1 + 1[1] 2Tram vs Brt
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# Problem Statement
Swiss transport planners want to determine whether people use the tram more frequently than the Bus Rapid Transit (BRT). Understanding this difference helps guide infrastructure investment and sustainable transport policy decisions.
# Hypotheses
Null Hypothesis (Ho): There is no difference in average daily trips between tram and BRT users.
Alternative Hypothesis (H,): Tram users make more daily trips than BRT users.
# Description of the data
Sample data of daily trips per person:
| Mode | Trip per person |
Tram 1 5, 6, 5, 8, 7, 6, 4 I
BRT \ 1 4, 3, 5, 4, 3, 4, 6 |
Variable: trips (numeric)
Groups: mode (tram, BRT)
Sample size: 7 for each mode
# Justification of the chosen test
Two independent groups (tram vs BRT)
Numeric outcome variable (trips)
Two-sample t-test is appropriate to compare means
Assumptions:
Approximately normally distributed (Shapiro-Wilk test)
Equal variance (var.equal = TRUE)
If assumptions fail + non-parametric alternative: Mann-Whitney U test
# Create sample data
my_tram <- c(5, 6, 5, 8, 7, 6, 4) >
my_brt <- c(4, 3, 5, 4, 3, 4, 6) >
# Tram trips from Monday to Friday >
tram_vector <- c(5, 6, 5, 8, 7, 6, 4) >
# brt trips from Monday to Friday >
brt_vector <- c(4, 3, 5, 4, 3, 4, 6) >
# The variable days_vector >
days_vector <- c("Monday", "Tuesday", "Wednesday", "Thursday", "Friday") >
# Assign the names of the day to tram_vector and brt_vector >
names(tram_vector) <- c("Monday", "Tuesday", "Wednesday", "Thursday", "Friday") >
names(brt_vector) <- c("Monday", "Tuesday", "Wednesday", "Thursday", "Friday")# Shapiro-wilk normality test
data: tram W = 0.96664, p-value = 0.8733
data: brt
W = 0.89358, p-value = 0.2939
Two Sample t-test
data: tram and brt
t = 2.6396, df = 12, p-value = 0.01079 alternative
hypothesis: true difference in means is greater than 0 95 percent confidence interval:
0.5568018 Inf
sample estimates:
mean of x mean of y
5.857143 4.142857
> mean(tram)
[1] 5.857143
> mean(brt) [1] 4.142857
> trips_data <- data.frame( + trips = c(5, 6, 5, 8, 7, 6, 4, 4, 3, 5, 4, 3, 4, 6), +mode = c(rep("tram",7), rep("BRT",7)) + )
# Mean trips with tram
mean_tram <- mean(tram)
# Mean trips with brt
mean_brt <- mean(brt)
# mean trips overall > trip_week
mean(tram_vector) + mean(brt_vector)
# Print out trip_week
trip_week [1] 10
# Calculate mean trips for tram and brt >
mean_tram <- mean(tram_vector)
mean_brt <- mean(brt_vector)
# Check if you realized higher mean trips in tram than in brt > mean_tram > mean_brt
[1] TRUE > # Business Conclusion
The results indicate that tram services are used more frequently than BRT in Switzerland. From a practical perspective, this suggests that tram infrastructure may provide greater value to commuters, possibly due to higher reliability, coverage, or convenience.
Transport planners can use this information to:
Prioritize maintenance or expansion of tram networks
Investigate ways to increase BRT usage (e.g., improve coverage, frequency, or connectivity)
Make data-driven decisions for sustainable transport planning and resource allocation
Overall, the evidence supports focusing investment on modes that maximize commuter benefit while identifying opportunities to improve underused services.