Quantitative Methods for
International Politics
IPOL 3270 • Fall 2026
October 25, 2026
Populations and samples
M&Ms
Sampling variation
Uncertainty
Choosing a sample size
But often we can’t measure everything we care about
Some people decide to leave their countries to live somewhere else. Have you ever thought about emigrating from [your country]?
42% of all Tunisians? From only 2,403 people?
Population
A collection of things in the world
People, countries, voters, MPs, exports, aid projects, legislation
All adults in Tunisia
Population parameter
Something we want to know about the population
Proportion of all Tunisian adults who have thought about emigrating (p)
Parameters are what you measure to answer research questions
Also called “estimands”—the thing you care about

Census
Measure every single thing
in the population
Expensive, slow, often impossible
Sampling
Only measure a part of the population
Interview 2,403 Tunisians
Sample statistic
(or point estimate)
The population parameter (or estimand), but calculated from a sample of the population
Uses a hat! \widehat{p} = 0.425
Same hat as \widehat{y} in regression: hat = estimate
| Population parameter | Sample statistic | |
|---|---|---|
| Proportion | p | \widehat{p} |
| Mean | \mu | \bar{x} |
| Regression slope | \beta_1 | b_1 or \widehat{\beta}_1 |
Everything you calculate from a sample is an estimate of some unknown population value
Random sampling fixes all these
What if we want to save money
and we just run an online survey?
Even if we surveyed a million Tunisians online,
our parameter would be wrong!

Population
Every plain M&M made in Janaszówek
Population parameter
The proportion of each color in the vat
p_\text{blue}, p_\text{brown}, p_\text{green},
p_\text{orange}, p_\text{red}, p_\text{yellow}
Census
Count every single M&M made in Poland (ever? on a specific day?)
Impossible!
Sample
Count the M&Ms in one bag
\widehat{p}_\text{green} = green M&Ms in your bag ÷ all M&Ms in your bag
Don’t eat anything yet!
Count how many M&Ms of each color are in your bag
Calculate \widehat{p} for each color (count ÷ total)
Fill out the form here:
andhs.co/mms
| Color | Count | \widehat{p} |
|---|---|---|
| Blue | ||
| Brown | ||
| Green | ||
| Orange | ||
| Red | ||
| Yellow | ||
| Total | 1.00 |
Every sample is a little different:
sampling variation
Bigger samples → better estimates
Let’s look at the data!
Sampling variation
Uncertainty
Choosing a sample size
Today: why that works and how much better bigger samples are
Computers to the rescue!
Make a pretend population where we know the truth, then take lots of random samples from it
Centered on the truth, but lots of bags are way off—just like our real bags!
Same idea, but with real survey data
Pretend the 2,403 Tunisian respondents are the entire population
We know the true parameter: p = 0.425
What happens if we take random samples of 100 people from this population?
| Sample | Yes | \widehat{p} |
|---|---|---|
| 1 | 40 | 0.4 |
| 2 | 40 | 0.4 |
| 3 | 43 | 0.43 |
| 4 | 44 | 0.44 |
| 5 | 39 | 0.39 |
This is a sampling distribution
The distribution of a statistic (like \widehat{p}) across lots of random samples of the same size
Population
All 2,403 Tunisians’ answers
One sample
100 people’s answers
(\widehat{p} = 0.4)
Sampling distribution
1,000 samples’ \widehat{p}s
Statistics have distributions too!
More samples → smoother picture of the same distribution; bigger samples → narrower distribution
With random samples, the sampling distribution of a sample proportion (or mean) is approximately normal
\widehat{p} \sim \mathcal{N}(p, \text{SE})
Standard deviation
How much do individual values vary?
How different is one Tunisian from another?
For yes/no data:
\sqrt{p(1 - p)} = 0.494
Standard error
How much do sample statistics vary?
How different is one sample’s \widehat{p} from another’s?
The standard deviation of the sampling distribution
\text{SE}(\widehat{p}) = \frac{\text{SD}}{\sqrt{n}} = \sqrt{\frac{p (1 - p)}{n}}
| Sample size | SE from 1,000 samples | SE from formula |
|---|---|---|
| 25 | 0.1 | 0.099 |
| 100 | 0.047 | 0.049 |
| 400 | 0.022 | 0.025 |
Random sampling puts the center in the right place; sample size controls the width
Trump’s approval rating is 34.9% (+/-2.1% pts.)


Our 1,000 samples of 100: 70% within 1 SE of p; 96.5% within 1.96 SEs
Margin of error
How far to reach
\text{MOE} = 1.96 \times \text{SE}
(≈ 2 × SE)
Confidence interval
The range you get
\widehat{p} \pm \text{MOE}
(a 95% confidence interval)
\text{SE} \approx \sqrt{\frac{\widehat{p} (1 - \widehat{p})}{n}}
40 out of 100 Tunisians have thought about emigrating
The true p (42.5%) is in there!
\text{SE} \approx \sqrt{\frac{\widehat{p} (1 - \widehat{p})}{n}}
400 out of 1,000 Tunisians have thought about emigrating
The true p (42.5%) is still in there!
(but now the estimate is narrower and more precise)
94 of these 100 intervals include the true p
Much more about confidence intervals in ModernDive chapter 8!
Trump’s approval rating is 34.9% (+/-2.1% pts.)
Approval is plausibly anywhere from about 33% to 37%
The margin of error only measures sampling variation, not these:
Back to Positron!
It depends on three things
Notice what’s not on the list: how big the population is!
\text{MOE} = 1.96 \times \sqrt{\frac{p (1 - p)}{n}} \quad \rightarrow \quad n = \frac{1.96^2 \times p (1 - p)}{\text{MOE}^2}
That’s why so many polls survey ≈1,000 people
1.96^2 \times 0.5 \times 0.5 \approx 1, so:
Margin of error from n
\text{MOE} \approx \frac{1}{\sqrt{n}}
n = 1,000 → \frac{1}{\sqrt{1000}} = ± 3.2 points
n from margin of error
n \approx \frac{1}{\text{MOE}^2}
± 3 points → \frac{1}{0.03^2} ≈ 1,111 people
| Margin of error | People needed | Cost at QAR 50 per interview |
|---|---|---|
| ± 10 points | 97 | QAR 4,850 |
| ± 5 points | 385 | QAR 19,250 |
| ± 3 points | 1,068 | QAR 53,400 |
| ± 2 points | 2,401 | QAR 120,050 |
| ± 1 points | 9,604 | QAR 480,200 |
Cut the margin of error in half → 4× the people
Most national surveys stop at ≈1,000–2,500 people
Plan your sample size around the smallest group you care about
Surveys often oversample small groups (and use weights to fix the overall results)
Assume p_\text{green} ≈ 0.17 (our class’s estimate) and each bag has ≈50 M&Ms
n = \frac{1.96^2 \times p (1 - p)}{\text{MOE}^2}
One bag can’t tell the factories apart—but a whole class can!