Lecture 6 — Practice

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Lecture 6 — Collected Practice Questions

Below are practice problems tagged for Lecture 6 (rendered directly from the original exam/quiz sources).

Problem 1


Problem 1.1

Which type of visualization should we use to visualize the distribution of "Range"?

Answer: Histogram

"Range" is a numerical (i.e. quantitative) variable, and we use histograms to visualize the distribution of numerical variables.

  • A bar chart couldn’t work here. Bar charts can show the distribution of a categorical variable, but "Range" is not categorical.
  • A scatter plot visualizes the relationship between two numerical variables, but we are only dealing with one numerical variable here ("Range").
  • Similarly, a line plot visualizes the relationship between two numerical variables, but we only have one here.
Difficulty: ⭐️⭐️⭐️

The average score on this problem was 63%.


Problem 1.2

Teslas, on average, tend to have higher "Range"s than BMWs. In which of the following visualizations would we be able to see this pattern? Select all that apply.

Answer:

  • A bar chart that shows the average "Range" for each "Brand"
  • An overlaid histogram showing the distribution of "Range" for each "Brand"

Let’s look at each option more closely.

  • Option 1: A bar chart showing the distribution of "Brand" would only show us how many cars of each "Brand" there are. It would not tell us anything about the average "Range" of each "Brand".

  • Option 2: A bar chart showing the average range for each "Brand" would help us directly visualize how the average range of each "Brand" compares to one another.

  • Option 3: An overlaid histogram, although perhaps a bit messy, would also give us a general idea of the average range of each "Brand" by giving us the distribution of the "Range" of each brand. In the scenario mentioned in the question, we’d expect to see that the Tesla distribution is further right than the BMW distribution.

  • Option 4: A scatter plot of "TopSpeed" against "Range" would only illustrate the relationship between "TopSpeed" and "Range", but would contain no information about the "Brand" of each EV.

Difficulty: ⭐️

The average score on this problem was 91%.


Source: fa22-midterm — Q2

Problem 2


Problem 2.1

Which type of visualization should we use to visualize the distribution of "Range"?

Answer: Histogram

"Range" is a numerical (i.e. quantitative) variable, and we use histograms to visualize the distribution of numerical variables.

  • A bar chart couldn’t work here. Bar charts can show the distribution of a categorical variable, but "Range" is not categorical.
  • A scatter plot visualizes the relationship between two numerical variables, but we are only dealing with one numerical variable here ("Range").
  • Similarly, a line plot visualizes the relationship between two numerical variables, but we only have one here.
Difficulty: ⭐️⭐️⭐️

The average score on this problem was 63%.


Problem 2.2

Teslas, on average, tend to have higher "Range"s than BMWs. In which of the following visualizations would we be able to see this pattern? Select all that apply.

Answer:

  • A bar chart that shows the average "Range" for each "Brand"
  • An overlaid histogram showing the distribution of "Range" for each "Brand"

Let’s look at each option more closely.

  • Option 1: A bar chart showing the distribution of "Brand" would only show us how many cars of each "Brand" there are. It would not tell us anything about the average "Range" of each "Brand".

  • Option 2: A bar chart showing the average range for each "Brand" would help us directly visualize how the average range of each "Brand" compares to one another.

  • Option 3: An overlaid histogram, although perhaps a bit messy, would also give us a general idea of the average range of each "Brand" by giving us the distribution of the "Range" of each brand. In the scenario mentioned in the question, we’d expect to see that the Tesla distribution is further right than the BMW distribution.

  • Option 4: A scatter plot of "TopSpeed" against "Range" would only illustrate the relationship between "TopSpeed" and "Range", but would contain no information about the "Brand" of each EV.

Difficulty: ⭐️

The average score on this problem was 91%.


Problem 2.3

Gabriel thinks "Seats" is a categorical variable because it can be used to categorize EVs by size. For instance, EVs with 4 seats are small, EVs with 5 seats are medium, and EVs with 6 or more seats are large.

Is Gabriel correct?

Justify your answer in one sentence. Your answer must fit in the box below.

Answer: No

"Seats" is a numerical variable, since it makes sense to do arithmetic with the values. For instance, we can find the average number of "Seats" that a group of cars has. Gabriel’s argument could apply to any numerical variable; just because we can place numerical variables into “bins” doesn’t make them categorical.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 51%.


