# understand the data
eggs = data.frame(size=c("A","B","A","B"),piercer=c("N","N","Y","Y"), total=c(54,200,60,70), broken=c(4,15,4,1), cracked=c(8,28,9,7))
eggs## size piercer total broken cracked
## 1 A N 54 4 8
## 2 B N 200 15 28
## 3 A Y 60 4 9
## 4 B Y 70 1 7# make a dataframe in a grid.
eggs_grid = expand.grid(size=c("A","B"),piercer=c("N","Y"), result=c("broken","cracked","ok"))
# expland.grid: create a dataframe from all combinations of the supplied vectors or factors
eggs_grid$n = c(4,15,4,1, 8,28,9,7, 42, 157, 47, 62) # add numbers to the grid.
eggs_grid## size piercer result n
## 1 A N broken 4
## 2 B N broken 15
## 3 A Y broken 4
## 4 B Y broken 1
## 5 A N cracked 8
## 6 B N cracked 28
## 7 A Y cracked 9
## 8 B Y cracked 7
## 9 A N ok 42
## 10 B N ok 157
## 11 A Y ok 47
## 12 B Y ok 62## does the piercer affect "broken-or-cracked" regardless of egg size
## get subset of dataframe by multiple conditions
notPiercer_broken = eggs_grid[which(eggs_grid$piercer=="N" & eggs_grid$result=="broken"),]
notPiercer_cracked = eggs_grid[which(eggs_grid$piercer=="N" & eggs_grid$result=="cracked"),]
notPiercer_ok = eggs_grid[which(eggs_grid$piercer=="N" & eggs_grid$result=="ok"),]
yesPiercer_broken = eggs_grid[which(eggs_grid$piercer=="Y" & eggs_grid$result=="broken"),]
yesPiercer_cracked = eggs_grid[which(eggs_grid$piercer=="Y" & eggs_grid$result=="cracked"),]
yesPiercer_ok = eggs_grid[which(eggs_grid$piercer=="Y" & eggs_grid$result=="ok"),]
#make a matrix
piercerOkBrokenCracked = matrix(c(sum(notPiercer_ok$n), sum(yesPiercer_ok$n),
sum(notPiercer_broken$n), sum(yesPiercer_broken$n),
sum(notPiercer_cracked$n), sum(yesPiercer_cracked$n)), nrow=2)
dimnames(piercerOkBrokenCracked) <- list(c("notPiercer", "yesPiercer"), c("ok", "broken", "cracked"))
piercerOkBrokenCracked## ok broken cracked
## notPiercer 199 19 36
## yesPiercer 109 5 16# marginal analysis for pierced vs. ok or broken or cracked
fisher.test(piercerOkBrokenCracked)##
## Fisher's Exact Test for Count Data
##
## data: piercerOkBrokenCracked
## p-value = 0.3211
## alternative hypothesis: two.sidedchisq.test(piercerOkBrokenCracked)##
## Pearson's Chi-squared test
##
## data: piercerOkBrokenCracked
## X-squared = 2.3623, df = 2, p-value = 0.3069Discussion
In both fisher.test and chisq.test, the p-values are 0.3211 and 0.3069, respectively. These p-values are lareger than 0.05. Therefore, we accept the null hypothesis; it seems that the piercer doesn’t affect how many eggs are broken or cracked during boiling.
One cell in the table has the frequency 5. This frequency is so small, so Fisher Exact test will be more appropriate than chisq test. Therefore, p-values between fisher.test and chisq.test are a little different, 0.3211 and 0.3069
# marginal analysis for pierced vs. ok or not_ok
notPiercer_brokencracked=sum(notPiercer_broken$n)+sum(notPiercer_cracked$n)
yesPiercer_brokencracked=sum(yesPiercer_broken$n)+sum(yesPiercer_cracked$n)
piercerOk_notOk = matrix(c(notPiercer_brokencracked, sum(notPiercer_ok$n),
yesPiercer_brokencracked, sum(yesPiercer_ok$n)), nrow=2)
dimnames(piercerOk_notOk) <- list(c("Broken&Cracked", "Ok"), c("notPiercer", "yesPiercer"))
piercerOk_notOk## notPiercer yesPiercer
## Broken&Cracked 55 21
## Ok 199 109fisher.test(piercerOk_notOk)##
## Fisher's Exact Test for Count Data
##
## data: piercerOk_notOk
## p-value = 0.2247
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
## 0.8029372 2.6337180
## sample estimates:
## odds ratio
## 1.433252chisq.test(piercerOk_notOk)##
## Pearson's Chi-squared test with Yates' continuity correction
##
## data: piercerOk_notOk
## X-squared = 1.3103, df = 1, p-value = 0.2523Discussion
This time, I considered broken and cracked as one category of damaged eggs. And, I examined whether the piercer has an effect on egg’s damage during boiling. The odds ratio is 1.433; it seems that the piercer has some effect to prevent egg’s damage during boiling. Let’s see the p-value of fisher.test and chisq.test. The p-values in both tests decreased compared to the previous test that I considered ‘broken’ and ‘cracked’ separately; 0.2247 in fisher.test; 0.2523 in chisq.test. But, still, the p-values are larger than 0.05. So, I still accept the Null hypothesis; piercer doesn’t seem to affect to prevent egg’s damage during boiling.
This time, I consider egg size A and B are different factors when examining the piercer’s effect on egg’s breaking&cracking.
