Sampling & inference drills

Use samples to make inferences and predictions about populations.

Free, timed SHSAT practice sets on sampling & inference— every question includes a full worked explanation, and there's no account required.

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How to solve sampling and population problems

A random sample is assumed to look like the whole population, so the same fraction applies to both. Set up a proportion.

Scale up a survey: \(\dfrac{\text{yes in sample}}{\text{sample size}} \times \text{population}\). If 12 of 40 sampled students walk to school, then of 600 students about \(\dfrac{12}{40} \times 600 = 180\) walk.

Tag and recapture: biologists tag 30 fish and release them. Later, 5 of 50 caught fish are tagged. The tagged fraction in the catch matches the tagged fraction in the lake:

\(\dfrac{5}{50} = \dfrac{30}{N}, \quad N = \dfrac{30 \times 50}{5} = 300\)

Common traps: flipping the fraction, adding the difference instead of scaling, stopping at the scale factor (population \(\div\) sample size), and rounding the fraction to a coarse decimal too early. Keep the fraction exact and simplify before you multiply.

Worked examples

Example 1.
In a random sample of 80 students, 60 said they walk to school. Based on this sample, about how many of the 3040 students in the full population would be expected to say the same?
Show answer and explanation
Answer:
B

Sample proportion = 60 \(\div\) 80. Apply it to the full population: 60 \(\div\) 80 \(\times\) 3040 = 2280. (Adding the difference instead of scaling (\(60 + (3040 - 80)\)) gives 3020. Rounding 60 \(\div\) 80 to 0.8 before multiplying (\(0.8 \times 3040\)) gives 2432. Stopping at the scale factor (\(3040 \div 80\)) gives 38.)

Example 2.
To estimate a fish population, biologists tagged 30 fish and released them back into a lake. Later, they caught a new random sample of 30 fish and found that 10 of them were tagged. Based on this, about how many fish are in the lake?
Show answer and explanation
Answer:
A

The fraction of the second sample that was tagged, 10 \(\div\) 30, should match the fraction of the whole lake that is tagged, 30 \(\div\) (population). So population = 30 \(\times\) 30 \(\div\) 10 = 90. (Setting up the proportion upside down (\(30 \times 10 \div 30\)) gives 10. Counting the different fish caught (\(30 + 30 - 10\)) gives 50. Multiplying without dividing by 10 (\(30 \times 30\)) gives 900.)