Source: fa23-final — Q4

Problem 3

The following code block produces a bar chart, which is shown directly beneath it.

    (txn.groupby("browser").mean()
        .sort_values(by="is_fraud", ascending=False)
        .take(np.arange(10))
        .plot(kind="barh", y="is_fraud"))


Problem 3.1

Based on the above bar chart, what can we conclude about the browser "icedragon"? Select all that apply.

Answer: C

First, let’s take a look at what the code is doing. We start by grouping by browser and taking the mean, so each column will have the average value of that column for each browser (where each browser is a row). We then sort in descending order by the "is_fraud" column, so the browser which has the highest proportion of fraudulent transactions will be first, and we take first the ten browsers, or those with the most fraud. Finally, we plot the "is_fraud" column in a horizontal bar chart. So, our plot shows the proportion of fraudulent transactions for each browser, and we see that icedragon has a proportion of 1.0, implying that every transaction is fraudulent. This makes the third option correct. Since we don’t have enough information to conclude any of the other options, the third option is the only correct one.

Difficulty: ⭐️⭐️

The average score on this problem was 83%.


Problem 3.2

How can we modify the provided code block so that the bar chart displays the same information, but with the bars sorted in the opposite order (i.e. with the longest bar at the top)?

Answer: C

Let’s analyze each option A: This isn’t correct, because we must remember that np.take(np.arange(10)) takes the rows indexed 0 through 10. And if we change ascending = False to ascending = True, the rows indexed 0 through 10 won’t be the same in the resulting DataFrame (since now they’ll be the 10 browsers with the lowest fraud rate). B: This will have the same effect as option A, since it’s being applied before the np.take() operation C: Once we have the 10 rows with the highest fraud rate, we can sort them in ascending order in order to reverse the order of the bars. Since we already have the 10 rows from the original plot, this option is correct.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 59%.


Problem 4


Problem 4.1

Suppose we run the three lines of code below.

families = living_cost.groupby("family_type").median()
sorted_families = families.sort_values(by="avg_housing_cost")
result = sorted_families.get("avg_childcare_cost").iloc[0]

Which of the following does result evaluate to?

Answer: The median "avg_childcare_cost" of the "family_type" with the lowest median "avg_housing_cost".

When we grouped living_cost by "family_type", families is a DataFrame with one row per "family_type". Using the .median() aggregation method takes the median of all numerical columns per "family_type".

sorted_families is the families DataFrame, but sorted in ascending order based on the "avg_housing_cost" column. The first row of sorted_families is the "family_type" with the lowest median "avg_housing_cost", and the last row of sorted_families is the "family_type" with the highest median "avg_housing_cost".

In the last line of code, we’re getting the "avg_childcare_cost" column from the sorted_families DataFrame. We then use iloc to get the first entry in the "avg_childcare_cost" column. Since sorted_families is sorted in ascending order, this means that we’re getting the lowest median in the column. Therefore, result evaluates to the median "avg_childcare_cost" of the "family_type" with the lowest median "avg_housing_cost".

Difficulty: ⭐️⭐️

The average score on this problem was 82%.


Problem 4.2

Suppose we define another_result as follows.

another_result = (living_cost.groupby("state").count()
                  .sort_values(by="median_income", ascending=False)
                  .get("median_income").index[0])

What does another_result represent?

Answer: The state with the most counties.

The living_cost DataFrame is being grouped by the "state" column, so there is now one row per "state". By using the .count() aggregation method, the columns in the DataFrame will contain the number of rows in living_count for each "state". All of the columns will also be the same after using .count(), so they will all contain the distribution of "state". Since living_cost has data on every county in the US, the grouped DataFrame represents the number of counties that each state has.

We then sort the DataFrame in descending order, so the state with the most counties is at the top of the DataFrame. The last line of the expression gets a column and uses .index to get the state corresponding to the first entry, which happens to be the state with the most counties and the value that gets assigned to another_result.

Since all the columns are the same, it doesn’t matter which column we pick to use in the .sort_values() method. In this case, we used the "median_income" column, but picking any other column will produce the same result.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 65%.


Problem 4.3

Which of the following DataFrames has exactly four columns?

