Question: Carry out a two-way fixed-effects analysis of variance for male, femal

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Question: Carry out a two-way fixed-effects analysis of variance for male, female, and interaction effects on offspring weight.

Could you please solve it with R Programming and kindly provide the code? Many thanks!

Problem 3: Mean offspring weights (milligrams) of n = 6 female mosquito fish (Gambusia affinis) from female and female parents drawn from populations in North Carolina and Iinois had these values: Illinois female North Carolina female Illinois male 1.43 1.15 1.10 1.11 0.96 1.01 6.76 0.1337 North Carolina male 1.25 1.17 1.29 1.09 1.07 1.01 6.88 Illinois North Carolina male 1.94 male 1.52 1.82 1.70 2.21 1.58 1.50 1.59 1.38 1.35 1.65 9.32 0.2521 Sum Within-cell sum 10.33 0.0609 0.3585 of squares SOURCE: Data reproduced with the kind permission of David Reznick.

Explanation / Answer

As directed, I shall use R programming to solve this problem and will accordingly paste the code here.
The R codes will be accompanied with comment lines, so that I can give a brief account of my methodology using them.

# DECLARING THE NECESSARY LIBRARIES
library(Rmisc)
library(ggpubr)
library(car)

# IMPORTING THE DATA FROM EXCEL, WHERE I HAD STORED THE DATA FIRST
data=read.csv("C:\Users\LAPTOP\Desktop\a.csv")
data

# ACCESSING THE VARIABLES IN THE DATASET
attach(data)
# SUMMARY STATISTICS
summarySE(data,measurevar="Weight",groupvars=c("Female_Loc","Male_Loc"))
# GENERATING FREQUENCY TABLE
table(Female_Loc,Male_Loc)

# VISUALIZING DATA USING BOX PLOTS AND LINE PLOTS WITH MULTIPLE GROUPS
ggboxplot(data,x="Female_Loc",y="Weight",color="Male_Loc")
ggline(data,x="Female_Loc",y="Weight",color="Male_Loc")

# ANOVA TABLE 1
# TWO-WAY ANOVA WITH INTERACTION EFFECT
res.aov1 <- aov(Weight ~ Female_Loc * Male_Loc,data)
summary(res.aov1)
# INTERACTION EFFECT COMES OUT TO BE SIGNIFICANT, HERE

# ANOVA TABLE 2
# SINCE INTERACTION IS INSIGNIFICANT,
# WE CARRY OUT TWO-WAY ANOVA WITHOUT INTERACTION EFFECT
res.aov2 <- aov(Weight ~ Female_Loc + Male_Loc,data)
summary(res.aov2)

# NORMALITY OF RESIDUALS
shapiro.test(residuals(object=res.aov2))
# FROM P-VALUE, WE CONCLUDE THAT RESIDUALS ARE NORMAL

# HOMOGENEITY OF VARIANCES
plot(res.aov1,1)
# FROM PLOT, WE CAN ASSUME HOMOGENEITY OF VARIANCES

From the ANOVA Table 1 output, we see that the p-value for the interaction between Female_Loc (Illinois or North Carolina) and Male_Loc (Illinois or North Carolina) is 0.264 (insignificant), which indicates that there is NO interaction between the two variables.

Hence, we rerun the ANOVA without any interaction in ANOVA Table 2. Here, we see that the p-value of Female_Loc is 5.07e-06 (significant), which indicates that the levels of Female_Loc are associated with significantly different Weights. Also, we see that that the p-value of Male_Loc is 0.379 (insignificant), which indicates that the levels of Male_Loc are NOT associated with significantly different Weights.
Also, the residuals satisfy the normality assumption and the homogeneity of variances is also satisfied.

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