Hotels data - NULL Hypotheses
Reshu
2024-02-25
Initial setup and Configure the data set. Load the data set file
in variable hotel_data files.
Data set - Hotels : This data comes
from an open hotel booking demand data-set from Antonio, Almeida and
Nunes.
NULL Hypotheses
Ask :-
Devise at least two different null hypotheses based on two
different aspects (e.g., columns) of your data. For each hypothesis:
- Come up with an alpha level, power level, and minimum effect size, and explain why you chose each value.
- Determine if you have enough data to perform a Neyman-Pearson hypothesis test. If you do, show your sample size calculation, perform the test, and interpret results. If not, explain why there isn’t enough data.
- Perform a Fisher’s style test for significance on the same hypothesis, and interpret the p-value.
- (In the end, you should have two hypothesis tests for each hypothesis, equating two four total tests.)
Summery of data set
summary(hotel_data)## hotel is_canceled lead_time arrival_date_year
## Length:119390 Min. :0.0000 Min. : 0 Min. :2015
## Class :character 1st Qu.:0.0000 1st Qu.: 18 1st Qu.:2016
## Mode :character Median :0.0000 Median : 69 Median :2016
## Mean :0.3704 Mean :104 Mean :2016
## 3rd Qu.:1.0000 3rd Qu.:160 3rd Qu.:2017
## Max. :1.0000 Max. :737 Max. :2017
##
## arrival_date_month arrival_date_week_number arrival_date_day_of_month
## Length:119390 Min. : 1.00 Min. : 1.0
## Class :character 1st Qu.:16.00 1st Qu.: 8.0
## Mode :character Median :28.00 Median :16.0
## Mean :27.17 Mean :15.8
## 3rd Qu.:38.00 3rd Qu.:23.0
## Max. :53.00 Max. :31.0
##
## stays_in_weekend_nights stays_in_week_nights adults
## Min. : 0.0000 Min. : 0.0 Min. : 0.000
## 1st Qu.: 0.0000 1st Qu.: 1.0 1st Qu.: 2.000
## Median : 1.0000 Median : 2.0 Median : 2.000
## Mean : 0.9276 Mean : 2.5 Mean : 1.856
## 3rd Qu.: 2.0000 3rd Qu.: 3.0 3rd Qu.: 2.000
## Max. :19.0000 Max. :50.0 Max. :55.000
##
## children babies meal country
## Min. : 0.0000 Min. : 0.000000 Length:119390 Length:119390
## 1st Qu.: 0.0000 1st Qu.: 0.000000 Class :character Class :character
## Median : 0.0000 Median : 0.000000 Mode :character Mode :character
## Mean : 0.1039 Mean : 0.007949
## 3rd Qu.: 0.0000 3rd Qu.: 0.000000
## Max. :10.0000 Max. :10.000000
## NA's :4
## market_segment distribution_channel is_repeated_guest
## Length:119390 Length:119390 Min. :0.00000
## Class :character Class :character 1st Qu.:0.00000
## Mode :character Mode :character Median :0.00000
## Mean :0.03191
## 3rd Qu.:0.00000
## Max. :1.00000
##
## previous_cancellations previous_bookings_not_canceled reserved_room_type
## Min. : 0.00000 Min. : 0.0000 Length:119390
## 1st Qu.: 0.00000 1st Qu.: 0.0000 Class :character
## Median : 0.00000 Median : 0.0000 Mode :character
## Mean : 0.08712 Mean : 0.1371
## 3rd Qu.: 0.00000 3rd Qu.: 0.0000
## Max. :26.00000 Max. :72.0000
##
## assigned_room_type booking_changes deposit_type agent
## Length:119390 Min. : 0.0000 Length:119390 Length:119390
## Class :character 1st Qu.: 0.0000 Class :character Class :character
## Mode :character Median : 0.0000 Mode :character Mode :character
## Mean : 0.2211
## 3rd Qu.: 0.0000
## Max. :21.0000
##
## company days_in_waiting_list customer_type adr
## Length:119390 Min. : 0.000 Length:119390 Min. : -6.38
## Class :character 1st Qu.: 0.000 Class :character 1st Qu.: 69.29
## Mode :character Median : 0.000 Mode :character Median : 94.58
## Mean : 2.321 Mean : 101.83
## 3rd Qu.: 0.000 3rd Qu.: 126.00
## Max. :391.000 Max. :5400.00
##
## required_car_parking_spaces total_of_special_requests reservation_status
## Min. :0.00000 Min. :0.0000 Length:119390
## 1st Qu.:0.00000 1st Qu.:0.0000 Class :character
## Median :0.00000 Median :0.0000 Mode :character
## Mean :0.06252 Mean :0.5714
## 3rd Qu.:0.00000 3rd Qu.:1.0000
## Max. :8.00000 Max. :5.0000
##
## reservation_status_date
## Length:119390
## Class :character
## Mode :character
##
##
##
## NULL Hypotheses 1: Upon “LEAD TIME/Cancellation”
- Lead_time : Number of days between booking and arrival.Cancellation : Cancel the booking for his/her stay in hotel.