Answer: living_cost.groupby("family_type").sum()

Since we can’t take the sum of columns with categorical data, all of the columns in living_cost that contain non-numerical data are dropped after we use the .sum() aggregation method. There are four columns in living_cost that have numerical data ("is_metro", "avg_housing_cost", "avg_childcare_cost", and "median_income"). Since Python can take the sum of these numerical columns, these four columns are kept. Therefore, the resulting DataFrame has exactly four columns.

Although "is_metro" contains Boolean values, Python can still calculate the sum of this column. The Boolean value True corresponds to 1 and False corresponds to 0.

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 35%.


Problem 4.4

Suppose we define the Series three_columns to be the concatenation of three columns of the living_cost DataFrame as follows.

three_columns = (living_cost.get("state") + " " +
                 living_cost.get("county") + " " + 
                 living_cost.get("family_type"))

For example, the first element of three_columns is the string "CA San Diego County 1a2c" (refer back to the first row of living_cost provided in the data overview).

What does the following expression evaluate to?

(living_cost.assign(geo_family=three_columns)
            .groupby("geo_family").count()
            .shape[0])

Answer: 31430, the number of rows in the living_cost DataFrame.

The first line of the expression creates a new column in living_cost, called "geo_family" that represents the concatenation of the values in "three_columns". When we group the DataFrame by "geo_family", we create a new DataFrame that contains a row for every unique value in "three_columns". "three_columns" has various combinations of "state", "country", and "family_type". Since it’s given in the DataFrame description that each of the 31430 rows of the DataFrame represents a different combination of "state", "country", and "family_type", this means that the grouped DataFrame has 31430 unique combinations as well. Therefore, when we use .shape[0] to get the number of rows in the grouped DataFrame in the last line of the expression, we get the same value as the number of rows in the living_cost DataFrame, 31430.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 74%.


Source: fa23-midterm — Q1

Problem 5


Problem 5.1

You’re interested in comparing the "avg_housing_cost" across different "family_type" groups for San Diego County, CA specifically. Which type of visualization would be most appropriate?

Answer: Bar chart

"family_type" is a categorical variable, and we use bar charts to visualize the distribution of categorical variables.

  • A scatter plot visualizes the relationship between two numerical variables, but we are only using one numerical variable here ("avg_housing_cost").
  • A line plot is used to visualize the trend between two numerical variables, but we are only using one numerical variable.
  • A histogram is used to visualize the distribution of numerical variables, but we want to see the distribution of the categorical variable "family_type".
Difficulty: ⭐️⭐️

The average score on this problem was 89%.


Problem 5.2

Suppose we run the three lines of code below.

families = living_cost.groupby("family_type").median()
sorted_families = families.sort_values(by="avg_housing_cost")
result = sorted_families.get("avg_childcare_cost").iloc[0]

Which of the following does result evaluate to?

Answer: The median "avg_childcare_cost" of the "family_type" with the lowest median "avg_housing_cost".

When we grouped living_cost by "family_type", families is a DataFrame with one row per "family_type". Using the .median() aggregation method takes the median of all numerical columns per "family_type".

sorted_families is the families DataFrame, but sorted in ascending order based on the "avg_housing_cost" column. The first row of sorted_families is the "family_type" with the lowest median "avg_housing_cost", and the last row of sorted_families is the "family_type" with the highest median "avg_housing_cost".

In the last line of code, we’re getting the "avg_childcare_cost" column from the sorted_families DataFrame. We then use iloc to get the first entry in the "avg_childcare_cost" column. Since sorted_families is sorted in ascending order, this means that we’re getting the lowest median in the column. Therefore, result evaluates to the median "avg_childcare_cost" of the "family_type" with the lowest median "avg_housing_cost".

Difficulty: ⭐️⭐️

The average score on this problem was 82%.


Problem 5.3

Suppose we define another_result as follows.

another_result = (living_cost.groupby("state").count()
                  .sort_values(by="median_income", ascending=False)
                  .get("median_income").index[0])

What does another_result represent?

Answer: The state with the most counties.

The living_cost DataFrame is being grouped by the "state" column, so there is now one row per "state". By using the .count() aggregation method, the columns in the DataFrame will contain the number of rows in living_count for each "state". All of the columns will also be the same after using .count(), so they will all contain the distribution of "state". Since living_cost has data on every county in the US, the grouped DataFrame represents the number of counties that each state has.

