Answer Key

How to use this key

Work the colab through with your partner first, then come here. The code below is one correct answer, not the only one. This was a 45 minute pair exercise with very little scaffolding, so there are several reasonable ways to write almost every cell. If your code looks different but produces the same numbers, you were right.

The written answers matter more than the code. You can already tell whether your code ran. What you cannot check on your own is whether you read the result correctly, and that is what the green Answer boxes are for. Compare your markdown cells against them.

⬅️ Back to the colab

Setup

import pandas as pd
import numpy as np

url = 'https://eds-217-essential-python.github.io/data/marine_microplastics.csv'
df = pd.read_csv(url)

df = df.dropna(subset=['Measurement'])
samples = df[df['Unit'] == 'pieces/m3'].copy()
samples['ocean'] = samples['Oceans'].str.replace(' Ocean', '')
positive = samples[samples['Measurement'] > 0].copy()

positive.shape

(7091, 23)

Part 1: All four oceans at once

1. Last night you computed the mean Measurement for the Atlantic and the Pacific with two filters. Do all four oceans in one line, and rank the result from dirtiest to cleanest.

positive.groupby('ocean')['Measurement'].mean().sort_values(ascending=False)

ocean
Pacific     595.199821
Atlantic    282.729748
Arctic        2.279608
Southern      0.075703
Name: Measurement, dtype: float64

✅ Answer

Pacific 595.2, Atlantic 282.7, Arctic 2.28, Southern 0.076 pieces/m3. One line replaces the two filters, two subsets and two .describe() calls you wrote last night, and it also gives you the two oceans you did not ask about. The spread is the first thing to notice: the Pacific mean is nearly eight thousand times the Southern mean, which is a larger difference than any plausible difference in ocean water, and should make you suspicious before you go any further.

2. Do the same with the median instead of the mean. The order changes. Which two oceans swap, and by how much does the Pacific-to-Atlantic ratio differ between the two versions?

positive.groupby('ocean')['Measurement'].median().sort_values(ascending=False)

ocean
Pacific     0.116665
Arctic      0.044000
Atlantic    0.019440
Southern    0.011500
Name: Measurement, dtype: float64

means = positive.groupby('ocean')['Measurement'].mean()
medians = positive.groupby('ocean')['Measurement'].median()

print('mean ratio:  ', means['Pacific'] / means['Atlantic'])
print('median ratio:', medians['Pacific'] / medians['Atlantic'])

mean ratio:   2.1051899432448007
median ratio: 6.001286008230453

✅ Answer

The Arctic and the Atlantic swap. By mean the Atlantic is second at 282.7 and the Arctic third at 2.28; by median the Arctic is second at 0.0440 and the Atlantic third at 0.0194 pieces/m3.

The Pacific-to-Atlantic ratio is 2.11 on the means and 6.00 on the medians, so the choice of average changes the size of the claim by a factor of about three. Note also how far the two averages are from each other within a single ocean: the Atlantic mean is 282.7 and the Atlantic median is 0.0194, roughly fifteen thousand times smaller. A gap that large between mean and median always means a few enormous values, and it is a warning that the mean is describing those values rather than the ocean.

3. How many samples went into each of those four numbers? One line.

positive.groupby('ocean')['Measurement'].count().sort_values(ascending=False)

ocean
Atlantic    6216
Pacific      799
Arctic        58
Southern      18
Name: Measurement, dtype: int64

✅ Answer

Atlantic 6,216, Pacific 799, Arctic 58, Southern 18. The four numbers you have been ranking rest on wildly different amounts of evidence: the Atlantic alone is 88% of the file, and the Arctic and Southern together are 76 rows, about 1% of the 7,091. .count() costs one line and tells you which rows of a grouped table you are allowed to take seriously.

4. In a markdown cell: two of the four oceans have numbers you should refuse to report. Name them and say why in one sentence.

✅ Answer

The Arctic (58 samples) and the Southern Ocean (18 samples). Both are too thin to describe an ocean: the Southern figure comes from 18 samples collected by three organizations, 14 of them by one method, so it describes a handful of cruises rather than a body of water covering twenty million square kilometres. A grouped table prints a number for every group it finds, whether or not that group has enough rows to support one, so reading the count column is part of reading the result.