- Aspect of data : The chance of cancellation id not greatly impacted by the lead time, or the number of days between the reservation and the arrival.
- NULL Hypothesis : There is no difference in cancellation rates between bookings with different lead times.
- Alpha Level (is the probability of rejecting the null hypothesis when it is true): 0.05. if the p-value is less than 0.05 then the null hypothesis is rejected.
- Power Level ( is the probability of rejecting the NULL Hypothesis when NULL hypothesis is false): 0.80 or grater. if it is more than or equal 80% then NULL hypothesis can rejected.
-
Minimum Effect Size:0.1
1. The minimum effect size implies the smallest difference or association that is practically meaningful or relevant in the context of the testing.
2. Choosing a minimum effect size of 0.1 implies that the testing aims to detect relatively small but still meaningful effects.
3. Selecting a minimum effect size of 0.1 indicates that it is sufficiently powered to detect even minute effects in the testing result. A smaller effect size necessitates a larger date set to achieve sufficient statistical power.
-
Explanation : Alpha level is a critical threshold used in null
hypothesis analysis. It represent maximum acceptable risk of making a
Type I error - false positive. 0.05 value is used commonly (balance
choose between the risk of Type I errors -false positives and Type II
errors -false negatives) for avoiding false positive.
Power is the probability of correctly rejecting the null hypothesis when there is a true effect ( avoiding a Type II error). 0.80 or higher is to ensure that their testing can detect real effects.A power of 0.80 means an 80% chance of detecting a true effect.
The minimum effect size quantifies the magnitude of the difference or association between groups or variables and value of Minimum Effect Size - 0.1 signifies a small but meaningful effect. In our case, Let’s say we have a new hotel reservation system and we are comparing it to the old one.Even a small reduction in cancellations matters for hotel revenue and customer satisfaction.
- Meal booked : The type of meal booked by guest like Veg or non-veg / breakfast,lunch,dinner etc. and does not significantly affect the total amount of special requests made by guests. Small change in option may have big impact on reservation.
- Null Hypothesis: There is no difference in the total amount of special requests between guests booking different meal types.
- Alpha Level:0.01
- Power Level:0.9 , A higher value of power level will help to determine a true effect.
- Minimum Effect Size:0.5 A moderate effect size value represents a noticeable difference in the total amount of special requests.
-
Explanation:
1. Alpha level is a critical threshold used in null hypothesis analysis.It represent maximum acceptable risk of making a Type I error - false positive.
2. 0.01 value is more stringent. it reduces the risk positive and more significant for rejecting the null hypothesis means meal booking does not play major role in booking.