We then sort the DataFrame in descending order, so the state with the most counties is at the top of the DataFrame. The last line of the expression gets a column and uses .index to get the state corresponding to the first entry, which happens to be the state with the most counties and the value that gets assigned to another_result.

Since all the columns are the same, it doesn’t matter which column we pick to use in the .sort_values() method. In this case, we used the "median_income" column, but picking any other column will produce the same result.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 65%.


Problem 5.4

Which of the following DataFrames has exactly four columns?

Answer: living_cost.groupby("family_type").sum()

Since we can’t take the sum of columns with categorical data, all of the columns in living_cost that contain non-numerical data are dropped after we use the .sum() aggregation method. There are four columns in living_cost that have numerical data ("is_metro", "avg_housing_cost", "avg_childcare_cost", and "median_income"). Since Python can take the sum of these numerical columns, these four columns are kept. Therefore, the resulting DataFrame has exactly four columns.

Although "is_metro" contains Boolean values, Python can still calculate the sum of this column. The Boolean value True corresponds to 1 and False corresponds to 0.

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 35%.


Problem 5.5

Suppose we define the Series three_columns to be the concatenation of three columns of the living_cost DataFrame as follows.

three_columns = (living_cost.get("state") + " " +
                 living_cost.get("county") + " " + 
                 living_cost.get("family_type"))

For example, the first element of three_columns is the string "CA San Diego County 1a2c" (refer back to the first row of living_cost provided in the data overview).

What does the following expression evaluate to?

(living_cost.assign(geo_family=three_columns)
            .groupby("geo_family").count()
            .shape[0])

Answer: 31430, the number of rows in the living_cost DataFrame.

The first line of the expression creates a new column in living_cost, called "geo_family" that represents the concatenation of the values in "three_columns". When we group the DataFrame by "geo_family", we create a new DataFrame that contains a row for every unique value in "three_columns". "three_columns" has various combinations of "state", "country", and "family_type". Since it’s given in the DataFrame description that each of the 31430 rows of the DataFrame represents a different combination of "state", "country", and "family_type", this means that the grouped DataFrame has 31430 unique combinations as well. Therefore, when we use .shape[0] to get the number of rows in the grouped DataFrame in the last line of the expression, we get the same value as the number of rows in the living_cost DataFrame, 31430.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 74%.


Source: fa24-final — Q1

Problem 6


Problem 6.1

Notice that bookstore has an index of "ISBN" and sales does not. Why is that?

Answer: The bookstore can sell multiple copies of the same book.

In the sales DataFrame, each row represents an individual sale, meaning multiple rows can have the same "ISBN" if multiple copies of the same book are sold. Therefore we can’t use it as the index because it is not a unique identifier for rows of sales.

Difficulty: ⭐️⭐️

The average score on this problem was 87%.


Problem 6.2

Is "ISBN" a numerical or categorical variable?

Answer: categorical

Even though "ISBN" consists of numbers, it is used to identify and categorize books rather than to quantify or measure anything, thus it is categorical. It doesn’t make sense to compare ISBN numbers like you would compare numbers on a number line, or to do arithmetic with ISBN numbers.

Difficulty: ⭐️⭐️

The average score on this problem was 75%.


Problem 6.3

Which type of data visualization should be used to compare authors by median rating?

Answer: bar chart

A bar chart is best, as it visualizes numerical values (median ratings) across discrete categories (authors).

Difficulty: ⭐️⭐️

The average score on this problem was 88%.


Source: fa24-midterm — Q12

Problem 7


Problem 7.1

What would be the best type of plot to visualize the distribution of "neighborhood" among the houses represented in treat?

Answer: bar chart

Difficulty: ⭐️⭐️

The average score on this problem was 76%.


Problem 7.2

Suppose we had access to historical data about the price of fun-sized candies over time. If we wanted to compare the prices of Milky Way and Skittles over time, which would be the best type of visualization to plot?

Answer: overlaid line plot

Difficulty: ⭐️

The average score on this problem was 90%.


Problem 8

The expression below evaluates to True.

(
classify_artist('Michelle Branch')=='outdated' 
and 
classify_artist('Drake')=='trending'
)

Consider the scatterplot created by the code below.

merged.plot(kind='scatter', x='Year', y='Top_Hit_Year');

The point for Drake is somewhere in this scatterplot. Relative to the point for Drake, there are four quadrants of the scatterplot, as shown below.