Part 2: One call instead of four

5. Replace questions 1 through 3 with a single .agg() call that reports the count, median, mean and max of Measurement for each ocean.

positive.groupby('ocean')['Measurement'].agg(['count', 'median', 'mean', 'max'])

	count	median	mean	max
ocean
Arctic	58	0.044000	2.279608	63.000000
Atlantic	6216	0.019440	282.729748	110480.000000
Pacific	799	0.116665	595.199821	21156.558533
Southern	18	0.011500	0.075703	0.499663

6. Look at the Atlantic row. Its median is 0.019 and its maximum is 110,480. In a markdown cell, say what a single number that large does to a mean computed from 6,216 values, and which of the two averages you would send to the journalist.

✅ Answer

That one value contributes about 17.8 to the Atlantic mean of 282.7, which is 6% of it, and the ten largest values together contribute 38%. A mean adds every value and divides by the count, so dividing a value 5.7 million times the median by 6,216 reduces it only a little, so most of it remains in the mean. The result is a mean that only 294 of the 6,216 Atlantic samples, under 5%, actually exceed. An average that is larger than 95% of the data is not describing a typical sample.

Send the median, 0.0194 pieces/m3, and say how many samples it came from. If the journalist wants the extremes, give them the maximum separately as a maximum, which is an honest way to report a large number without letting it stand in for the ocean.

7. Use the dictionary form of .agg(), grouped by ocean, to report the mean Measurement, the number of distinct Sampling Method values, and the number of distinct Organization values in each ocean.

positive.groupby('ocean').agg({
    'Measurement': 'mean',
    'Sampling Method': 'nunique',
    'Organization': 'nunique'
})

	Measurement	Sampling Method	Organization
ocean
Arctic	2.279608	4	3
Atlantic	282.729748	10	20
Pacific	595.199821	7	9
Southern	0.075703	2	3

8. In a markdown cell: the two columns you just counted are not measurements. Why did you ask for them? What would it mean if one ocean had been sampled by one organization using one method?

✅ Answer

They describe how the number was produced rather than what was measured, and that is exactly what you need in order to decide whether two oceans can be compared. The Atlantic figure comes from 20 organizations using 10 methods; the Southern figure comes from 3 organizations using 2.

If an ocean had one organization and one method, its mean would be a property of that study, not of that ocean: any habit of that group, any bias in that instrument, would be inseparable from the result, and there would be no second measurement to check it against. The Southern Ocean is close enough to that situation to be worth the caution. nunique on a non-numeric column is a cheap way to ask “how many independent sources agree with this?” before you quote it.

Part 3: Where the answer falls apart

9. Group by Sampling Method instead of by ocean, and report the count and median of Measurement for each method. Rank by median, largest first.

by_method = positive.groupby('Sampling Method')['Measurement'].agg(['count', 'median'])
by_method.sort_values('median', ascending=False)

	count	median
Sampling Method
Stainless steel spoon	50	21840.000000
Van Dorn sampler	84	2000.000000
Aluminum bucket	57	1710.000000
PVC cylinder	292	1410.437236
CTD rosette sampler	4	21.000000
Intake seawater pump	115	2.500000
Plankton net	74	0.840000
Hand picking	12	0.777778
Manta net	884	0.252943
AVANI net	18	0.047533
Neuston net	4552	0.015120
Grab sample	949	0.006000

10. Write down the largest and smallest medians in that table. How many orders of magnitude separate them?

medians_by_method = positive.groupby('Sampling Method')['Measurement'].median()
print(medians_by_method.max(), medians_by_method.min())
print(np.log10(medians_by_method.max() / medians_by_method.min()))

21840.0 0.006
6.561101383649056

✅ Answer

21,840 pieces/m3 for the stainless steel spoon and 0.0060 for the grab sample: 6.56 on a log10 scale, so more than six orders of magnitude, a factor of about 3.6 million.

