Chi-Square test of independence

A/B Testing in R

Lauryn Burleigh

Data Scientist

Frequency tables

Whether nominal variables are related
- Eating the pizza topping again
Frequency table - number of times each of the named conditions co-occurs
- One-topping pizza: variable 1
- Would eat again: variable 2

table(Pizza$Topping, Pizza$EatAgain)

             Maybe  No Yes
  Cheese       207  90 703
  Pepperoni    147 101 752

Hypotheses

Null hypothesis: Topping and interest in eating again have no relationship
Alternative hypothesis: Topping and interest in eating again have a relationship

Cannot conclude cause

Photograph of pizza slices.

Sample size

library(pwr)
pwr.chisq.test(w = 0.1, df = 2, 
               power = 0.80, 
               sig.level = 0.05)

     Chi squared power calculation 
              w = 0.1
              N = 963.4689
             df = 2
      sig.level = 0.05
          power = 0.8
NOTE: N is the number of observations

w:
- Small: 0.10
- Medium: 0.30
- Large: 0.50
Degrees of freedom:
- (Conditions-of-variable-1 - 1) * (Conditions-of-variable-2 - 1)
- (2-1) * (3-1) = 2

Minimum of five values at each variable combination

Chi-square test

freqtbl <- table(Pizza$Topping, 
                 Pizza$EatAgain) 
chisq.test(freqtbl)

    Pearson's Chi-squared test

data:  freqtbl
X-squared = 12.453, df = 2, 
p-value = 0.001976

Large sample sizes: over 1000

Effect size

Effect size: w
Probabilities

probtbl <- freqtbl/sum(freqtbl) 

library(pwr)
ES.w2(probtbl)

Small: 0.10
Medium: 0.30
Large: 0.50

[1] 0.07890872

Power

pwr.chisq.test(w = 0.079, df = 2, 
               N = 1000, 
               sig.level = 0.002)

     Chi squared power calculation 

              w = 0.079
              N = 1000
             df = 2
      sig.level = 0.002
          power = 1

NOTE: N is the number of observations

Let's practice!

A/B Testing in R