There is one quadrant in which the point for Michelle Branch cannot appear. Which is it?

Answer: northwest

The scatterplot shows an artist’s year of performance at Sun God on the x-axis, and their top hit year on the y-axis.

Since Drake is considered trending, we know his top hit came out within the five years prior to his Sun God performance. In other words, the time gap between his top hit release and his Sun God performance is at most five years.

If Michelle Branch were in the northwest quadrant, that would mean her x-coordinate was smaller than Drake’s, and her y-coordinate was larger. In other words, that would mean she performed at Sun God before Drake and her top hit was released after Drake’s. This means the time gap between her top hit release and her Sun God performance is less than five years, so she could not be considered outdated.

We could also answer this question through process of elimination. This would entail finding scenarios that show each of the other three quadrants is possible.

Michelle Branch could wind up in the northeast quadrant if, say, she performed 10 years after Drake and released her top hit 1 year after Drake. She’d be considered outdated because the time gap between her top hit and her Sun God performance would exceed five years.

Similarly, Michelle Branch could be in the southwest quadrant if, say, she performed 1 year before Drake and released her top hit 10 years before Drake. Again, the time gap between her top hit and her Sun God performance would exceed five years, making her outdated.

She could also be in the southeast quadrant if, say, she performed 10 years after Drake and released her top hit 10 years before Drake. This would also make the time gap between her top hit and Sun God performance more than five years.

So since all three other quadrants are possible and we are told that one quadrant is impossible, the northwest quadrant must be impossible.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 65%.

Problem 9

Choose the best visualization to answer each of the following questions.


Problem 9.1

What kinds of dogs cost the most?

Answer: Bar plot

Here, there is a categorical variable (the breed of the dog) and a quantitative variable (the price of the dog). A bar plot is most suitable for this scenario when trying to compare quantitative values accross different categorical values. A scatter plot or histogram wouldn’t be appropriate here since both of those charts typically aim to visualize a comparison between two quantitative variables.

Difficulty: ⭐️⭐️

The average score on this problem was 89%.


Problem 9.2

Do toy dogs have smaller 'size' values than other dogs?

Answer: None of the above

We are comparing two categorical variables ('size' and 'kind'). None of the graphs here are suitable for this scenario since bar plots are used to compare a categorical variable against a quantitative variable, while scatter plots and histograms compare a quantitative variable against quantitative variable.

Difficulty: ⭐️⭐️⭐️⭐️⭐️

The average score on this problem was 26%.


Problem 9.3

Is there a linear relationship between weight and height?

Answer: Scatter plot

When looking for a relationship between two continuous quantitative variables, a scatter plot is the most appropriate graph here since the primary use of a scatter plot is to observe the relationship between two numerical variables. A bar chart isn’t suitable since we have two quantitative variables, while a bar chart visualizes a quantitative variable and a categorical variable. A histogram isn’t suitable since it’s typically used to show the frequency of each interval (in other words, the distribution of a numerical variable).

Difficulty: ⭐️

The average score on this problem was 100%.


Problem 9.4

Do dogs that are highly suitable for children weigh more?

Answer: Bar plot

We’re comparing a categorical variable (dog suitability for children) with a quantitative varible (weight). A bar plot is most approriate when comparing a categorical variable against a quantitative variable. Both scatter plot and histogram aim to compare a quantitative variable against another quantitative variable.

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 40%.


Source: su24-midterm — Q4

Problem 10

Consider the scatterplot generated by the following expression:

kart.plot(kind="scatter, x="Total Points", y="Races Won")

Which of the following questions would you be able to answer from this scatterplot? Select all that apply.

Answer:: Option 2 and Option 3

  • Option 1 is incorrect since the scatter plot shows all the observations based on the x and y values and they are not labeled by team. Thus, team information is not shown on this scatterplot.
  • Option 4 is incorrect since the scatter plot does not label the teams and thus you would not know this information.
  • Option 2 is correct. Since the x axis is total points, you can count the number of observations that fall at x≥6000.
  • Option 3 is correct. The phrase “100 total points per race that they won” indicates a relationship between total points and races won. Since the x-axis denotes total points and the y-axis represents races won, this relationship is simply the line x=100y. To answer this question, since we are interested in the number of teams that scored less than this, you can count the number of points below this line.
Difficulty: ⭐️

The average score on this problem was 95%.