The interval for a proportion

What you will be able to do

• find a 95% confidence interval for a proportion;
• check the three conditions before you use it;
• work out the sample size needed for a given margin of error.

The formula

Put the pieces of the last lesson together. The 95% confidence interval for a proportion is

A made-up survey asks $400$ students, chosen at random, how they come to school, and $92$ say they walk. Then

1. Estimate. $\hat p = \dfrac{92}{400} = 0.23$.
2. Standard error. $\sqrt{\dfrac{0.23 \times 0.77}{400}} \approx 0.0210$.
3. Margin. $1.96 \times 0.0210 \approx 0.041$.
4. Interval. From about $0.189$ to about $0.271$.

So we are $95\%$ confident that between about $19\%$ and $27\%$ of all the students walk.

Three conditions

The formula leans on the normal curve and on the standard error rule. Both need conditions.

1. Random. The data come from a random sample, or from a randomised experiment. No formula can repair a biased sample.
2. Large enough. There are at least $10$ successes and at least $10$ failures in the sample. Then the sample shares are close to a normal curve.
3. Independent. The sample is less than about a tenth of the population, so that one choice hardly changes the next.

A survey of $150$ people in which only $6$ own a kite fails the second condition. The sample shares of such a survey are skewed, and the interval would be misleading. Some books use $5$ in place of $10$ in the second condition.

How many to ask

Before a survey, choose the margin $m$ you want. Then solve $1.96\sqrt{\dfrac{\hat p(1 - \hat p)}{n}} \le m$ for $n$:

You do not know $\hat p$ yet. Use a guess from an earlier study, or use $0.5$, which gives the largest value of $\hat p(1 - \hat p)$ and so the safest $n$. For a margin of $3$ points with no guess, $n \ge \left(\dfrac{1.96}{0.03}\right)^2 \times 0.25 \approx 1067.1$. So ask $1068$ people. This is why so many national polls ask about a thousand people.

The picture shows the margin of error against the sample size for two expected shares. Both curves fall steeply at first and then slowly. Halving the margin always takes four times the sample.

Practise with new numbers

Try it in Python

The program checks the conditions and then gives the interval for three made-up surveys.

Common mistakes

• Skipping the conditions. Check random, at least $10$ successes and $10$ failures, and less than a tenth of the population.
• Rounding the sample size down. A sample must be a whole number of people, and rounding down gives too large a margin.
• Using the margin of the whole sample for a small group in it. A group of $100$ inside a survey of $1000$ has a much larger margin.

Remember these

• The interval is $\hat p \pm 1.96\sqrt{\dfrac{\hat p(1 - \hat p)}{n}}$.
• Conditions: a random sample, at least $10$ successes and $10$ failures, a sample under a tenth of the population.
• Sample size: $n \ge \left(\dfrac{1.96}{m}\right)^2 \hat p(1 - \hat p)$, with $\hat p = 0.5$ when nothing is known.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A made-up survey of $200$ people finds $50$ who cycle to work. Find the 95% interval for the share of cyclists, to three decimal places.

2. A survey of $80$ people finds $7$ who own an electric car. Can the formula be used?

3. With no guess for the share, how many people must you ask for a margin of $5$ points?

4. A team wants to halve its margin of error. By what factor must it multiply its sample size?

5. A random sample of $500$ is taken from a club of $2000$ members. Which condition is in doubt?

References

• OpenStax, Introductory Statistics, Chapter 8, section on a population proportion, including the sample size needed.