The methods sort into two families, and the reason is physical. The spoon, the Van Dorn sampler (2,000), the aluminum bucket (1,710) and the PVC cylinder (1,410) all scoop a small, fixed volume of surface water and count everything in it, including fibres a few tenths of a millimetre long. The nets, Manta at 0.253, Neuston at 0.0151, AVANI at 0.0475, tow a mesh through hundreds of cubic metres and keep only what the mesh cannot pass. Both report pieces per cubic metre, and the unit column says they are comparable, but they are counting different objects. No difference between oceans in this file is anywhere near six orders of magnitude, so method is the largest single influence on Measurement in the dataset.

11. In a markdown cell, before you write any more code: if the oceans were not sampled with the same mix of methods, what does that do to your answer from question 1?

✅ A model answer

It makes question 1 partly a comparison of sampling methods labelled as a comparison of oceans. If one ocean happens to have more bucket and cylinder samples, its mean will be higher no matter what is in the water, because those methods report values thousands of times larger than the nets do. The ocean means would then be measuring the composition of the sampling effort as much as the plastic, and I cannot tell which until I count the methods within each ocean.

Full marks here are for writing the prediction before running question 12 and for naming the mechanism, not just the risk. Writing the prediction down and then testing it is how you find a confound, rather than only naming one.

12. Find out. Group by both ocean and Sampling Method at once and count the samples. Then call .reset_index() on the result and look at the Atlantic and Pacific rows.

ocean_method = positive.groupby(['ocean', 'Sampling Method'])['Measurement'].count()
ocean_method.reset_index().head(15)

	ocean	Sampling Method	Measurement
0	Arctic	CTD rosette sampler	4
1	Arctic	Grab sample	21
2	Arctic	Intake seawater pump	6
3	Arctic	Manta net	27
4	Atlantic	Aluminum bucket	57
5	Atlantic	Grab sample	748
6	Atlantic	Hand picking	12
7	Atlantic	Intake seawater pump	106
8	Atlantic	Manta net	631
9	Atlantic	Neuston net	4390
10	Atlantic	PVC cylinder	104
11	Atlantic	Plankton net	34
12	Atlantic	Stainless steel spoon	50
13	Atlantic	Van Dorn sampler	84
14	Pacific	AVANI net	18

ocean_method.reset_index()

	ocean	Sampling Method	Measurement
0	Arctic	CTD rosette sampler	4
1	Arctic	Grab sample	21
2	Arctic	Intake seawater pump	6
3	Arctic	Manta net	27
4	Atlantic	Aluminum bucket	57
5	Atlantic	Grab sample	748
6	Atlantic	Hand picking	12
7	Atlantic	Intake seawater pump	106
8	Atlantic	Manta net	631
9	Atlantic	Neuston net	4390
10	Atlantic	PVC cylinder	104
11	Atlantic	Plankton net	34
12	Atlantic	Stainless steel spoon	50
13	Atlantic	Van Dorn sampler	84
14	Pacific	AVANI net	18
15	Pacific	Grab sample	166
16	Pacific	Intake seawater pump	3
17	Pacific	Manta net	226
18	Pacific	Neuston net	158
19	Pacific	PVC cylinder	188
20	Pacific	Plankton net	40
21	Southern	Grab sample	14
22	Southern	Neuston net	4

✅ Answer

The two oceans were sampled with almost opposite method mixes. The full table is 23 rows, one per ocean and method pair that actually occurs, and .head(15) stops one row into the Pacific, which is why the second cell prints all of it.

The Atlantic is 4,390 Neuston net out of 6,216, then 748 grab samples and 631 Manta net: three low-reading methods account for 93% of it. The Pacific is spread across Manta 226, PVC cylinder 188, grab 166 and Neuston 158. The PVC cylinder row is the one to stop on: 188 of the Pacific’s 799 samples used a method whose median across the whole file is 1,410 pieces/m3, and the Pacific mean of 595 follows from that fact rather than from the Pacific. In the other direction, the stainless steel spoon and the Van Dorn sampler, the two highest medians in the file, appear in the Atlantic and in no other ocean, and so does the aluminum bucket.

