# Binomial proportion confidence interval lower bound

I'm not sure how to do this:

It was claimed that 15% of drivers on a particular road exceeded the speed limit. In a random sample of 600 drivers, 74 were observed to exceed the speed limit.

Give a one-sided lower 95% confidence bound for the proportion of drivers speeding

Let $\hat{p}$ be the proportion of drivers in your sample that exceeded the limit.

Let $p$ be the true proportion of drivers that exceed the limit.

Let $\hat{\sigma} = \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}}$ be the estimated standard deviation of $\hat{p}$

Let $n$ be your sample size.

Using a Normal approximation:

$$Z = \frac{\hat{p} - p}{\hat{\sigma}} \sim N(0,1)$$

From there you can calculate a one sided confidence bound.

Let $z_{0.95}$ be the value such that $P(Z \leq z_{0.95}) = 0.95$.

Substitute your expression for $Z$:

$$0.95 = P(Z \leq z_{0.95}) = P\left(\frac{\hat{p} - p}{\hat{\sigma}} \leq z_{0.95}\right) = P\left(\hat{p} + z_{0.95}\hat{\sigma} \leq p\right)$$

So your lower 95% confidence bound is $(\hat{p} + z_{0.95}\hat{\sigma}, \infty)$