Introduction

In the global athletic shoe market Nike and Adidas have been maintaining the world championship in sport’s shoes for a few decades because of their good designs, quality and the different features which they have. In present they produce apparel, accessories, equipment apart from the athletic footwear under their competitive brands. The objective of this report is to compare the cost price of Nike and Adidas products on a selected range to see who is dealing better with regard to purchasing power.

Problem Statement

Cost price is one of the main factor of deciding the selling rice of a product. Many companies try to keep cost price down all the time. Although the cost price of a product is not a important factor for consumers, it ultimately effects the price the of the products. We are looking at cost prices of brands of Nike and Adidas said period of the particular state and category to see, which brand is doing better with the cost price.

Data

Data Source: https://www.kaggle.com/australiastats/australia-top-sport-chains
About data file: The dataset consists of data about Australia top sports brand - Nike and Adidas. It contains postcode wise information about their sales statistics in Australia. Also, it includes the information about their cost price and sale price, number of units were sold under each category such as men, women & kids etc. and each of the transactions done day by day for the period of 2016-2018.

The original dataset includes the nine variables and 76466 rows. But for our statistical purpose, we have selected Kids category in the state of Victoria which included 2479 records.

Description of Variables

The original dataset includes the nine variables and 76466 rows. The variables are as follows:
Date- The sales dates represent from 01/01/2016 to 01/08 /2017.
Chain – Represent the two brand names Nike and Adidas.
Post code – Post code numbers relevant to the states in Australia.
Category -All the products classify under ten categories (Accessories, Equipment , Groceries, Home, Hosiery, Juniors, Kids , Men , Shoes ,Women).
Total unit - number of units were sold during each day under each category and each postcode.
Sale price – Price of both brands in Australian dollars.
Cost price - Cost of unit prices of each product in Australian dollars.

Types of Variables

str(VICKids)

## 'data.frame':    2479 obs. of  9 variables:
##  $ Date      : Date, format: "2016-01-01" "2016-01-01" ...
##  $ Chain     : Factor w/ 2 levels "Nike","Adidas": 1 1 1 1 2 2 2 2 2 2 ...
##  $ Postcode  : num  3550 3018 3550 3018 3429 ...
##  $ Category  : Factor w/ 1 level "Kids": 1 1 1 1 1 1 1 1 1 1 ...
##  $ Units     : num  65 68 50 17 19 1 1 2 2 1 ...
##  $ Sale_Price: num  2.19 1.95 2.74 2.47 3.47 1 1 1 1 4 ...
##  $ Cost_Price: num  3.92 2.75 3.18 3.28 3.95 3.25 3.25 2.11 1.8 5.4 ...
##  $ State     : chr  "VIC" "VIC" "VIC" "VIC" ...
##  $ Suburb    : chr  "Bendigo" "Altona" "Bendigo" "Altona" ...

head(VICKids)

##         Date  Chain Postcode Category Units Sale_Price Cost_Price State
## 1 2016-01-01   Nike     3550     Kids    65       2.19       3.92   VIC
## 2 2016-01-01   Nike     3018     Kids    68       1.95       2.75   VIC
## 3 2016-01-01   Nike     3550     Kids    50       2.74       3.18   VIC
## 4 2016-01-01   Nike     3018     Kids    17       2.47       3.28   VIC
## 5 2016-01-01 Adidas     3429     Kids    19       3.47       3.95   VIC
## 6 2016-01-01 Adidas     3630     Kids     1       1.00       3.25   VIC
##       Suburb
## 1    Bendigo
## 2     Altona
## 3    Bendigo
## 4     Altona
## 5    Sunbury
## 6 Shepparton

sum(is.na(VICKids))

## [1] 0

Descriptive Statistics and Visualisation

VICKids %>% group_by(Chain) %>% summarise(Min = min(Cost_Price, na.rm = TRUE),
                                               Q1 = quantile(Cost_Price, probs = 0.25, na.rm = TRUE),
                                               Median = median(Cost_Price, na.rm = TRUE),
                                               Q3 = quantile(Cost_Price, probs = 0.75, na.rm = TRUE),
                                               Max = max(Cost_Price, na.rm = TRUE),
                                               IQR = Q3 - Q1,
                                               Mean = mean(Cost_Price, na.rm = TRUE),
                                               SD = sd(Cost_Price, na.rm = TRUE),
                                               n = n(),
                                               Missing = sum(is.na(Cost_Price)))

