A pollster runs a survey and gets a margin of error of ±6 percentage points. She wants to cut the margin of error to ±3 percentage points. By what factor must she increase the sample size?
A2 — she needs twice as many respondents
B3 — she needs three times as many respondents
C4 — she needs four times as many respondents
D6 — she needs six times as many respondents
The margin of error scales as 1/√n. To cut ME in half (from ±6 to ±3), you need to multiply 1/√n by 1/2, which means increasing n by a factor of 4. In general, halving the margin of error requires quadrupling the sample size. This counterintuitive result — the 'square root penalty' — is one of the most practically important facts in survey design. Option A (doubling) is the common misconception.
Question 2 Multiple Choice
A researcher is designing a survey to estimate the proportion of voters who support a candidate. She has no prior data on what the proportion might be. What value of p should she use in the sample size formula, and why?
Ap = 0.5, because it maximizes p(1−p) and gives the largest, most conservative required sample size
Bp = 0.5, because polls always assume 50-50 races for fairness
Cp = 0.1, because underestimating support is safer than overestimating it
DShe should use any p between 0 and 1 — the sample size formula is insensitive to the choice of p
The sample size formula for proportions is n = z²p(1−p)/m². The factor p(1−p) is maximized when p = 0.5, giving p(1−p) = 0.25. Using p = 0.5 therefore produces the largest n — the most conservative estimate — guaranteeing sufficient precision regardless of the true proportion. Option B gives the right answer but the wrong reason; it has nothing to do with fairness and everything to do with worst-case coverage.
Question 3 True / False
Doubling the sample size reduces the margin of error by a factor of √2 (approximately 1.41).
TTrue
FFalse
Answer: True
Since ME = z*(σ/√n), doubling n replaces √n with √(2n) = √2 · √n. The margin of error becomes ME/√2 — a reduction by the factor √2 ≈ 1.41. For example, a study with n=100 and ME=±5% that doubles to n=200 would have ME ≈ ±5%/√2 ≈ ±3.5%. This is the precise quantitative expression of the 1/√n relationship.
Question 4 True / False
Doubling the sample size will cut the margin of error in half.
TTrue
FFalse
Answer: False
This is the most common misconception about sample size planning. Because ME ∝ 1/√n, doubling n reduces ME by only 1/√2 ≈ 29%, not by 50%. To cut the margin of error in half, you need to multiply n by 4. The confusion likely comes from thinking the relationship is linear (double the input, double the output), when it is actually a square-root relationship.
Question 5 Short Answer
Why does reducing margin of error from ±4% to ±2% require four times the sample size rather than twice the sample size?
Think about your answer, then reveal below.
Model answer: The margin of error is proportional to 1/√n. Halving ME means setting 1/√(n_new) = (1/2) × 1/√(n_old), which gives √(n_new) = 2√(n_old), so n_new = 4 × n_old. The square-root relationship means sample size must grow as the square of the desired precision improvement. Doubling n only improves precision by √2, not by 2.
This derivation is worth tracing explicitly. Start from ME = z*(σ/√n). If you want ME_new = ME_old/2, you need σ/√n_new = (1/2)(σ/√n_old), so √n_new = 2√n_old, meaning n_new = 4n_old. The practical upshot is that high precision is disproportionately expensive: going from ±4% to ±2% to ±1% requires sample sizes in the ratio 1:4:16.