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Confidence intervalsStatistics and probabilityCollegeabout 1:39

A sample of 200 has a sample proportion of 0.42. Find the 99% confidence interval for the proportion: animated lesson

A narrated, step-by-step confidence intervals video lesson with captions. Final answer: 99% CI: (0.330104, 0.509896); margin of error 0.0898961.

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Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.7 MB) · Captions (WebVTT)

Steps on screen

1. The problem
Worked example
A sample of 200 has a sample proportion of 0.42. Find the 99% confidence interval for the proportion.
2. Steps, part 1
1. $\hat p = 0.42$
  Sample proportion
2. $SE = \sqrt{\frac{\hat p(1-\hat p)}{n}} = \sqrt{\frac{0.42(1-0.42)}{200}} \approx 0.0348999$
  Standard error
3. $z^* = 2.57583$
  Critical value for 99% confidence
4. $E = z^* \times SE = 2.57583 \times 0.0348999 \approx 0.0898961$
  Margin of error
3. Steps, part 2
1. $0.42 \pm 0.0898961 = (0.330104,\ 0.509896)$
  Interval: estimate ± margin
4. The answer
$(0.330104,\ 0.509896)$
0.42 ± 0.0898961
5. Check
- The interval is centred on the estimate: (0.330104 + 0.509896) / 2 = 0.42.
- Large-sample condition met: np̂ ≈ 84 and n(1 − p̂) ≈ 116 are both at least 10.

Transcript

The problem. Here is the problem: A sample of 200 has a sample proportion of 0.42. Find the 99 percent confidence interval for the proportion. Build a 99 percent confidence interval for the population proportion p.

Steps, part 1. Sample proportion. Standard error. Critical value for 99 percent confidence. The middle 99 percent leaves 0.5 percent in each tail. Large-sample condition met: np̂ approximately 84 and n(1 minus p̂) approximately 116 are both at least 10. Margin of error.

Steps, part 2. Interval: estimate plus or minus margin.

The answer. So the answer is 99 percent CI: (0.330104, 0.509896); margin of error 0.0898961. We are 99 percent confident that the population proportion p lies between 0.330104 and 0.509896. The confidence describes the method: about 99 percent of intervals built this way from repeated samples would capture the true value.

Check. The interval is centred on the estimate: (0.330104 plus 0.509896) / 2 equals 0.42. Large-sample condition met: np̂ approximately 84 and n(1 minus p̂) approximately 116 are both at least 10.

About confidence intervals

Build confidence intervals for a mean or a proportion.

A sample of 200 has a sample proportion of 0.42. Find the…