3. Power is the probability of correctly rejecting the null hypothesis when there is a true effect ( avoiding a Type II error). High value of power 0.9 is sensitive enough to identify real effects. In another word, 0.9 or 90% indicate that there are 90% chance of correctly detecting an effective of meal booking in hotel reservation.
4. The minimum effect size quantifies the magnitude of the difference or association between groups or variables and value of Minimum Effect Size. 0.5 value of Minimum Effect Size is moderate value which will impact practically significant. if the impact of meal booking on special request exceeds 0.5 standard deviations,It is considered practically meaningful.small effect size may be significant but may not have substantial practical implications. larger effect size must have more tangible impact.
Neyman-Pearson hypothesis test
Neyman-Pearson hypothesis test, we need to ensure that we have
sufficient sample size and that the data meet certain assumptions of the
test. The Neyman-Pearson is often used in the context of binary
hypothesis testing with known distributions and parameters.
To
perform a Neyman-Pearson hypothesis test.We need:
1. Knowledge of
the underlying distributions of the data.
2. Specific parameters of
the distributions under the null and alternative hypotheses.
3. A
large sample size.
# Find the number of record in hotel_data set.
total_no_of_record <- nrow(hotel_data)
#Assumption value for alpha
alpha <- 0.05
margin_of_error <- 0.02
# Determine the Z-Score(confidence level) when alpha is 0.05
z_score <- qnorm(1 - alpha/2)
#To display z_score
z_score## [1] 1.959964# Assumption value for sigma
sigma<-0.5
# Calculate sample size for proportions
sample_size <- ((z_score*z_score) * (sigma*sigma) )/(margin_of_error*margin_of_error)
# Round up to nearest integer
sample_size <- ceiling(sample_size)
#To display sample_size
sample_size## [1] 2401# Display the result
paste('Required sample size for Neyman-Pearson hypothesis test in hotel_data set :', sample_size )## [1] "Required sample size for Neyman-Pearson hypothesis test in hotel_data set : 2401"Perform a Fisher’s style test for significance on the same hypothesis, and interpret the p-value.
To perform a Fisher’s style test for significance, we can use a
contingency table and conduct a chi-squared test of independence.This
test will help us determine if there’s a significant association between
lead time (categorized) and cancellation status.
Hypothesis 1
(with visualizations ): Lead time does not significantly affect the
likelihood of cancellation.
# First, categorize lead time (e.g., into quadrilles)
lead_time_categories <- cut(hotel_data$lead_time, breaks = quantile(hotel_data$lead_time, probs = c(0, 0.25, 0.5, 0.75, 1)), labels = c("Q1", "Q2", "Q3", "Q4"))
# Create a contingency table
contingency_table_hypothesis1 <- table(lead_time_categories, hotel_data$is_canceled)
# Perform chi-squared test
chi_squared_test_hypothesis1 <- chisq.test(contingency_table_hypothesis1)
# Print the results
chi_squared_test_hypothesis1##
## Pearson's Chi-squared test
##
## data: contingency_table_hypothesis1
## X-squared = 8676.7, df = 3, p-value < 2.2e-16# Create 2X2 matrix from contingency_table_hypothesis1
# Note - contingency_table_hypothesis1_dataset_with_limit dataset has been limited with two row and 2 column as rstudio is given error - workspace memory error in window.
contingency_table_hypothesis1_dataset_with_limit <- matrix(contingency_table_hypothesis1, nrow = 2,ncol = 2)## Warning in matrix(contingency_table_hypothesis1, nrow = 2, ncol = 2): data
## length differs from size of matrix: [8 != 2 x 2]# Determine fisher result over data set - contingency_table_hypothesis1_dataset_with_limit
fisher_result_test_1 <- fisher.test(contingency_table_hypothesis1_dataset_with_limit)
#Display fisher result
fisher_result_test_1##
## Fisher's Exact Test for Count Data
##
## data: contingency_table_hypothesis1_dataset_with_limit
## p-value < 2.2e-16
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
## 0.7928541 0.8423255
## sample estimates:
## odds ratio
## 0.8172328barplot(contingency_table_hypothesis1, beside = TRUE, legend = TRUE,
main = "Lead Time vs. Cancellation Status",
xlab = "Lead Time",
ylab = "Frequency",
col = c("lightblue", "salmon"),
args.legend = list(title = "Cancellation Status"))
- Contingency Table : Categorize lead time and create a contingency
table with lead time categories as rows and cancellation status as
columns.
