Problem 1
Read in the data and print the first 10 observations.
ne_work_commute <- read.csv("travel-time-31109.csv")
head(ne_work_commute, n=10)
RESPONDENT TRANTIME DEPARTS ARRIVES
1 1 7 647 654
2 2 8 1305 1309
3 3 10 1445 1454
4 4 10 1605 1614
5 5 10 1605 1614
6 6 15 1515 1534
7 7 15 1435 1449
8 8 7 1545 1554
9 9 12 1535 1544
10 10 15 1435 1449
Problem 2
Use dim() to determine the dimensions of the dataset.
The dataset contains 861 rows and 4 columns.
Problem 3
Compute the mean of TRANTIME.
mean(ne_work_commute$TRANTIME)
The mean travel time is 18.35 minutes.
Problem 4
Compute the median of TRANTIME.
median(ne_work_commute$TRANTIME)
The median travel time is 15 minutes.
Problem 5
Compute the variance of TRANTIME.
var(ne_work_commute$TRANTIME)
The variance of travel time is 193.33 minutes squared.
Variance measures the amount of variability in travel time using squared deviations from the mean.
Problem 6
Compute the standard deviation of TRANTIME.
sd(ne_work_commute$TRANTIME)
The standard deviation of travel time is 13.9 minutes.
Problem 7
Use summary() on TRANTIME.
summary(ne_work_commute$TRANTIME)
Min. 1st Qu. Median Mean 3rd Qu. Max.
1.00 10.00 15.00 18.35 20.00 129.00
The values returned by summary() have the following interpretations:
- Minimum: The smallest observed travel time.
- 1st Quartile: The value below which approximately 25% of observations fall.
- Median: The middle travel time.
- Mean: The arithmetic average travel time.
- 3rd Quartile: The value below which approximately 75% of observations fall.
- Maximum: The largest observed travel time.
Problem 8
Create a histogram of TRANTIME.
hist(
ne_work_commute$TRANTIME,
main = "Distribution of Travel Time to Work",
xlab = "Travel Time (minutes)",
ylab = "Number of Respondents"
)
The data is right-skewed with large outliers.
Problem 9
Would the mean or median be a better description of travel time?
The median would be a better description since there are large values that would increase the mean.
Problem 10
Create filtered_travel_time by removing outliers using the IQR method.
An observation is considered an outlier if
\[
x < Q_1 - 1.5(IQR)
\]
or
\[
x > Q_3 + 1.5(IQR).
\]
First calculate the quartiles and IQR.
ne_summary <- summary(ne_work_commute$TRANTIME)
iqr <- IQR(ne_work_commute$TRANTIME)
q1 <- as.numeric(ne_summary[2])
q3 <- as.numeric(ne_summary[5])
iqr
Calculate the lower and upper boundaries:
lower_bound <- q1 - 1.5 * iqr
upper_bound <- q3 + 1.5 * iqr
lower_bound
Now retain only observations within these boundaries:
filtered_travel_time <- ne_work_commute$TRANTIME[
ne_work_commute$TRANTIME >= lower_bound &
ne_work_commute$TRANTIME <= upper_bound
]
Problem 11
Create a histogram of filtered_travel_time.
hist(
filtered_travel_time,
main = "Travel Time to Work After Removing Outliers",
xlab = "Travel Time (minutes)",
ylab = "Number of Respondents"
)
After removing the outliers, the data is somewhat symmetrical with some right-skew.
Problem 12
Repeat Problems 3–6 using filtered_travel_time and arrange the results into a formatted table.
filtered_summary <- data.frame(
Statistic = c(
"Mean",
"Median",
"Variance",
"Standard Deviation"
),
Value = c(
mean(filtered_travel_time),
median(filtered_travel_time),
var(filtered_travel_time),
sd(filtered_travel_time)
)
)
knitr::kable(
filtered_summary,
digits = 2,
col.names = c("Statistic", "Value"),
caption = "Summary Statistics After Removing Outliers"
)
Summary Statistics After Removing Outliers
| Mean |
16.24 |
| Median |
15.00 |
| Variance |
54.74 |
| Standard Deviation |
7.40 |
Problem 13
Describe what changed after removing the outliers.
Calculate the original and filtered summary statistics:
original_stats <- c(
Mean = mean(ne_work_commute$TRANTIME),
Median = median(ne_work_commute$TRANTIME),
Variance = var(ne_work_commute$TRANTIME),
SD = sd(ne_work_commute$TRANTIME)
)
filtered_stats <- c(
Mean = mean(filtered_travel_time),
Median = median(filtered_travel_time),
Variance = var(filtered_travel_time),
SD = sd(filtered_travel_time)
)
rbind(
Original = original_stats,
Filtered = filtered_stats
)
Mean Median Variance SD
Original 18.34727 15 193.32926 13.904289
Filtered 16.23601 15 54.74082 7.398704
The median stayed the same. The mean slightly decreased, and the variance / standard deviation greatly decreased.
Problem 14
Let \(X\) be normally distributed with the mean and variance calculated from filtered_travel_time.
Thus,
\[
X \sim N(\mu,\sigma^2),
\]
where \(\mu\) and \(\sigma^2\) are the mean and variance of filtered_travel_time.
mu <- mean(filtered_travel_time)
sigma <- sd(filtered_travel_time)
mu
Note that pnorm() requires the standard deviation, not the variance, so sigma is supplied to the sd argument.
(a) Probability of less than 10 minutes
p_less_10 <- pnorm(
10,
mean = mu,
sd = sigma
)
p_less_10
(b) Probability of between 10 and 20 minutes
We want
\[
P(10 < X < 20).
\]
Using cumulative probabilities,
\[
P(10 < X < 20)
=
P(X < 20)-P(X < 10).
\]
p_10_20 <- pnorm(
20,
mean = mu,
sd = sigma
) -
pnorm(
10,
mean = mu,
sd = sigma
)
p_10_20
(c) Probability of greater than 20 minutes
We want
\[
P(X > 20).
\]
Using the complement,
\[
P(X > 20)=1-P(X\leq20).
\]
p_greater_20 <- 1 - pnorm(
20,
mean = mu,
sd = sigma
)
p_greater_20
