Two-way ANOVA using R

ANOVA CHEATSHEET (Applies to both One-way and Two-way)
  • BUSINESS PROBLEM – Test of Means across groups(1 or more factors)
  • H0: Mean1 = Mean2 ….= Mean(One-way – only 1 Mean, Two-way – max 3 Mean if interaction, or min 2 Means if no interaction)
  • ASSUMPTIONS:
    • Samples are independent and random
    • Groups/Levels are normally distributed using Shapiro test 
      • For Shapiro test, 
        • H0: Data/Group/Levels is normal
        • Ha: Data is not normal
        • Favourable: P > 0.05 because we do not want to reject NULL hypothesis
      • For n levels, we do Shapiro test n times
    • Homogenity of variances using LEVENE’S test
      • For LEVENE’s test, 
        • H0: Sigma1pow2 = Sigma2pow2… = Sigmanpow2
        • Ha = Atleast one is different
        • Favourable: P > 0.05
      • For n factors, we do LEVENE’s test n times
  • F-stat > F-crit => P-value < 0.05 => REJECT H0
  • F-stat < F-crit => P-value > 0.05 => DO NOT REJECT H0
  • If H0 is rejected, TukeyHSD test is done. 
  • In TukeyHSD test, pairwise testing of means is done between each levels. 
    • H0: For three levels, Mu1 = Mu2, Mu1 = Mu3, Mu2 = Mu3
  •  ADDITIONAL Conditions
    • With Interaction: ‘*’
    • Without Interaction: ‘+’
  • For Anova, the first condition is data/levels should be balanced
SCENARIO
Consider the following sample Dataset showing sales figures of different stores using combination of coupon and promotion:

NULL Hypothesis H0: 
  • For Coupon: Level1 Mean = Level2 Mean and 
  • For Promotion: Level1 Mean = Level2 Mean = Level3 Mean
  • NULL hypothesis is that there is no interaction effect
SOLUTION
Open R Studio and enter the following code selecting dataset stored in .csv file

library(psych)
library(car)

#install.packages("psych")

## Summary Statistics
## Coupon and Instore Promotion Data
df<-read.csv(file.choose())

df$promotion; df$coupon;

#Step 2: Clearly identify the factors in the data
df$promotion<-factor(df$promotion, labels=c("1","2","3"))
df$coupon<-factor(df$coupon, labels=c("1","2"))

df$promotion; df$coupon;


###Calculate the means sales for promotion and coupon
tapply(df$sales, list(df$promotion, df$coupon), mean)

## Note: High Promotion and high coupon gives relatively higher sales values
## As the intensity of promotion and coupons are reduced, sales reduces accordingly

###Create a plot to identify the interaction effects between promotion & coupon
interaction.plot(df$promotion, df$coupon, df$sales)

##Both lines are almost parallel hence we can say interaction effect is negligible

## Test for normality for all the 3 groups in promotion
cat("Normality p-values by Factor Place: ")
for (i in unique(factor(df$promotion))){
  cat(shapiro.test(df[df$promotion==i, ]$sales)$p.value," ")
}
####P values are greater than 0.05, hence we do not reject the null hypothesis

###Run the normality assumption test for both groups in coupon
cat("Normality p-values by Factor Place: ")
for (i in unique(factor(df$coupon))){
  cat(shapiro.test(df[df$coupon==i, ]$sales)$p.value," ")
}
####P values are greater than 0.05, hence we do not reject the null hypothesis

###Create a qqplot to check the normality of entire dataset visually
qqnorm(df$sales, pch=19, cex=0.6)
qqline(df$sales, col = 'red')
###From the image, we can see the entire dataset is normally distributed

######## Levene's test for variance
## Test for homogeneity of variance
####Testing variance visually using a boxplot
boxplot(df$sales~df$promotion)
###From the boxplot, we see that 2, 3 have almost same variance .

####Check for variances in groups using Levene's test
leveneTest(df$sales~df$promotion)
leveneTest(df$sales~df$coupon)
####As all the p values are greater than 0.05, we do not reject the null hypothesis

##################################################################################
####ANOVA based on promotion
aov1 <- aov(df$sales~df$promotion)
summary(aov1)
##p value is less than 0.05, hence we reject the null hypothesis

TukeyHSD(aov1)

plot(TukeyHSD(aov1))
###We reject the null hypothesis as all the intervals do not contain zero within them

###Business Intuiton
###High, medium & low are boosting sales in some manner
###Jump from low to high gives a sales boost of 4.6
###Jump from medium to high gives a sales boost of 2.1
###If the costs of doing a high promotion are less than the above boosted sales values
### We can conclude saying that all stores can run the high promotion for better sales

###However, if the costs of doing high promotion are more than the boosted sales figures
### We can actually drop the high promotion stores to medium or even low

##################################################################################
##################################################################################
####ANOVA based on coupon
aov1 <- aov(df$sales~df$coupon)
summary(aov1)
####p value is less than 0.05, hence we reject the null hypothesis

TukeyHSD(aov1)
plot(TukeyHSD(aov1))

###Since the intervals do not contain zero, it means there is a significant difference
###In Business terms, I can offer higher coupons in all stores if my net profit increases
###We would need to calculate the net increase in profit using the sales boost of 2.666667


REFERENCES

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