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Instructor(s): Peter Chi, Sam Lau

This exam was administered in-person. Students were allowed one page of double-sided handwritten notes. No calculators were allowed. Students had 3 hours to take this exam.

This exam covered material from the Winter 2026 offering of DSC 10.

Note (groupby / pandas 2.0): Pandas 2.0+ no longer silently drops columns that can’t be aggregated after a groupby, so code written for older pandas may behave differently or raise errors. In these practice materials we use .get() to select the column(s) we want after .groupby(...).mean() (or other aggregations) so that our solutions run on current pandas. On real exams you will not be penalized for omitting .get() when the old behavior would have produced the same answer.

We will work with a DataFrame called zoo containing information about animals at the San Diego Zoo. Each row corresponds to one animal. The DataFrame includes columns such as:

• "species" (str): The species name (e.g. "Giant Panda").
• "exhibit" (str): The exhibit where the animal is housed.
• "kind" (str): The animal’s class (e.g. "Reptile", "Mammal", "Bird").
• "weight_lb" (float): The animal’s weight in pounds.
• "age" (float): The animal’s age.
• "status" (str): Conservation status (e.g. "Endangered", "Vulnerable").
• "daily_food_lb" (float): Estimated pounds of food consumed per day.
• Location or area of the zoo where the animal’s habitat is (e.g. Asian Passage), as used in the exam.

A preview of zoo is shown below.

Assume we have already run import babypandas as bpd and import numpy as np.

Problem 1

A silly capybara has scrambled up Jeffrey’s Python code! Unscramble the code so that after it runs, jeffrey is a single string containing the name of the exhibit with the largest mean reptile weight.

Available lines (each may be used once, more than once, or not at all):

• A. jeffrey = jeffrey.get(['exhibit', 'weight_lb'])
• B. jeffrey = jeffrey.get(['kind', 'weight_lb'])
• C. jeffrey = jeffrey[jeffrey.get('kind') == 'Reptile']
• D. jeffrey = jeffrey[jeffrey.get('species') == 'Reptile']
• E. jeffrey = zoo
• F. jeffrey = jeffrey.loc[0]
• G. jeffrey = jeffrey.iloc[0]
• H. jeffrey = jeffrey.index[0]
• I. jeffrey = jeffrey.max()
• J. jeffrey = jeffrey.mean()
• K. jeffrey = jeffrey.reset_index()
• L. jeffrey = jeffrey.sort_values(by='weight_lb', ascending=False)
• M. jeffrey = jeffrey.sort_values(by='weight_lb', ascending=True)
• N. jeffrey = jeffrey.groupby('exhibit')

For each of Lines 1–9, select a line from A–N such that the code produces the desired result when run in order. Your code may not need all 9 lines — use “Not used” for any remaining lines at the end. Line 1 has been filled in for you (E). Not all of the code above will be used, and some lines may be used more than once.

State the correct sequence of line letters for Lines 2–7 (and indicate which of Lines 8–9 are not used).

jeffrey = zoo
jeffrey = jeffrey[jeffrey.get('kind') == 'Reptile']
jeffrey = jeffrey.get(['exhibit', 'weight_lb'])
jeffrey = jeffrey.groupby('exhibit')
jeffrey = jeffrey.mean()
jeffrey = jeffrey.sort_values(by='weight_lb', ascending=False)
jeffrey = jeffrey.index[0]

Problem 2

Kate is the San Diego Zoo’s newest zookeeper! She creates the DataFrames first_5 and first_7 using the code below.

first_5 = zoo.take(np.arange(5))
first_7 = zoo.take(np.arange(7))

Problem 2.1

(a) Kate’s first merge call is shown below.

merged_a = first_7.merge(first_7, on='species')

How many rows does merged_a have? Write your answer as a single number in the box, or select Not Enough Information.

• 17

• Not Enough Information

Problem 2.2

(b) Kate’s second merge call is shown below.

merged_b = merged_a.merge(first_5, on='species')

How many rows does merged_b have? Write your answer as a single number in the box, or select Not Enough Information.

• 35

• Not Enough Information

Problem 3

For this problem, let weight and age be defined by:

weight = np.array(zoo.get('weight_lb'))
age = np.array(zoo.get('age'))

Problem 3.1

(a) Avi wants to quickly estimate each animal’s weight in kilograms instead of pounds using the shortcut ( ).

