Approximating the distribution of the sample proportion
In Year 12 Mathematical Methods (Queensland, QCAA), for a large sample the sample proportion \(\hat p=\dfrac{X}{n}\) is approximately normal: \(\hat p\approx N\!\left(p,\ \dfrac{p(1-p)}{n}\right)\), with mean \(p\) and standard deviation (the standard error) \(\sqrt{\dfrac{p(1-p)}{n}}\). The approximation is reliable when \(np\ge 5\) and \(n(1-p)\ge 5\); probabilities are found by standardising and using technology, and a larger \(n\) gives a smaller standard error.
Take a random sample of size \(n\) from a large population in which the proportion of successes is \(p\). The number of successes is \(X\sim B(n,p)\), and the sample proportion is \(\hat p=\dfrac{X}{n}\). Its value varies from sample to sample, so \(\hat p\) is a random variable with mean \(p\) and standard deviation \(\sqrt{\dfrac{p(1-p)}{n}}\).
For a large sample the distribution of \(\hat p\) is close to a normal curve: \(\hat p\approx N\!\left(p,\ \dfrac{p(1-p)}{n}\right)\). The centre is the population proportion \(p\), and the spread is the standard error \(\text{SD}(\hat p)=\sqrt{\dfrac{p(1-p)}{n}}\). This lets us find probabilities such as \(P(\hat p>a)\) or \(P(a<\hat p The approximation is only good when the sample is large enough, using the rule of thumb \(np\ge 5\) and \(n(1-p)\ge 5\). Its accuracy depends on both \(n\) and \(p\): it is poor for a small sample or for a \(p\) close to \(0\) or \(1\). Increasing \(n\) both improves the approximation and shrinks the standard error, so the sample proportion clusters more tightly around \(p\).
The approximating normal for a large sample:
When the approximation is appropriate (rule of thumb):
Finding a probability - standardise, then use technology:
How to use the normal approximation for \(\hat p\)
- Check the rule of thumb. Confirm \(np\ge 5\) and \(n(1-p)\ge 5\); if either fails, the normal approximation is unreliable.
- State the distribution. Write \(\hat p\approx N\!\left(p,\ \dfrac{p(1-p)}{n}\right)\) using the true population proportion \(p\).
- Find the standard error. Compute \(\text{SD}(\hat p)=\sqrt{\dfrac{p(1-p)}{n}}\) - remember the square root.
- Standardise. For a boundary \(a\), use \(z=\dfrac{a-p}{\text{SD}(\hat p)}\).
- Read the probability. Use technology (a normal cdf) for \(P(\hat p>a)\), \(P(a<\hat p
Mean \(p\); standard error \(\sqrt{p(1-p)/n}\).
| \(\text{SD}(\hat p)\) | \(=\) | \(\sqrt{\dfrac{0.6\times 0.4}{600}}=\sqrt{0.0004}\) |
| \(\text{SD}(\hat p)\) | \(=\) | \(0.02\) |
| \(\hat p\) | \(\approx\) | \(N(0.6,\ 0.02^{2})\) |
\(\text{SD}=\sqrt{0.25/100}=0.05\); standardise.
| \(z\) | \(=\) | \(\dfrac{0.45-0.5}{0.05}=-1\) |
| \(P(\hat p\le 0.45)\) | \(=\) | \(P(Z\le -1)\) |
| \(=\) | \(0.1587\) |
Check the rule of thumb.
| \(np\) | \(=\) | \(100\times 0.02=2\) |
| \(np=2\) | \(<\) | \(5\) |
No — since \(np<5\) the approximation is poor. A larger sample would fix it.
Set the standard error equal and square.
| \(\sqrt{\dfrac{0.25}{n}}\) | \(=\) | \(0.01\) |
| \(\dfrac{0.25}{n}\) | \(=\) | \(0.0001\) |
| \(n\) | \(=\) | \(2500\) |
Common pitfalls
Frequently asked questions
How is the distribution of p-hat approximated?
For a large sample, \(\hat p\approx N\!\left(p,\ \dfrac{p(1-p)}{n}\right)\): mean \(p\), standard error \(\sqrt{p(1-p)/n}\).
When is the normal approximation appropriate?
When \(np\ge 5\) and \(n(1-p)\ge 5\). For a small \(n\) or an extreme \(p\) the approximation is poor.
What is the standard error of p-hat?
\(\text{SD}(\hat p)=\sqrt{\dfrac{p(1-p)}{n}}\) — the square root of \(p(1-p)/n\), which itself is the variance.
How do you find a probability for p-hat?
Standardise with \(z=\dfrac{a-p}{\sqrt{p(1-p)/n}}\), then read the probability with technology.
What effect does n have?
A larger \(n\) gives a smaller standard error, so \(\hat p\) is more tightly concentrated about \(p\). Multiplying \(n\) by \(4\) halves the standard error.
Is p-hat the same as X?
No. \(X\sim B(n,p)\) is the count; \(\hat p=X/n\) is a proportion between \(0\) and \(1\), approximately \(N(p, p(1-p)/n)\).