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Inference from Sample Statistics, Margin of Error

What this skill is

Inference means using a sample to say something about a whole population. The SAT asks you to scale a sample result up to a population estimate, to read and compare margins of error and confidence intervals, and to judge which sampling method gives trustworthy results. Most questions need careful reading more than calculation.

Key ideas

  • A random sample represents the population it was drawn from. That is what lets you scale its results up.
  • To estimate a count in the population, multiply the sample proportion by the population size.
  • A sample result is an estimate. The margin of error gives a range of plausible values: estimate ±\pm margin of error.
  • Larger random samples give smaller margins of error. Sample size does not fix bias, though. A large biased sample is still biased.
  • A confidence interval is about a population parameter (such as the mean or the percent), not about individual members.
  • If two intervals overlap, the data do not clearly show which population value is larger. If they do not overlap, there is convincing evidence of a difference.

Formulas and rules

population estimate=count in samplesample size×population size\text{population estimate} = \frac{\text{count in sample}}{\text{sample size}} \times \text{population size}

GivenFind
estimate EE, margin MMinterval from E−ME - M to E+ME + M
interval from LL to UUestimate =L+U2= \frac{L + U}{2}, margin =U−L2= \frac{U - L}{2}
  • To estimate how many do not have a trait, use the complement: 1−sample proportion1 - \text{sample proportion}.

Worked example 1 (easy)

An inspector tests a random sample of 250250 phones from a shipment of 15,00015{,}000 and finds 99 defective. Estimate the number of defective phones in the shipment.

  1. Sample proportion: 9250=0.036\frac{9}{250} = 0.036.
  2. Scale up: 0.036×15,000=5400.036 \times 15{,}000 = 540.

Check with a proportion: 9250=x15,000\frac{9}{250} = \frac{x}{15{,}000} gives x=9×60=540x = 9 \times 60 = 540. Correct.

Worked example 2 (SAT-level)

In a random survey of households in Town A, the plausible values for the percent that own an electric bike are from 41.2% to 46.8%. In Town B, a similar survey estimated 49% with a margin of error of 22 percentage points. What was Town A's estimate and margin of error, and what can be concluded?

  1. Town A's estimate is the midpoint: 41.2+46.82=44\frac{41.2 + 46.8}{2} = 44, so 44%.
  2. Its margin of error is half the width: 46.8−41.22=2.8\frac{46.8 - 41.2}{2} = 2.8 percentage points.
  3. Town B's interval runs from 47% to 51%.
  4. The intervals do not overlap (A tops out at 46.8%), so there is convincing evidence that a greater percent of households in Town B own an electric bike.

If Town B's margin had been 33 points, its interval would start at 46%, overlapping Town A's, and no clear conclusion could be drawn.

Common traps

  • Applying an interval to individuals. "The mean nightly sleep is plausibly between 6.86.8 and 7.47.4 hours" does not mean most students sleep that long.
  • Thinking a bigger sample removes bias. A survey of 5,0005{,}000 volunteers can still misrepresent the population.
  • Scaling to the wrong population. A sample of seniors at one school tells you about seniors at that school, not about all students.
  • Forgetting the complement. If the question asks how many are not infected, subtract from the total or use 1−p1 - p.
  • Doubling the margin incorrectly. The margin of error is half the interval's width, not the full width.
Practice questions