Which of the following expressions output an array of the animal weights in kilograms using this shortcut? Select all that apply.

• A. weight * 0.45

• B. weight - 0.55

• C. weight * 0.90 / 2

• D. weight / 0.45

Problem 3.2

(b) Michelle wants to find the range of age, i.e., the age of the oldest zoo animal minus the age of the youngest zoo animal.

Which of the following expressions incorrectly computes this value?

• A. age.max() - age.min()

• B. (age - age.max()).min()

• C. (age.max() - age).max()

• D. (age - age.min()).max()

Problem 4

Bianca is writing helper functions for the San Diego Zoo’s animal tracking tools. Complete the code below so that mystery returns a string made of the first letter of each word in a species name. Then, Bianca can use blank (b) to create bianca, a copy of zoo with one extra column called initials. For example, mystery("Giant Panda") should return "GP".

def mystery(value):
    _____(a)_____
    return result

bianca = zoo.assign(initials=_____(b)_____)

Problem 4.1

(a) Which snippets could replace blank (a)? Select all that apply.

Snippet 1.

result = ""
for word in value:
    result = result + word[0]

Snippet 2.

words = value.split()
result = ""
for word in words:
    result = result + word[0]

Snippet 3.

temp = bpd.DataFrame().assign(word=value.split())
result = temp.get('word').get(0)

Snippet 4.

words = np.array(value.split())
for i in np.arange(len(words)):
    words[i] = words[i][0]
result = ''.join(words)

• Snippet 1

• Snippet 2

• Snippet 3

• Snippet 4

Problem 4.2

(b) Which snippets could replace blank (b)? Select all that apply.

• zoo.get('species').apply(mystery)

• zoo.apply(mystery).get('species')

• mystery(zoo.get('species'))

• zoo.mystery('species')

Problem 5

Sam has subsetted zoo to include only the animals whose status is "Endangered". Suppose that in Sam’s subset:

• () of the animals are mammals,
• () of the animals live in the Asian Passage, and
• () of the animals are mammals that live in the Asian Passage.

Problem 5.1

(a) What is the probability that a randomly selected animal from Sam’s sample lives in the Asian Passage but is not a mammal? Give your answer as a simplified fraction, or select Not Enough Information.

• ()

• Not Enough Information

Problem 5.2

(b) Sam chooses two animals uniformly at random with replacement from his sample. What is the probability that at least one of the two animals is a mammal living in the Asian Passage? Give your answer as a simplified fraction, or select Not Enough Information.

• ()

• Not Enough Information

Problem 6

Sofia gets a summer internship at the San Diego Zoo. Her first task is to study the relationship between conservation status and how much food an animal eats each day. She treats zoo as an SRS of animals from zoos worldwide. Complete the code below so that bootstrap(zoo) returns a 75% bootstrap confidence interval for the statistic:

median daily food of Endangered animals − median daily food of Vulnerable animals.

def bootstrap(df):
    vulnerable = df[df.get('status') == 'Vulnerable']
    endangered = df[df.get('status') == 'Endangered']
    median_diffs = []
    for i in range(1000):
        _____(a)_____
    return np.percentile(median_diffs, _____(b)_____)

Problem 6.1

(a) Which snippets could replace blank (a)? Select all that apply.

• Resample row counts from the full df for each group size, then take medians (samples from wrong population)

• Resample vulnerable and endangered separately with replacement, append difference of sample medians of daily_food_lb

• Build NumPy arrays of daily_food_lb per group, np.random.choice each with replacement, append np.median(sample_e) - np.median(sample_v)

• None of the above

Problem 6.2

(b) Which expression should replace blank (b)?

• 75

• 87.5

• [25, 75]

• [12.5, 87.5]

• [75, 25]

Problem 7

Assume Peter ran bootstrap code similar to lecture to bootstrap the age column from zoo 5,000 times and stored the median of each bootstrap sample in ages5000. He then bootstraps the median age 1,000 more times and stores results in ages1000:

>>> ages1000
array([5, 9, 12, ..., 16, 19, 12], shape=(1000,))
>>> ages5000
array([12, 7, 13, ..., 11, 11, 16], shape=(5000,))

Problem 7.1

(a) Suppose Peter creates a 95% bootstrap confidence interval using ages5000. Select all true statements about this interval.

