Math Detours · Grade 11 · Sharpen · 16

Plus or minus three points

A poll of 1,000 people reports 52% ± 3%. Where the 3 comes from, why it barely depends on the size of the country, and what it does not cover.

Name ____________________Date ____________

Tonight's idea

Ask \(n\) people chosen at random, and a fraction \(\hat p\) ("p hat") say yes. The true fraction in the whole population is \(p\). A second random sample would give a slightly different \(\hat p\). How different?

Score each answer 1 for yes and 0 for no. That is a random variable with mean \(p\) and variance \(p(1 - p)\). Averaging \(n\) independent answers divides the variance by \(n\), so \(\hat p\) has standard deviation \(\sqrt{p(1 - p)/n}\). For large \(n\) its spread is bell-shaped, and 95% of a bell lies within 1.96 standard deviations of the middle.

Margin of error \(\approx 1.96\sqrt{\dfrac{p(1 - p)}{n}}\). At \(p = 0.5\) this is \(\dfrac{0.98}{\sqrt n}\), about \(\dfrac{1}{\sqrt n}\).

So about 95% of random samples land within the margin of the truth. Missing from the formula: the size of the population. A spoonful tells you how the soup tastes whether the pot is small or huge, as long as the pot is stirred.

500 1,000 1,500 2,000 2,500 ±5 ±10 ±15 n = 100: ±9.8 400: ±4.9 1,000: ±3.1 2,500: ±2.0 people asked, n margin, percentage points
The margin at \(p = 0.5\), which is \(0.98/\sqrt n\). Four times the people buys only half the margin.

The margin covers the luck of the draw and nothing else. It says nothing about who was asked (a sample that was not random) or how the question was worded. Those errors do not shrink as \(n\) grows.

Watch one: 52% of 1,000

Nobody knows \(p\), so use \(\hat p = 0.52\) in its place:

\[1.96\sqrt{\frac{0.52 \times 0.48}{1000}} = 1.96 \times 0.0158 = 0.031.\]

That is 3.1 percentage points: the "± 3". The rule of thumb, \(1/\sqrt{1000} = 0.032\), gives 3.2 points: close enough. The poll supports any true value from about 48.9% to 55.1%. That range includes 50%, so this poll alone does not show a majority.

In the demo, 36% of a town of 5,000 would say yes. Ask 200 samples of 1,000 and most land within 3 points of 36%. Then tick "Lopsided sample" and the whole pile slides away from the truth: mathdetours.com/demos/sample.html.

Your turn

  1. Use the rule of thumb \(1/\sqrt n\) to find the margin of error, in percentage points, for random polls of 100, 400 and 2,500 people.
  2. A random poll of 600 people finds 30% in favour. Use the full formula to find the margin of error, and give the range of values the poll supports.
  3. How many people must you ask for a margin of 1 percentage point when \(p\) is near 0.5? Use the full formula. Roughly how many times bigger is that than a poll of 1,000?
  4. A news website invites readers to click yes or no. It gets 100,000 clicks and 70% say yes. The same week a random poll of 1,000 people asks the same question and finds 52% yes. Compute the formula's margin for each. The two ranges do not overlap. Which result do you believe, and what has the formula left out?

Stretch: why 0.5 is the worst case

Complete the square to show that \(p(1 - p) = \tfrac14 - \left(p - \tfrac12\right)^2\). Explain why this makes the margin largest at \(p = 0.5\). Then find the margin for a random poll of 1,000 in which 2% say yes.

Answers

  1. \(1/\sqrt{100} = 0.10\), \(1/\sqrt{400} = 0.05\), \(1/\sqrt{2500} = 0.02\): margins of 10, 5 and 2 points. Four times the people halves the margin.
  2. \(1.96\sqrt{\dfrac{0.3 \times 0.7}{600}} = 1.96 \times 0.0187 = 0.037\), so 3.7 points. The poll supports values from about 26.3% to 33.7%.
  3. \(\dfrac{0.98}{\sqrt n} = 0.01\) gives \(\sqrt n = 98\), so \(n = 9{,}604\). That is nearly ten times the poll of 1,000, for a margin about a third the size. This is why most polls stop near 1,000.
  4. Website: \(1.96\sqrt{\dfrac{0.7 \times 0.3}{100000}} = 0.0028\), about 0.3 points. Random poll: 3.1 points, as in the example. Believe the random poll. The formula assumes every person was equally likely to be asked. The website's readers chose themselves: they are that site's audience, and the ones who cared enough to click. That error is not in the formula, and 100,000 clicks do not shrink it. A tiny margin on a lopsided sample is a precise answer to the wrong question.

Stretch. \(p(1 - p) = -(p^2 - p) = -\left(\left(p - \tfrac12\right)^2 - \tfrac14\right) = \tfrac14 - \left(p - \tfrac12\right)^2\). The squared term is never negative and is zero only at \(p = \tfrac12\), so \(p(1 - p)\) is at most \(\tfrac14\), and the margin is largest there. At \(\hat p = 0.02\): \(1.96\sqrt{\dfrac{0.02 \times 0.98}{1000}} = 0.0087\), about 0.9 points. A poll that says ± 3 is quoting its worst case.

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