Confidence Intervals for Proportions
A free Statistics and Data Analysis lesson from the “Inference and Conclusions” unit, with a worked example and practice problems including step-by-step solutions.
For a population proportion, calculate p-hat=x/n and the estimated SE=√[p-hat(1−p-hat)/n]. A large-sample interval is p-hat ± z*SE. Verify random sampling/assignment as relevant, independence, and at least 10 observed successes and failures for this normal approximation. When sampling without replacement, check the sample is at most about 10% of the population. A 95% procedure covers the fixed parameter in about 95% of repeated samples. For independent proportions, estimate p1−p2 using SE=√[p1-hat(1−p1-hat)/n1+p2-hat(1−p2-hat)/n2]. Report differences in percentage points, and do not confuse an interval SE with the pooled SE used for an equality test.
What you'll learn
- Build a confidence interval for a proportion
- Interpret confidence level
- Write a conclusion in context
Worked example
Problem. A random sample has n=400 and 220 successes. Assuming the normal-approximation conditions hold, use z*=1.96 to choose a 95% population-proportion interval.
- Estimate=0.55; SE≈0.0249.
- Margin≈0.0488.
- Estimate ± margin gives 0.5012 to 0.5988.
Answer: 0.5012 to 0.5988
Lesson preview: this page shows 5 practice items. The interactive activity contains 8 practice items and 8 quiz items (typed response and multiple choice). Counts include repeated prompts and variants.
Practice problems
1. In a simple random sample of 400 voters, 220 support a proposal. Find the sample proportion.
Show solution
- p-hat=220/400=0.55.
Answer: 0.55
2. With p-hat=0.55 and n=400, calculate SE=√[p-hat(1−p-hat)/n] to four decimals.
Show solution
- SE=√[0.55×0.44999999999999996/400]≈0.0249.
Answer: 0.0249
3. Using z*=1.96 and unrounded SE=√[0.55×0.44999999999999996/400], calculate the 95% margin of error to four decimals.
Show solution
- Margin=1.96×SE≈0.0488.
- Keep unrounded intermediate values.
Answer: 0.0488
4. A random sample has n=400 and 220 successes. Assuming the normal-approximation conditions hold, use z*=1.96 to choose a 95% population-proportion interval.
Choices: 0.5012 to 0.55 · 0.5012 to 0.5988 · 0.5256 to 0.5744 · 0.55 to 0.5988
Show solution
- Estimate=0.55; SE≈0.0249.
- Margin≈0.0488.
- Estimate ± margin gives 0.5012 to 0.5988.
Answer: 0.5012 to 0.5988
5. A sample has 4 successes and 16 failures. Is the ordinary large-count normal-approximation proportion interval justified by these counts?
Choices: No; the success count is below 10 · Yes; the total is at least 20 · Yes; percentages are always normal · No; intervals cannot estimate proportions
Show solution
- The usual success/failure check requires at least 10 of each for this approximation.
- Use an appropriate alternative rather than applying it blindly.
Answer: No; the success count is below 10
Explain and apply
Complete these tasks on paper. Use the rubric to review your reasoning with a teacher or study partner; these tasks are not automatically scored.
Two populations: estimate a difference in support
Independent simple random samples of 200 people from each of two large towns find support from 120 in Town A and 90 in Town B. Construct and interpret a 95% interval for pA−pB using z*=1.96. State the checks that justify this approximation.
Review rubric and solution
- Use estimates .60 and .45; difference=.15.
- Check independent sampling, small sampling fractions and at least 10 successes/failures per sample.
- Calculate SE=√(.6×.4/200+.45×.55/200)≈.04937.
- Construct .15±.09677≈(.0532,.2468).
- Interpret the difference as about 5.3 to 24.7 percentage points, not a causal effect.
- The estimated difference is .15.
- SE≈.04937; margin≈.09677.
- We are 95% confident Town A's support proportion exceeds Town B's by about 5.3 to 24.7 percentage points.
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