Grouping by two keys is what made this visible. Either key on its own averages the other one away.

13. Simpler and more direct: run .value_counts() on Sampling Method for the Atlantic samples and again for the Pacific samples. Which method dominates the Atlantic? What fraction of the Pacific samples used it?

atlantic = positive[positive['ocean'] == 'Atlantic']
pacific = positive[positive['ocean'] == 'Pacific']

print(atlantic['Sampling Method'].value_counts())
print(pacific['Sampling Method'].value_counts())

Sampling Method
Neuston net              4390
Grab sample               748
Manta net                 631
Intake seawater pump      106
PVC cylinder              104
Van Dorn sampler           84
Aluminum bucket            57
Stainless steel spoon      50
Plankton net               34
Hand picking               12
Name: count, dtype: int64
Sampling Method
Manta net               226
PVC cylinder            188
Grab sample             166
Neuston net             158
Plankton net             40
AVANI net                18
Intake seawater pump      3
Name: count, dtype: int64

✅ Answer

The Neuston net dominates the Atlantic, 4,390 of 6,216 samples, or 71%. In the Pacific it is 158 of 799, or 20%. The Atlantic is effectively one method with a few others attached; the Pacific is four methods of comparable size, and its most common one, the Manta net at 226, is not the Atlantic’s most common one.

That is enough to explain the difference you reported last night without any reference to the amount of plastic in the water. The Neuston net has the second lowest median of any method in the file at 0.0151, so an ocean measured mostly by Neuston net will read low, and the Atlantic is that ocean.

14. Now control for it. Filter positive to the rows collected with a 'Neuston net', end the line with .copy(), then group that by ocean and report count, median and mean.

neuston = positive[positive['Sampling Method'] == 'Neuston net'].copy()
neuston.groupby('ocean')['Measurement'].agg(['count', 'median', 'mean'])

	count	median	mean
ocean
Atlantic	4390	0.015120	0.085767
Pacific	158	0.030680	0.396195
Southern	4	0.306271	0.278664

neuston[neuston['ocean'] == 'Pacific']['Organization'].value_counts()

Organization
5 Gyres Institute                                                         82
Sea Education Association                                                 69
Scripps Institution of Oceanography-University of California San Diego     7
Name: count, dtype: int64

15. In a markdown cell, three or four sentences. Compare this table with your answer to question 5. Does the Pacific still come out higher than the Atlantic? By how much, and is it the same “how much” you reported last night? What happened to the number of Pacific samples you are now relying on?

✅ Answer

Yes, the Pacific is still higher, but the claim shrinks from a factor of 6.0 to a factor of 2.0 and the evidence for it shrinks from 799 samples to 158. Within Neuston net samples only, the Pacific median is 0.0307 pieces/m3 against the Atlantic’s 0.0151. That is a real difference, it points the same way it did last night, and it is now a comparison of like with like.

The means change the most. They fall from 595.2 and 282.7 to 0.396 and 0.0858, a drop of three orders of magnitude in both oceans, because the extreme values behind the question 5 means came from the scoop methods and not from the nets. The 2.1-fold difference in means you reported last night did not describe seawater. It described 188 PVC cylinder samples in the Pacific against 4,390 Neuston tows in the Atlantic.

So the effect survives the control and changes size, which is a different claim from either “the Pacific is dirtier” or “the difference was an artefact”. Both halves have to be said. And the 158 Pacific samples that carry the surviving claim come from only three organizations, 82 of them from one, so this is not 158 independent lines of evidence. It is a difference worth reporting with its sample size attached and worth checking against another single method, for example the Manta net, before anyone builds on it.

Part 4: Write it again

16. The journalist replies: “Thanks. My editor wants one sentence with a number in it.” Write that sentence in a markdown cell. It must contain a number, must name the sampling method you controlled for, and must not overstate what 158 samples can support.

✅ A model answer

Comparing only samples taken with a Neuston net, the one method used in both oceans in reasonable numbers, the median Pacific sample contains about twice as much floating plastic as the median Atlantic sample, 0.031 against 0.015 pieces per cubic metre, though the Pacific figure rests on 158 samples from three research groups and should be treated as provisional.

