero

0

9

4

5

0.444

10

Zero

0

7

2

5

0.286

17

Zero

0

7

3

4

0.429

22

Zero

0

6

1

5

0.167

36

Zero

0

7

5

2

0.714

39

Zero

0

3

2

1

0.667

44

Zero

0

4

3

1

0.75

51

Zero

0

10

10

0

1

63

Zero

0

3

0

3

0

68

Zero

0

4

0

4

0

73

Zero

0

7

5

2

0.714

84

Zero

0

6

0

6

0

94

Zero

0

8

6

2

0.75

106

Zero

0

12

2

10

0.167

109

Zero

0

4

1

3

0.25

121

Zero

0

6

1

5

0.167

132

Zero

0

8

0

8

0

137

Zero

0

6

1

5

0.167

142

Zero

0

12

6

6

0.5

154

Zero

0

13

2

11

0.154

157

Zero

0

8

7

1

0.875

168

Zero

0

4

3

1

0.75

173

Zero

0

6

4

2

0.667



I ran the following codes to have my results:

###Proportion estimate:
p<-Data$Success/(Data$Success+Data$Failure)
plot(Data$Group.ID,p,ylab="Proportion of success")

###Response variable:
y<-cbind(Data$Success,Data$Failure)
model1 <- glm(y~Diet.quality, data=Data, family=binomial)
summary(model1)
plot(model1)# gives Q-Q plots
###The residual deviance is 152.66  on 44 d.f. so the model is quite badly
overdispersed:
#152.66/44 where The overdispersion factor is almost 3.46 (unbelievable).

## model with logit link functions and weights:
model2<-glm(cbind(Success,Failure)~Group.ID,data=Data,
family="binomial"(link="logit"),weights=Success+Failure)
summary(model2)
###The residual deviance is 1196.1  on 46 d.f. so the model is quite badly
overdispersed:
#1192.1/44 where The overdispersion factor is almost 27.09 (unbelievable).

#The simplest way to take this into account is to use what is called an
#‘empirical scale parameter’ to reflect the fact that the errors are not
#binomial as we assumed, but were larger than this (overdispersed) by a
factor of 3.38.

model3<-glm(y ~ Group.ID,data=Data,family="quasibinomial")
summary(model3)

###Note that the ratio of the residual deviance and the degrees of freedom
is still
#larger than 1, but that is no longer a problem as we now allow for
overdispersion.

 Each models gives me different results with overdispersion. So, can anyone
help me to give me some valuable suggesions to solve this problem. I'll
really appreciate your kind assistance and will be grateful to you forever.

With kind regards,

Moshi
mrahmankuf...@gmail.com

-- 
MD. MOSHIUR RAHMAN
PhD Candidate
School of Animal Biology/Zoology (M092)
University of Western Australia
35 Stirling Hwy, Crawley, WA, 6009
Australia.
Mob.: 061-425205507

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