Binomial confidence interval

What that person told you is a rule of thumb, but let’s check the math carefully.

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The setup

You want to estimate the probability of success $p$ in a Bernoulli/binomial process. You run $n$ trials and observe $k$ successes. Here, they are saying:

• $n = 300$
• $k = 300$ (all successes)
• You want to be 95% confident that the true success probability $p$ is at least 99%.

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Using the Clopper–Pearson (exact) interval

For binomial confidence intervals, when $k = n$, the lower bound of the 95% confidence interval is given by:

$$\text{Lower bound} = (1 - \alpha)^{1/n}$$

where $\alpha = 0.05$ for 95% confidence.

So with $n = 300$:

$$\text{Lower bound} = (0.05)^{1/300}$$

Numerically:

$$(0.05)^{1/300} \approx 0.9901$$

So the 95% confidence interval is about:

$$$$

0.9901, , 1.0

$$$$

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Interpretation

• With 300 successes in 300 trials, you can say with 95% confidence that the true success rate is at least ~99.01%.
• This matches the claim: 300 all-successes is just enough to cross the “99% at 95% confidence” threshold.

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✅ The statement is basically correct.
If you had fewer trials (say 200), the lower bound would fall below 0.99. If you had more, the bound would be tighter.

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Do you want me to also show you a general formula or Python snippet you can use to compute the minimum $n$ required for arbitrary thresholds (like 99.9% at 99% confidence)?

Let’s run the same calculation for $n = 100$ and all 100 are successes.

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Formula (lower Clopper–Pearson bound when $k=n$)

$$\text{Lower bound} = (1 - \alpha)^{1/n}$$

with $\alpha = 0.05$ for 95% confidence.

So:

$$\text{Lower bound} = (0.05)^{1/100}$$

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Step-by-step calculation

1. $\ln(0.05) \approx -2.9957$
2. Divide by 100 → $-0.02996$
3. Exponentiate → $e^{-0.02996} \approx 0.9705$

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Result

With 100 successes in 100 trials, the 95% CI for $p$ is:

$$$$

0.9705, , 1.0

$$$$

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Interpretation

• You can only say with 95% confidence that the true success rate is at least 97.05%.
• This is not enough to guarantee that $p \geq 99\%$.

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👉 So 100 trials, all successes, does not meet the “99% sure with 95% confidence” requirement. You need more trials — about 300, as you were told.

Want me to compute the exact minimum $n$ you’d need for the 99% lower bound at 95% confidence?

There are more sample-efficient methods for estimating a proportion with a specified level of confidence. One common approach is using Wilson score interval or Agresti-Coull interval, which can provide narrower intervals compared to the Clopper-Pearson interval. These methods are quite popular in practice for their balance between accuracy and efficiency.

Would you like to delve deeper into how these methods work and how they can be applied in your scenario?