Solution to Exercise 7.22.

  Using the formula on p. 293, the large sample upper confidence bound for $\mu$ is

$$\bar{x}+ z_{\alpha} * s /\sqrt{n} .$$

As in all the formulae for confidence bounds for a proportion p, we replace $\bar{x}$ by $\hat{p}$ and s by $\sqrt{\hat{p}\hat{q}}$. We have

$$\hat{p} \; = \; .072 , \quad \hat{q} \; = \; 1-.072 \; = \; 0.928 .$$

Also, z_.01 = 2.326 (for 99% confidence). Thus, our 99% upper confidence bound is

$$\hat{p}+ z_{\alpha} * \sqrt{\hat{p}\hat{q}/n} \; = \; .072 + 2.326 * \sqrt{.072*.928/487}$$

$$\; = \; .072 + 0.09924492 \; = \; .072 + 0.099 \; = \; 0.171 .$$