Inference for a proportion is really similar to inference for a mean!

• We can apply the Central Limit Theorem to the sampling distribution for a proportion.
  • But… isn’t our Central Limit Theorem only for means?

• A binomial experiment is made up of a series of Bernoulli trials, which result in 0s and 1s.
• If we add up these values, we get the number of successes \(x\).
• If we take the mean of these successes, we get the proportion of successes.
• That is, \(\bar{x} = \hat{p}\) and we can work with the sampling distribution for a sample mean!

By the Central Limit Theorem, \(\hat{p}\) is approximately normally distributed with mean \[\mu_{\hat{p}} = p\] and standard error \[\sigma_{\hat{p}} = \frac{\sqrt{p(1-p)}}{\sqrt{n}} = \sqrt{\frac{p(1-p)}{n}}\]

• Confidence intervals all use the same basic formula: \[\text{estimate }\pm\text{ critical value }\times\text{ standard error }\]
• We do not know the true value of \(p\) for the standard error \[\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\] so we will plug in \(\hat{p}\).

A \(100(1-\alpha)\%\) confidence interval for \(p\):

\[\hat{p}\pm z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]

To use this formula, we need to check that \(n\hat{p} > 10\) and \(n(1-\hat{p})>10\).

Suppose we take a random sample of 27 US households and find that 15 of them have dogs. Find a 95% confidence interval for the proportion of US households with dogs.