Exercise: make it board-ready, then catch the tricks

Session 6 — Applied Data Science 2026

Author

Affiliation

Published

02 11 2026

The setting

The board liked your charts from last week — but they were charts for you. Now they want one they can put in front of their investors, and they want to know whether the charts a consultant sent them can be trusted.

Alone or in pairs. Two parts, as in the lecture: A and B, then C, a short exchange with a fellow student.

Start here

Install the one new package, if you have not already:

install.packages("gghighlight")

Create session06.qmd in your project, copy both CSV files from the session page into data/, and start with this setup chunk:

```{r}
#| label: setup
#| include: false
library(tidyverse)
library(scales)
library(gghighlight)
library(here)

i_am("session06.qmd")

prices   <- read_csv(here("data", "hotels-europe_price.csv"))
features <- read_csv(here("data", "hotels-europe_features.csv"))

hotels <- prices |>
  filter(year == 2017, month == 11, weekend == 0) |>
  inner_join(features, by = "hotel_id") |>
  filter(city == city_actual, accommodation_type == "Hotel")

theme_board <- function(base_size = 13) {
  theme_minimal(base_size = base_size) +
    theme(
      plot.title            = element_text(face = "bold"),
      plot.title.position   = "plot",
      plot.caption          = element_text(colour = "grey45", hjust = 0),
      plot.caption.position = "plot",
      panel.grid.minor      = element_blank(),
      legend.position       = "top",
      legend.justification  = "left"
    )
}
```

Part A — Communication

A1 · The finding first

Board: “Where does Vienna stand among Europe’s cities — is it expensive?”

Before you write any code, compute what you need and write the one sentence your chart will say. It goes into the title later.

cities <- hotels |>
  group_by(city) |>
  summarise(n = n(), median_price = median(price), .groups = "drop")

europe_median <- median(hotels$price)

Where is Vienna’s median, compared with the European median? How many cities are more expensive?

A2 · The makeover

A colleague sent you this:

cities_mean <- hotels |>
  group_by(city) |>
  summarise(mean_price = mean(price))

ggplot(data = cities_mean, mapping = aes(x = city, y = mean_price, fill = city)) +
  geom_col()

Turn it into a chart that answers the board’s question in ten seconds. Work through these in order and render after each one:

The right number: median, not mean (session 5).
Readable: cities on the y axis, ordered by price.
Emphasis: highlight Vienna with gghighlight(); the other 43 cities stay, in grey.
A reference: a dashed line at the European median, labelled with annotate("label") on a white background, so that no line runs through the text.
Say it: a title with your sentence from A1, a subtitle with what is measured, a caption with the source and n.
House style: + theme_board().

Hint for steps 2–4

ggplot(data = cities, mapping = aes(x = median_price,
                             y = fct_reorder(city, median_price))) +
  geom_col() +
  gghighlight(city == "Vienna") +
  geom_vline(xintercept = europe_median, linetype = "dashed") +
  annotate("label", x = europe_median, y = 3, label = "European median",
           hjust = -0.05, colour = "grey30", fill = "white", label.size = 0)

fill = city is gone: with gghighlight() the highlighted bar keeps its colour and you no longer need a legend. If you want a particular colour, put it in geom_col(fill = "#00395B").

Hint for step 5

labs(title = …, subtitle = …, caption = …, x = NULL, y = NULL). For n in the caption, compute it and paste it in: caption = paste0("Data: Békés & Kézdi (2021). ", nrow(hotels), " hotels in ", nrow(cities), " cities.")

A3 · Colour that means something

Take the weekend chart from session 5 — four cities, one line each — and replace ggplot’s default colours with the colour-blind-safe Okabe–Ito palette:

okabe_ito <- unname(palette.colors(palette = "Okabe-Ito"))
# then: + scale_colour_manual(values = okabe_ito[2:5])

The weekend chart from session 5

weekends <- prices |>
  filter(weekend == 1) |>
  inner_join(features, by = "hotel_id") |>
  filter(city == city_actual, accommodation_type == "Hotel",
         city %in% c("London", "Paris", "Rome", "Vienna")) |>
  group_by(city, year, month) |>
  summarise(n = n(), median_price = median(price), .groups = "drop") |>
  mutate(date = make_date(year, month))

ggplot(data = weekends, mapping = aes(x = date, y = median_price, colour = city)) +
  geom_line() +
  geom_point() +
  scale_x_date(date_breaks = "1 month", date_labels = "%b") +
  scale_y_continuous(labels = label_number(prefix = "€")) +
  labs(x = NULL, y = NULL, colour = NULL) +
  theme_board()

Is colour doing a categorical, sequential or diverging job here?

Part B — Deception

B1 · Spot the trick

A consultant sent the board these four charts. Every number on them is correct. For each one, write down:

(a) what the chart wants you to believe,
(b) the trick — name it with the checklist from the lecture,
(c) what an honest version would show.

Chart 1

Chart 2

Chart 3

Chart 4

The checklist, again

Claim · Baseline · Window · Axes · n · Summary · Encoding · Source

B2 · Fix two of them

Pick two of the four and build the honest version in your document. Give each the title it can actually support.

What you need for each

Chart 1: the median prices of the four cities; bars from zero.
Chart 2: Amsterdam’s median weekend price for every month in the data (session 5, Q6).
Chart 3: the median, and n — for example in the city label: paste0(city, " (n = ", n, ")").
Chart 4: the number of hotels per city, as bars or dots.

Part C — The checklist on a fellow student’s chart

Swap laptops with a fellow student. Run the eight checklist questions over their A2 chart. Write down the one thing you would change, and tell them.

The checklist

Before you hand in any chart — in Report 2 and after:

Claim: the title states a finding, and the chart shows it
Baseline: bars start at zero
Window: the time range is justified
Axes: one scale per axis; log scales labelled as such
n: visible wherever a group is small
Summary: the right one — median for prices
Encoding: amounts by position or length, not by area or colour
Source: data source and what was removed, in the caption

If you finish early

a) Diverging instead of highlighting

Redraw A2 as the difference from the European median, in percent, with scale_fill_gradient2(): blue below, red above, grey in the middle. Which version answers the board’s question better?

b) Make your own trick

Build one misleading chart from this data — honest numbers, one bad choice. Show it to a fellow student and let them find the trick.

c) Save it

Store your A2 chart as an object, a2 <- ggplot(…) + …, and save it twice:

as a PDF, 8 × 4.5 inches — the default for a report;
as a PNG for a PowerPoint deck: same size, dpi = 300, bg = "white".

Open both files and zoom in. Which one stays sharp? Does the text size still work at 8 × 4.5 inches — and if not, what do you change?

d) Two series, one axis

Lecture B3 put Vienna’s median weekend price and the number of hotels in the data on two y axes — and showed that the second axis can tell any story. Build the honest version: set both series to 100 in November 2017 and show the relative change from there, on one axis.

vienna_supply <- prices |>
  filter(weekend == 1) |>
  inner_join(features, by = "hotel_id") |>
  filter(city == "Vienna", city_actual == "Vienna", accommodation_type == "Hotel") |>
  group_by(year, month) |>
  summarise(n = n(), median_price = median(price), .groups = "drop") |>
  mutate(date = make_date(year, month))

What does the chart show in January — and what does its title not allow you to say?