Full marks for a sentence that carries all four things: the number, the method, the direction, and the limit. The common failure is dropping the last one, or quoting the mean because it sounds more impressive. The sentence must also survive the reader asking “compared with what?”, which is why the method is named inside it rather than in a footnote.

17. In a markdown cell of four or five sentences: what changed between last night’s answer and this one, and was it the data or the question? Name the specific line of code that made the difference visible.

✅ A model answer

The data did not change. It is the same 7,091 rows and the same Measurement column, and last night’s arithmetic was correct: the Pacific mean really is 595.2 and the Atlantic mean really is 282.7. What changed is the question. Last night I asked which ocean has the larger average, and this afternoon I asked whether the two averages were produced in comparable ways, which is a question about the sampling rather than about the water.

The line that made it visible is:

positive.groupby(['ocean', 'Sampling Method'])['Measurement'].count()

Grouping by two keys at once showed 4,390 Neuston tows in the Atlantic against 188 PVC cylinder samples in the Pacific, and once that table is on screen the ocean means cannot be read as ocean means. Question 14 then measured what was left after holding the method fixed: a two-fold difference in the medians, on 158 Pacific samples. The general lesson is that a groupby on one column silently averages over every column you did not group by, and if one of those columns drives the measurement, your comparison is partly about that column instead.

If you finish early

Group by Regions, report count and median, and rank by count to find the ten most-sampled regions. Look at which sampling methods were used in the extremes.

by_region = positive.groupby('Regions')['Measurement'].agg(['count', 'median'])
by_region.sort_values('count', ascending=False).head(10)

	count	median
Regions
Caribbean Sea	873	0.006480
Mediterranean Sea	340	0.557847
Gulf of Mexico	263	1000.000000
North Sea	158	0.022733
Inner Seas off the West Coast of Scotland	154	0.046132
Gulf of California	85	1410.437236
Coastal Waters of Southeast Alaska and British Columbia	22	0.003500
Celtic Sea	21	0.023467
Irish Sea and St. George's Channel	20	0.056972
Gulf of St. Lawrence	18	0.004546

print(positive[positive['Regions'] == 'Gulf of California']['Sampling Method'].value_counts())
print(positive[positive['Regions'] ==
               'Coastal Waters of Southeast Alaska and British Columbia']['Sampling Method'].value_counts())

Sampling Method
PVC cylinder    70
Grab sample      9
Neuston net      6
Name: count, dtype: int64
Sampling Method
Grab sample    22
Name: count, dtype: int64

✅ Answer

The Gulf of California has a median of 1,410 pieces/m3 and coastal Alaska and British Columbia has 0.0035, a ratio of about 403,000. The same explanation as Part 3 applies, only more starkly: 70 of the Gulf of California’s 85 samples were taken with a PVC cylinder, and all 22 Alaska samples were grab samples. Those are the extremes of the method table from question 9.

The Gulf of Mexico row makes the point a third time, with a median of exactly 1,000.0 across 263 samples. A median that comes out as an exact round number usually reflects a reporting convention or a detection threshold rather than the water, and it is worth investigating before it goes into any comparison. Finer geography does not remove a method confound. It only makes the groups smaller, so the confound is harder to see.

Where the marks are

If you compare your notebook against this key, look for these four things before you look at anything else.

1. Your question 15 answer says both halves. The Pacific is still higher after controlling for method, and the size of the difference falls from 6.0 to 2.0 while the sample count falls from 799 to 158. An answer that reports only the survival, or only the collapse, is half the finding.
2. You read the count column of every grouped table before quoting a number from it. The Southern Ocean’s 18 rows and the Pacific’s 158 Neuston samples are the two places this matters most in the colab.
3. You wrote your question 11 prediction before running question 12. The value of the exercise is in committing to an expectation and then testing it, not in the table itself.
4. Your interpretation answers say what the number means, not what the number is. “The Pacific mean is 595.2” restates the output. “The Pacific mean is 595.2, and 188 of its 799 samples used a method whose median is 1,410” reads it.

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