# Slice analysis when size==A
A_notPiercer_broken = eggs_grid[which(eggs_grid$size=="A" & eggs_grid$piercer=="N" & eggs_grid$result=="broken"),]
A_notPiercer_cracked = eggs_grid[which(eggs_grid$size=="A" & eggs_grid$piercer=="N" & eggs_grid$result=="cracked"),]
A_notPiercer_ok = eggs_grid[which(eggs_grid$size=="A" & eggs_grid$piercer=="N" & eggs_grid$result=="ok"),]
A_yesPiercer_broken = eggs_grid[which(eggs_grid$size=="A" & eggs_grid$piercer=="Y" & eggs_grid$result=="broken"),]
A_yesPiercer_cracked = eggs_grid[which(eggs_grid$size=="A" & eggs_grid$piercer=="Y" & eggs_grid$result=="cracked"),]
A_yesPiercer_ok = eggs_grid[which(eggs_grid$size=="A" & eggs_grid$piercer=="Y" & eggs_grid$result=="ok"),]
# marginal analysis in slice A for pierced vs. ok or not_ok
A_notPiercer_brokencracked=sum(A_notPiercer_broken$n)+sum(A_notPiercer_cracked$n)
A_yesPiercer_brokencracked=sum(A_yesPiercer_broken$n)+sum(A_yesPiercer_cracked$n)
A_piercerOk_notOk = matrix(c(A_notPiercer_brokencracked, sum(A_notPiercer_ok$n),
A_yesPiercer_brokencracked, sum(A_yesPiercer_ok$n)), nrow=2)
dimnames(A_piercerOk_notOk) <- list(c("A_Broken&Cracked", "A_Ok"), c("A_notPiercer", "A_yesPiercer"))
A_piercerOk_notOk## A_notPiercer A_yesPiercer
## A_Broken&Cracked 12 13
## A_Ok 42 47fisher.test(A_piercerOk_notOk)##
## Fisher's Exact Test for Count Data
##
## data: A_piercerOk_notOk
## p-value = 1
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
## 0.3837906 2.7608613
## sample estimates:
## odds ratio
## 1.032686chisq.test(A_piercerOk_notOk)##
## Pearson's Chi-squared test with Yates' continuity correction
##
## data: A_piercerOk_notOk
## X-squared = 5.0627e-31, df = 1, p-value = 1Discussion
In the egg size==A, the odds ratio is close to 1, and the p-values in fisher.test and chisq.test is 1. These results show that the piercer doesn’t look effective at all, and we need to strongly accept the null hypothesis.
# Slice analysis when size==B
B_notPiercer_broken = eggs_grid[which(eggs_grid$size=="B" & eggs_grid$piercer=="N" & eggs_grid$result=="broken"),]
B_notPiercer_cracked = eggs_grid[which(eggs_grid$size=="B" & eggs_grid$piercer=="N" & eggs_grid$result=="cracked"),]
B_notPiercer_ok = eggs_grid[which(eggs_grid$size=="B" & eggs_grid$piercer=="N" & eggs_grid$result=="ok"),]
B_yesPiercer_broken = eggs_grid[which(eggs_grid$size=="B" & eggs_grid$piercer=="Y" & eggs_grid$result=="broken"),]
B_yesPiercer_cracked = eggs_grid[which(eggs_grid$size=="B" & eggs_grid$piercer=="Y" & eggs_grid$result=="cracked"),]
B_yesPiercer_ok = eggs_grid[which(eggs_grid$size=="B" & eggs_grid$piercer=="Y" & eggs_grid$result=="ok"),]
# marginal analysis in slice B for pierced vs. ok or not_ok
B_notPiercer_brokencracked=sum(B_notPiercer_broken$n)+sum(B_notPiercer_cracked$n)
B_yesPiercer_brokencracked=sum(B_yesPiercer_broken$n)+sum(B_yesPiercer_cracked$n)
B_piercerOk_notOk = matrix(c(B_notPiercer_brokencracked, sum(B_notPiercer_ok$n),
B_yesPiercer_brokencracked, sum(B_yesPiercer_ok$n)), nrow=2)
dimnames(B_piercerOk_notOk) <- list(c("B_Broken&Cracked", "B_Ok"), c("B_notPiercer", "B_yesPiercer"))
B_piercerOk_notOk## B_notPiercer B_yesPiercer
## B_Broken&Cracked 43 8
## B_Ok 157 62fisher.test(B_piercerOk_notOk)##
## Fisher's Exact Test for Count Data
##
## data: B_piercerOk_notOk
## p-value = 0.07609
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
## 0.9154887 5.5159171
## sample estimates:
## odds ratio
## 2.11738chisq.test(B_piercerOk_notOk)##
## Pearson's Chi-squared test with Yates' continuity correction
##
## data: B_piercerOk_notOk
## X-squared = 2.807, df = 1, p-value = 0.09385Discussion
The fisher.test and chisq.test results in egg size==B are quite different from the result in egg size==A. The odds ratio is 2.117 and this is the biggest odds ratio that I got while performing several tests with the egg data. However, the p-value is still 0.076, which is larger than 0.05 and non-significant. Still we accept the Null hypothesis, and it seems that the piercer doesn’t affect preventing egg’s damage during boiling. However, if I experiment with more eggs of size B to increase sample size, I may be able to get significant p-value (< 0.05) and reject the Null hypothesis to conclude that the piercer seems to be effective to prevent egg’s damage during boiling.