# Predictive Analytics on Chocolate Data

### This script shows only the application of predictive statistics
### The datasets are the final chocolateData and summary dataframes

### Load datafiles into datasets

chocolateData<-read.csv("ChocolateDataAnalysis.csv",stringsAsFactors = TRUE)
summary<-read.csv("SummaryData.csv",stringsAsFactors = TRUE)





### Scatterplot and Correlation ###

# Create a scatterplot of number of chocolates eaten vs. minimum survival time

plot(x=summary$Eaten, y=summary$MinSurvive, xlab="Number eaten", ylab="Minimum survival time",
     main="Number eaten vs. Minimum survival time")

# Calculate correlation between number of chocolates eaten and minimum survival time

# Create a data frame to analyse

dataObs<-data.frame(NumberEaten=summary$Eaten, MinSurvival=summary$MinSurvive)

# Calculate Pearson's correlation

corData<-cor(dataObs,method="pearson")





### Statistical Analyses

# t-test: Survival time for different types of chocolates

# Extract chocolateData$SurvivalTime for each type of chocolate

Roses<-chocolateData$SurvivalTime[chocolateData$Chocolate=="Roses"]
QStreet<-chocolateData$SurvivalTime[chocolateData$Chocolate=="Quality Street"]

# Is each data sample normal? Use Shapiro test to find out

shapiro.test(Roses)
shapiro.test(QStreet)

# Both of these show as p<0.05, so they are not normally distributed
# Therefore create boxplots to see where the outliers are 

boxplot(Roses, main="Roses")
boxplot(QStreet, main="QStreet")

# Roses outliers are above 200, so filter the data sample to remove them
# NB: This may take several iterations to remove all outliers

Roses<-Roses[Roses<200]
boxplot(Roses, main="Roses")

# QStreet outliers are above 200, so filter the data sample to remove them
# NB: This may take several iterations to remove all outliers

QStreet<-QStreet[QStreet<200]
boxplot(QStreet, main="QStreet")

# Perform t-test for independent samples

# Now normal, so can do the t-test

# HO: There is no difference between the means, i.e. both chocolates have the same survival time
# H1: There is a difference between the means, i.e. one type lasts longer than the other

t.test(Roses, QStreet)

# The calculated p-value for this is 0.7967. This is more than 0.05,
# so we cannot reject the null hypothesis at the p=0.05 level.
# We therefore accept the null hypothesis that the type of chocolate 
# makes no difference to the mean survival time.

# Difference between chocolates eaten by doctors and those eaten by nurses

# t-test: Survival time for different types of chocolates

# Extract chocolateData$SurvivalTime for each type of chocolate

Doctors<-chocolateData$SurvivalTime[chocolateData$WhoAte=="Doctor"]
Nurses<-chocolateData$SurvivalTime[chocolateData$WhoAte=="Nurse"]

# Is each data sample normal? Use Shapiro test to find out

shapiro.test(Doctors)
shapiro.test(Nurses)

# Both of these show as p<0.05, so they are not normally distributed
# Therefore create boxplots to see where the outliers are 

boxplot(Doctors, main="Doctors")
boxplot(Nurses, main="Nurses")

# Doctors outliers are above 200, so filter the data sample to remove them
# NB: This may take several iterations to remove all outliers

Doctors<-Doctors[Doctors<200]
boxplot(Doctors, main="Doctors")

# Nurses outliers are above 300, so filter the data sample to remove them
# NB: This may take several iterations to remove all outliers

Nurses<-Nurses[Nurses<300]
boxplot(Nurses, main="Nurses")

# Perform t-test for independent samples

# Now normal, so can do the t-test

# HO: There is no difference between the means, i.e. doctors and nurses eat chocolates at similar times
# H1: There is a difference between the means, i.e. doctors and nurses eat chocolates at different times

t.test(Doctors,Nurses)

# The calculated p-value for this is 0.7624. This is more than 0.05,
# so we cannot reject the null hypothesis at the p=0.05 level.
# We therefore accept the null hypothesis that doctors and nurses eat chocolates
# at about the same times





# ANOVA

# Differences between doctors, nurses, and administrators

# H0: There is no difference between doctors, nurses, and administrators
# H1: There is a significant difference between at least two of the groups

# Filter the dataSet to create a new dataframe with just the survival time and the person who ate the chocolate

anovaData<-chocolateData[,c(5,10)]
anovaData

# Filter the new dataset to extract only chocolates eaten by doctors, nurses and administrators
anovaData<-anovaData[anovaData$WhoAte %in% c("Doctor","Nurse","Administrative"),]
anovaData

# Carry out ANOVA test

# Compute the analysis of variance
resaov <- aov(anovaData$SurvivalTime ~ anovaData$WhoAte, data = anovaData)

# Summary of the analysis
summary(resaov)

#                    Df    Sum Sq   Mean Sq   F value   Pr(>F)  
# anovaData$WhoAte     2    45814     22907     3.382   0.0378 *
# Residuals          103   697674      6774                 
#  ---
#  Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

# There is a significant difference between groups at the p=0.05 level
# We can reject the null hypothesis and accept the alternative hypothesis
# It is therefore necessary to perform post-hoc tests to find out where these differences occur

# Perform Tukey Pair-wise post-hoc analysis

TukeyHSD(resaov)

#Tukey multiple comparisons of means
#95% family-wise confidence level

#Fit: aov(formula = anovaData$SurvivalTime ~ anovaData$WhoAte, data = anovaData)

#$`anovaData$WhoAte`
#                       diff        lwr       upr     p adj
#Doctor-Administrative  58.65367   4.006131 113.30122 0.0323376
#Nurse-Administrative   24.73349 -23.998754  73.46574 0.4517825
#Nurse-Doctor          -33.92018 -78.978504  11.13815 0.1779281

# The Tukey test shows that there is a significant difference at the p=0.05 level
# between doctors and administrators, but not between other pairs of groups