Pricing Analysis Example
Yang Zhang
Analysis Purpose
(1) Which type of advertisement (natural vs family healthe caring) is more effective?
(2) What’s the price elasticity (the reaction of sales quantity of the grape juice to its own price change)?
(3) What’s the cross-price elasticity (the reaction of sales quantity of the grape juice to price change of other products)?
(4) What’s the unit price of the grape juice that can maximum the profit?
Load Data
setwd('~/Desktop/R/Marketing')
grape_sale.df <- read.csv('grapeJuice.csv', colClasses = c(ad_type = "factor"))
summary(grape_sale.df)## sales_grape price_grape ad_type price_apple price_cookies
## Min. :131.0 Min. : 8.200 0:15 Min. :7.300 Min. : 8.790
## 1st Qu.:182.5 1st Qu.: 9.585 1:15 1st Qu.:7.438 1st Qu.: 9.190
## Median :204.5 Median : 9.855 Median :7.580 Median : 9.515
## Mean :216.7 Mean : 9.738 Mean :7.659 Mean : 9.622
## 3rd Qu.:244.2 3rd Qu.:10.268 3rd Qu.:7.805 3rd Qu.:10.140
## Max. :335.0 Max. :10.490 Max. :8.290 Max. :10.580From the summary table, we could see that the mean sales of grape juice is 216.7 while the mean price of grape juice is $9.7
Check the outliers and distribution of sales data
par(mfrow=c(1,2))
boxplot(grape_sale.df$sales_grape,main="Boxplot for sales data", xlab="sales_grape")
hist(grape_sale.df$sales_grape,main='histogram plot for sales data',xlab='sales_grape',prob=T)
lines(density(grape_sale.df$sales_grape),lty='dashed',lwd=2.5)
From the boxplot, we don’t find outliers; For the histogram plot, the distribution for sales of grape juice is roughly normal, so we don’t need to transform the data
Ad Effectiveness Analysis
Determine the Ad effectiveness is simple, just compare the mean sales of two Ad and conducted t-tests
Sales_Ad_Nature <- grape_sale.df[grape_sale.df$ad_type==0,]
Sales_Ad_Family <- grape_sale.df[grape_sale.df$ad_type==1,]
mean(Sales_Ad_Nature$sales_grape)## [1] 186.6667mean(Sales_Ad_Family$sales_grape)## [1] 246.6667shapiro.test(Sales_Ad_Nature$sales_grape)##
## Shapiro-Wilk normality test
##
## data: Sales_Ad_Nature$sales_grape
## W = 0.94255, p-value = 0.4155shapiro.test(Sales_Ad_Family$sales_grape)##
## Shapiro-Wilk normality test
##
## data: Sales_Ad_Family$sales_grape
## W = 0.89743, p-value = 0.08695t.test(Sales_Ad_Nature$sales_grape,Sales_Ad_Family$sales_grape)##
## Welch Two Sample t-test
##
## data: Sales_Ad_Nature$sales_grape and Sales_Ad_Family$sales_grape
## t = -3.7515, df = 25.257, p-value = 0.0009233
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -92.92234 -27.07766
## sample estimates:
## mean of x mean of y
## 186.6667 246.6667Conclusion: the mean sale with Family Care Ad is higher than Nature Ad. Both shapiro test for Nature Ad sales and Family Care Ad sales show that the sales data are distrubuted normaly. We could use t-test and the t-test shows that the difference is significant (P<0.05)
Price Elasticity Analysis
Use multiple linear regression to determine the Sales Drivers
library(s20x)## Warning: package 's20x' was built under R version 3.2.5pairs(grape_sale.df, pch=20)
pairs20x(grape_sale.df)
From the above figure, we could see that the correlation coefficients between grape juice sales and its price, AD’s type, apple juice price, and cookies price are 0.85, 0.58, 0.37, and 0.37 respectively. They are likely to be the sales drivers, so we will add of the them as independent variabels.
sales_grape.lm <- lm(sales_grape~price_grape+ad_type+price_apple+price_cookies, grape_sale.df)
summary(sales_grape.lm)##
## Call:
## lm(formula = sales_grape ~ price_grape + ad_type + price_apple +
## price_cookies, data = grape_sale.df)
##
## Residuals:
## Min 1Q Median 3Q Max
## -36.290 -10.488 0.884 10.483 29.471
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 774.813 145.349 5.331 1.59e-05 ***
## price_grape -51.239 5.321 -9.630 6.83e-10 ***
## ad_type1 29.742 7.249 4.103 0.000380 ***
## price_apple 22.089 12.512 1.765 0.089710 .
## price_cookies -25.277 6.296 -4.015 0.000477 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 18.2 on 25 degrees of freedom
## Multiple R-squared: 0.8974, Adjusted R-squared: 0.881
## F-statistic: 54.67 on 4 and 25 DF, p-value: 5.318e-12We could conclude that grape juice price, ad_type, and cookies price all significantly affect the sales of grape juice.
For price elasticity (percent change in quantity divided by the percent change in price), PE = (ΔQ/Q) / (ΔP/P) = (ΔQ/ΔP) * (P/Q) = -51.24 * 0.045 = -2.3; it suggests that 1% decrease in price will increase the sales by 2.3%
For cross-price elasticity (percent change in quantity divided by the change in the price of some other product), CPEapple = (ΔQ/ΔPapple) * (Papple/Q) = 22.1 * ( 7.659 / 216.7) = 0.78, its suggests that 1% decrease in apple juice price will decrease the sales of grape juice by 0.78%; so the applie juice are substitutes of grape juice; CPEcookies = (ΔQ/ΔPcookies) * (Pcookies/Q) = -25.28 * ( 9.622 / 216.7) = – 1.12; it suggests that 1% decrease in cookies price will increase the grape juice sales by 1.12%, and cookies are compliments for grape juice.
Optimal Pricing
First, simplify the question by setting ad_type=1, price_apple as 7.66 (mean value), price_cookies=9.74 (mean value), then sale_grape=772.64 - 51.24price, assume that the cost puer unit of grape juice is $5, then the Profit (Y) Y= (price_grape-cost)(772.64-51.24price)
f <- function(x) {(-51.2)*x^2+1028.84*x-3863.2}
curve(f, from=0, to =20, ylab="Profit")
optimize(f, lower=0, upper=20, maximum=TRUE)## $maximum
## [1] 10.04727
##
## $objective
## [1] 1305.314We may conclude that the price of $10.04 will product the maximum profit
Reference http://www.dataapple.net/?p=19