- Perform the Chi-Squared Test:Calculate the chi-squared statistic and
its associated p-value.
- Interpret the Results: Based on the p-value which is less than
2.2e-16. In fact which is small value. i.e small p value indicate strong
enough evidence against the NULL Hypothesis.Since p value is closer to
zero means there is association between lead time and cancellation
status .
- Perform fisher on limited data set (due to workspace memory error in
rstudio):Determine fisher test on contingency_table_hypothesis1 with 2X2
matrix which is similar to our Chi-Squared Test for p-value. Explanation
: The p-value in a chi-squared test represents the probability of
observing the data if the null hypothesis are true.
if p-value is less than choose significance level (eg 0.05 like #1) we reject the null hypothesis, indicating that there is a significant association between lead time and cancellation status.
If the p-value is greater than the significance level, we fail to reject the null hypothesis, suggesting that there is no significant association between lead time and cancellation status.
Hypothesis 2 (with visualizations ): The type of meal booked does not significantly affect the total amount of special requests made by guests.
# Create a contingency table for meal type and total special requests
contingency_table_hypothesis2 <- table(hotel_data$meal, hotel_data$total_of_special_requests)
# Perform chi-squared test
chi_squared_test_hypothesis2 <- chisq.test(contingency_table_hypothesis2)## Warning in chisq.test(contingency_table_hypothesis2): Chi-squared approximation
## may be incorrect # Print the results
chi_squared_test_hypothesis2##
## Pearson's Chi-squared test
##
## data: contingency_table_hypothesis2
## X-squared = 1841.1, df = 20, p-value < 2.2e-16 contingency_table_hypothesis2_dataset_with_limit <- matrix(contingency_table_hypothesis2, nrow = 2,ncol = 2)## Warning in matrix(contingency_table_hypothesis2, nrow = 2, ncol = 2): data
## length differs from size of matrix: [30 != 2 x 2]# Determine fisher result over data set - contingency_table_hypothesis1_dataset_with_limit
fisher_result_test_2 <- fisher.test(contingency_table_hypothesis2_dataset_with_limit)
#Display fisher result
fisher_result_test_2##
## Fisher's Exact Test for Count Data
##
## data: contingency_table_hypothesis2_dataset_with_limit
## p-value < 2.2e-16
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
## 38.89655 46.05566
## sample estimates:
## odds ratio
## 42.33422 barplot(contingency_table_hypothesis2, beside = TRUE, legend = TRUE,
main = "Meal Type vs. Total Special Requests",
xlab = "Meal Type",
ylab = "Frequency",
col = rainbow(ncol(contingency_table_hypothesis2)),
args.legend = list(title = "Total Special Requests"))
Explanation of above Hypothesis :
- Contingency table - contingency_table_hypothesis2 in chi square test and contingency_table_hypothesis2_dataset_with_limit in fisher.
- P-value is less than 2.2e-16 which is indicating the strong enough evidence against the NULL Hypothesis.
- The alternative hypothesis states that the true odds ratio is not equal to 1.
- The 95% confidence interval for the odds ratio is between [38.89655 - 46.05566].
- The estimated odds ratio is 42.33422 This result indicate that there
is a significant association between Meal Type and Total Special
Requests.
Note : For fisher test, I have limit the table -“contingency_table” data for two rows and two column as rstudio is throwing error for memory workspace. I have tried to increase the memory but option did not work.
Thank you!!!