## # A tibble: 2 x 11
##   Chain    Min    Q1 Median    Q3   Max   IQR  Mean    SD     n Missing
##   <fct>  <dbl> <dbl>  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <int>   <int>
## 1 Nike    0.35  1.79   2.58  3.2     12  1.41  2.63  1.19  1524       0
## 2 Adidas  0.5   2.52   3.3   4.11    15  1.58  3.44  1.47   955       0

The mean cost price of Adidas(3.44) is higher than the mean cost price of Nike(2.63)
The Standard deviation of adidas (1.47) also higher than the Standard deviation of Nike(1.19)

Histograme of cost price for both brands

VICKids %>% histogram(~Cost_Price | Chain, data = .,layout=c(1,2), main= 'Comparison of Price Distribution')

Histograms of cost price is right skewed for both brands

Hitograme of cost price for Nike

NikeVICKids$Cost_Price %>% hist(col = "blue",
                                ylim = c(0,0.4),
                                xlim = c(0,20),
                                xlab = "Cost price($)",
                                main = "cost price of Nike kid's products",
                                breaks = 10,
                                density = 20,
                                prob = TRUE)
lines(density(NikeVICKids$Cost_Price, adjust = 2), col = "red", lwd = 2)

Histogram of cost price for Adidas

AdidasVICKids$Cost_Price %>% hist(col = "blue",
                                  ylim = c(0,0.4),
                                  xlim = c(0,20),
                                  xlab = "Cost price($)",
                                  main = "cost price of Adidas kid's products",
                                  breaks = 10,
                                  density = 20,
                                  prob = TRUE)
lines(density(AdidasVICKids$Cost_Price, adjust = 2), col = "red", lwd = 2)

Box plots for both brands

boxplot(VICKids$Cost_Price ~ VICKids$Chain, main="Cost price of Nike and Adidas Kid's Products", ylab = "Cost Price", xlab = "Chain")

Hypothesis Testing

In the Q-Q plot, the dotted arcs correspond to 95% CI for the normal quantiles, points should be falling inside the tails of the distribution.
Here, many points are falling outside the tails of the distribution.
Therefore, Q-Q plots show non-linear trends and strange ‘S’ -shape of Non-normality.
But sample size is very large 2479 records. So we will assuming it has normality.

leveneTest(Cost_Price ~ Chain, data = VICKids)

## Levene's Test for Homogeneity of Variance (center = median)
##         Df F value    Pr(>F)    
## group    1  29.544 6.002e-08 ***
##       2477                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Hypthesis Testing

Homogeneity of Variance using the Levene’s test , assumption of equal variance can be tested .
* statistical hypotheses: H0: Null Hypothesis: σ21= σ22

H1: Alternative Hypothesis:σ21 != σ22

Levene’s Test for Homogeneity of Variance (center = median) Df F value Pr(>F)
group 1 29.544 6.002e-08

The p-value for the Levene’s test of equal variance for cost price between Nike and Adidas is 6.002e-08.It is < 0.05 and equal variance assumption is violated.

*Therefore, Levene’s test is statistically significant and H0 reject

Welch two-sample t-test to check the population mean of two samples

t.test(
  Cost_Price ~ Chain,
  data = VICKids,
  var.equal = FALSE,
  alternative = "two.sided"
)

## 
##  Welch Two Sample t-test
## 
## data:  Cost_Price by Chain
## t = -14.361, df = 1714.6, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.9224804 -0.7007888
## sample estimates:
##   mean in group Nike mean in group Adidas 
##             2.626575             3.438209

H0: Null Hypothesis: μ1 - μ2= 0

H1: Alternative Hypothesis:μ1 - μ2 !=0

Welch Two Sample t-test

data: Cost_Price by Chain t = 14.361, df = 1714.6, p-value < 2.2e-16 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 0.7007888 0.9224804 sample estimates: mean in group Adidas mean in group Nike 3.438209 2.626575

According to the results of the Welch Two Sample t-test,P<0.05 (2.2e-16).It is statistically significant and reject H0
Therefore there is a significant difference between the mean values of kids products of Nike and Adidas.

Discussion

By doing analysis on the whole dataset containing different categories and products that we collected, there is not enough evidence to prove that either one of the Chain is costly or cheaper. Considering the dataset that we have collected summary statistics shows that both mean cost price and the standard deviation of the adidas kid’s products higher than the Nike kid’s products and histograms depicts that , cost price is right skewed for both brands Boxplot depicts that Adidas cost price is slighty expensive than Nike.

based on the results of the Welch Two Sample t-test, there is a significant difference between the mean values of kids products of Nike and Adidasand it is found that Adidas is slightly expensive than Nike

These results are based on the dataset that we selected for the investigation and it may vary if the categories and products are increased .

Cost Price competition of Nike and Adidas

RPubs link information