Hotel Pricing Project
Aditya Satheesh
29 December 2017
Reading the dataset and describing the dimensions
cities <- read.csv(paste("Cities42.csv", sep=""))
View(cities)
dim(cities)## [1] 13232 19Describing essesntial statistics
library(psych)
describe(cities)[,c(1:5)]## vars n mean sd median
## CityName* 1 13232 18.07 11.72 16
## Population 2 13232 4416836.87 4258386.00 3046163
## CityRank 3 13232 14.83 13.51 9
## IsMetroCity 4 13232 0.28 0.45 0
## IsTouristDestination 5 13232 0.70 0.46 1
## IsWeekend 6 13232 0.62 0.48 1
## IsNewYearEve 7 13232 0.12 0.33 0
## Date* 8 13232 14.30 2.69 14
## HotelName* 9 13232 841.19 488.16 827
## RoomRent 10 13232 5473.99 7333.12 4000
## StarRating 11 13232 3.46 0.76 3
## Airport 12 13232 21.16 22.76 15
## HotelAddress* 13 13232 1202.53 582.17 1261
## HotelPincode 14 13232 397430.26 259837.50 395003
## HotelDescription* 15 13224 581.34 363.26 567
## FreeWifi 16 13232 0.93 0.26 1
## FreeBreakfast 17 13232 0.65 0.48 1
## HotelCapacity 18 13232 62.51 76.66 34
## HasSwimmingPool 19 13232 0.36 0.48 0Variable Analysis
One way contingency tables
mytable <- with(cities,table(CityName))
mytable## CityName
## Agra Ahmedabad Amritsar Bangalore
## 432 424 136 656
## Bhubaneswar Chandigarh Chennai Darjeeling
## 120 336 416 136
## Delhi Gangtok Goa Guwahati
## 2048 128 624 48
## Haridwar Hyderabad Indore Jaipur
## 48 536 160 768
## Jaisalmer Jodhpur Kanpur Kochi
## 264 224 16 608
## Kolkata Lucknow Madurai Manali
## 512 128 112 288
## Mangalore Mumbai Munnar Mysore
## 104 712 328 160
## Nainital Ooty Panchkula Pune
## 144 136 64 600
## Puri Rajkot Rishikesh Shimla
## 56 128 88 280
## Srinagar Surat Thiruvanthipuram Thrissur
## 40 80 392 32
## Udaipur Varanasi
## 456 264 mytable1 <- with(cities,table(FreeWifi))
mytable1## FreeWifi
## 0 1
## 981 12251 mytable2 <- with(cities,table(FreeBreakfast))
mytable2## FreeBreakfast
## 0 1
## 4643 8589 mytable3 <-with(cities,table(HasSwimmingPool))
mytable3## HasSwimmingPool
## 0 1
## 8524 4708Two Way Contingency tables
mytable4<-xtabs(~StarRating+FreeWifi,data=cities)
mytable4## FreeWifi
## StarRating 0 1
## 0 0 16
## 1 0 8
## 2 80 360
## 2.5 104 528
## 3 336 5617
## 3.2 0 8
## 3.3 0 16
## 3.4 0 8
## 3.5 96 1656
## 3.6 0 8
## 3.7 0 24
## 3.8 0 16
## 3.9 0 32
## 4 231 2232
## 4.1 0 24
## 4.3 0 16
## 4.4 0 8
## 4.5 24 352
## 4.7 0 8
## 4.8 0 16
## 5 110 1298 mytable5<-xtabs(~StarRating+FreeBreakfast,data=cities)
mytable5## FreeBreakfast
## StarRating 0 1
## 0 16 0
## 1 0 8
## 2 216 224
## 2.5 296 336
## 3 1789 4164
## 3.2 0 8
## 3.3 8 8
## 3.4 0 8
## 3.5 661 1091
## 3.6 8 0
## 3.7 0 24
## 3.8 8 8
## 3.9 16 16
## 4 783 1680
## 4.1 0 24
## 4.3 16 0
## 4.4 0 8
## 4.5 224 152
## 4.7 8 0
## 4.8 0 16
## 5 594 814Boxplots
par(mfrow=c(1,3))
boxplot(cities$StarRating,beside=TRUE,ylab="Star Rating",col="pink")
boxplot(cities$Airport,beside=TRUE,ylab="Distance to airport",col="pink")
boxplot(cities$HotelCapacity,beside=TRUE,ylab="Hotel Capacity",col="pink")
Histograms and Plots of Variables
hist(cities$Population,col="blue",main="Population",xlab="population")
cities$IsMetroCity=factor(cities$IsMetroCity, levels=c(0,1), labels=c("No","Yes"))
plot(cities$IsMetroCity,col="green",main="Metro City")
hist(cities$StarRating,col="blue",main="Star Rating of Hotels",xlab="Star Rating")
cities$FreeWifi=factor(cities$FreeWifi, levels=c(0,1), labels=c("No","Yes"))
plot(cities$FreeWifi,col="green",main="Has Wifi?")