• The total width can be calculated using the CLT as in class, without bootstrapping.

• The midpoint equals np.median(zoo.get('age')) exactly.

• This interval estimates the median age of animals in all zoos even if San Diego Zoo is a nonrandom sample of that population.

• If Peter’s target population is only animals in the San Diego Zoo, this interval isn’t useful.

• None of the above

Problem 7.2

(b) Select all true statements about ages5000 and ages1000.

• Because ages5000 has more bootstrap medians, its variance should be smaller than ages1000.

• A 95% CI from ages5000 will be narrower than one from ages1000.

• The variances of the two arrays should be similar (same statistic, same original sample).

• The variances might not be exactly equal because bootstrap resampling is random.

• None of the above

Problem 8

Raymond wants to estimate the average daily food consumption (pounds) of animals at the San Diego Zoo. He runs my_sample = zoo.sample(30). His sample consumes 1200 pounds of food per day in total, with a sample standard deviation of approximately 88.

Use ( ) where hints suggest.

Problem 8.1

(a) A 95% CLT-based confidence interval for the average daily food can be written as ([x,, 9x]) where (x > 0). What is (x)? (Simplified fraction or whole number, or Not Enough Information.)

• 8

• Not Enough Information

Problem 8.2

(b) Let (z > 0) so that some normal tail fraction uses (z) SDs. With the same sample, Raymond makes a valid CLT-based CI whose right endpoint is () times the midpoint. What is (z)? (Simplified fraction or whole number, or Not Enough Information.)

(Hint: ( ).)

• ()

• Not Enough Information

Problem 8.3

(c) Ella takes an SRS at another zoo and builds a 95% CLT interval for average daily food. Raymond uses 90% with his SD Zoo sample. Ella’s interval is narrower than Raymond’s. Select the best explanation.

• No. A smaller confidence level results in a narrower confidence interval.

• Yes. A smaller confidence level results in a wider confidence interval.

• Yes. Ella might have a smaller sample size than Raymond.

• Yes. Ella’s sample might have a smaller standard deviation.

Problem 8.4

(d) Assume 12,000 animals at the SD Zoo. If a CLT interval for the average daily food is ([a, b]), what is the interval for total daily food?

• ([30a, 30b])

• ([,a,, ,b])

• ([12000a, 12000b])

• ([a + 12000,, b + 12000])

• Cannot be determined

Problem 9

Punch is a baby Japanese macaque; zookeepers recorded his last 50 interactions (20 friendly). For a typical macaque, 50% of interactions are friendly. Minchan tests whether Punch is treated worse (fewer friendly) using:

• Null: (P() = 0.5) for each interaction.
• Alternative: (P() < 0.5).

Assume total_variation_distance(dist1, dist2) computes TVD as in class. Code skeleton:

A. expected = 50 * 0.5
B. obs = abs(20 - expected)
C. stats = np.array([])
D. for i in np.arange(1000):
E.     n = np.random.multinomial(50, [0.5, 0.5])[0]
F.     stat = abs(n - expected)
G.     stats = np.append(stats, stat)
H. p_value = np.count_nonzero(stats >= obs) / 1000

Problem 9.1

(a) If Minchan runs the code unchanged, select all true statements.

• A histogram of stats would be centered at 0.

• A histogram of stats would have a peak at 5.

• p_value is approximately twice what it should be for the stated one-sided alternative.

• p_value is more than twice what it should be.

• None of the above

Problem 9.2

(b) Line B should be:

• Left as-is

• obs = 20 - expected

• obs = 20 / 50 - expected

• obs = abs(20 / 50 - expected / 50)

• obs = total_variation_distance([0.4, 0.6], [0.5, 0.5])

Problem 9.3

(c) Line E should be:

• Left as-is

• n = np.random.choice(['Friendly', 'Unfriendly'], 50)

• n = np.random.multinomial(20, [0.5, 0.5])[0]

• n = np.random.multinomial(50, [0.4, 0.6])[0]

• n = np.random.multinomial(50, [0.5, 0.5])

Problem 9.4

(d) Line F should be:

• Left as-is

• stat = obs

• stat = abs(n - 20)

• stat = total_variation_distance([n/50, 1 - n/50], [0.5, 0.5])

• stat = n - expected

Problem 9.5

(e) Line H should be:

• Left as-is

• np.count_nonzero(stats >= obs) / 50

• np.count_nonzero(stats <= obs) / 50

• np.count_nonzero(stats <= obs) / 1000

Problem 10

Austin tests whether mammals and birds have the same proportion of “fast” animals (max speed > 40 mph). He uses a DataFrame df with:

• kind: "Mammal" or "Bird".
• is_fast: True if max speed > 40 mph, else False.

df keeps only rows of zoo where kind is Mammal or Bird. Assume df is representative of all mammals and birds for inference where stated.