cities$FreeBreakfast=factor(cities$FreeBreakfast, levels=c(0,1), labels=c("No","Yes"))
plot(cities$FreeBreakfast,col="green",main="Has Free Breakfast?")
cities$HasSwimmingPool=factor(cities$HasSwimmingPool, levels=c(0,1), labels=c("No","Yes"))
plot(cities$HasSwimmingPool,col="green",main="Swimming pool?")
hist(cities$HotelCapacity,col="blue",main="Hotel Capacities",breaks=20,xlab="capacity")
Correlation Matrix
x<-cities[,c("Population", "RoomRent","StarRating","Airport","HotelCapacity")]
y<-cities[,c("Population", "RoomRent","StarRating","Airport","HotelCapacity")]
cor(x,y,method="pearson")## Population RoomRent StarRating Airport
## Population 1.00000000 -0.08872806 0.13413659 -0.25970102
## RoomRent -0.08872806 1.00000000 0.36937343 0.04965324
## StarRating 0.13413659 0.36937343 1.00000000 -0.06091918
## Airport -0.25970102 0.04965324 -0.06091918 1.00000000
## HotelCapacity 0.25998305 0.15787331 0.63743034 -0.11767207
## HotelCapacity
## Population 0.2599831
## RoomRent 0.1578733
## StarRating 0.6374303
## Airport -0.1176721
## HotelCapacity 1.0000000Corrgram
library(corrgram)
corrgram(cities, order=TRUE, lower.panel=panel.shade,
upper.panel=panel.pie, text.panel=panel.txt,
main="Hotel Pricing")
Scatter Plots
library(car)
scatterplot(RoomRent~StarRating, data=cities,
spread=FALSE, smoother.args=list(lty=2),
main="Scatter plot of Star Rating vs Room rent",
ylab="Room Rent",
xlab="Star Rating")
library(car)
scatterplot(HotelCapacity~StarRating, data=cities,
spread=FALSE, smoother.args=list(lty=2),
main="Scatter plot of Star Rating vs Hotel Capacity",
ylab="Room Rent",
xlab="Hotel Capacity")
Variance - Covariance Matrix
cc <-cbind(cities[,c(10,11,12,18)])
cov(cc)## RoomRent StarRating Airport HotelCapacity
## RoomRent 53774601.806 2048.3754792 8287.178584 88753.41284
## StarRating 2048.375 0.5718875 -1.048528 36.95522
## Airport 8287.179 -1.0485276 518.013328 -205.32017
## HotelCapacity 88753.413 36.9552206 -205.320172 5877.26810Relation between Pricing and various independent variables
cor.test(cities$RoomRent,cities$StarRating) ##
## Pearson's product-moment correlation
##
## data: cities$RoomRent and cities$StarRating
## t = 45.719, df = 13230, p-value < 2.2e-16
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## 0.3545660 0.3839956
## sample estimates:
## cor
## 0.3693734Since p value is 2.2e-16 we can reject the null hypothesis that the correlation between pricing and star rating is not 0.
cor.test(cities$RoomRent,cities$Airport)##
## Pearson's product-moment correlation
##
## data: cities$RoomRent and cities$Airport
## t = 5.7183, df = 13230, p-value = 1.099e-08
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## 0.03264192 0.06663581
## sample estimates:
## cor
## 0.04965324Since p value is 1.099e-08 we can reject the null hypothesis that the correlation between pricing and distance to airport is not 0.
cor.test(cities$RoomRent,cities$HotelCapacity)##
## Pearson's product-moment correlation
##
## data: cities$RoomRent and cities$HotelCapacity
## t = 18.389, df = 13230, p-value < 2.2e-16
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## 0.1412142 0.1744430
## sample estimates:
## cor
## 0.1578733Since p value is less than 2.2e-16 we can reject the null hypothesis that the correlation between pricing and hotel capacity is not 0.
T-test
Let the null hypothesis be that the RoomRent depends on the other factors(Star rating, Airport distance, Hotel Capcity)
t.test(cities$RoomRent,cities$StarRating)##
## Welch Two Sample t-test
##
## data: cities$RoomRent and cities$StarRating
## t = 85.813, df = 13231, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 5345.575 5595.491
## sample estimates:
## mean of x mean of y
## 5473.991838 3.458933Since p value lower than 0.05 we can reject the null hypothesis
t.test(cities$RoomRent,cities$Airport)##
## Welch Two Sample t-test
##
## data: cities$RoomRent and cities$Airport
## t = 85.535, df = 13231, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 5327.875 5577.792
## sample estimates:
## mean of x mean of y
## 5473.99184 21.15874Since p value lower than 0.05 we can reject the null hypothesis
t.test(cities$RoomRent,cities$HotelCapacity)##
## Welch Two Sample t-test
##
## data: cities$RoomRent and cities$HotelCapacity
## t = 84.882, df = 13234, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 5286.515 5536.445
## sample estimates:
## mean of x mean of y
## 5473.99184 62.51164Since p value lower than 0.05 we can reject the null hypothesis