Problem 10.1

(a) Select all valid statements of the null hypothesis for Austin’s permutation test.

• The distribution of is_fast is the same for mammals and birds.

• The proportion of fast animals is the same among mammals and birds.

• The proportion of fast animals among mammals and birds in df is the same.

• The average maximum speed is the same for mammals and birds.

• There is no association between kind and is_fast.

• The proportion of fast animals among both groups is 0.5.

• None of the above

Problem 10.2

(b) Select the valid alternative hypothesis.

• Proportion of fast mammals minus proportion of fast birds ().

• The distribution of kind differs for fast vs non-fast animals.

• The proportion fast among both groups is not 0.5.

• None of the choices above are valid.

Problem 10.3

(c) Select a valid test statistic for this test.

• Proportion of fast mammals minus proportion of fast birds (signed).

• TVD between distribution of kind among fast vs non-fast animals.

• Absolute difference between number of fast mammals and fast birds.

• None of the choices above are valid.

Problem 10.4

(d) Correct way to simulate under the null:

• Shuffle the kind column and recompute the statistic.

• Resample rows of df with replacement.

• For each row, flip a fair coin for is_fast.

• Separately resample mammals and birds with replacement.

Problem 10.5

(e) Consider

def stat(tbl):
    props = tbl.groupby("kind").mean().get("is_fast")
    return np.std(props)

Using stat as the test statistic with 1,000 null simulations, select all true statements.

• The empirical distribution is symmetric about 0.

• The distribution has a peak at 0.

• The distribution contains negative values.

• stat(df) is a valid observed statistic for this two-sided test.

• Same shuffles () same (p)-value as absolute difference in proportions.

• Same shuffles () same (p)-value as TVD between is_fast distributions by kind.

• None of the above

Problem 10.6

(f) Austin wants a 95% CLT interval for the proportion of mammals that are fast, with width at most 0.1 for any true proportion. Minimum number of mammals to sample?

• 400

• Not Enough Information

Problem 11

Hemanth fits a regression line predicting daily_food_lb from weight_lb:

[ = 0.04 + 2 ]

Given: (r = 0.8), mean of weight_lb is 200, SD of weight_lb is 100, SD of daily_food_lb is 5.

Problem 11.1

(a) Select all statements that must be true.

• About 80% of points fall on or near the regression line.

• Some other line could have a lower average squared residual than this line.

• No bootstrap resample can reproduce exactly this regression line.

• None of the above

Problem 11.2

(b) What is the mean of daily_food_lb? (Single number.)

Problem 11.3

(c) Convert both variables to kg ((1) lb () kg) and refit predicting daily_food_kg from weight_kg. Compared to the original line, does each increase, decrease, or stay the same?

• Slope: ( ) Increase ( ) Decrease ( ) Stay the same
  
• Intercept: ( ) Increase ( ) Decrease ( ) Stay the same
  
• r: ( ) Increase ( ) Decrease ( ) Stay the same

Problem 11.4

(d) Standardize weight_lb only, then refit predicting daily_food_lb. Compared to the original line, does each increase, decrease, or stay the same?

• Slope: ( ) Increase ( ) Decrease ( ) Stay the same
  
• Intercept: ( ) Increase ( ) Decrease ( ) Stay the same
  
• r: ( ) Increase ( ) Decrease ( ) Stay the same

Problem 11.5

(e) Bootstrap 1000 times; each time predict daily_food_lb for 150 lb and 400 lb animals. For which weight will the 1000 predictions vary more?

• 150 lb

• 400 lb

• Equal variation

• Need more information

Problem 12

Congratulations on finishing DSC 10! Before you turn in your exam:

1. Make sure you’ve written your PID on each page where required.
2. Fill bubbles and squares completely.

If you’d like, you may draw a picture about DSC 10 in the space provided on the paper exam.

(This question is not graded.)

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