Answer
The administrator should not be concerned (at the 5% confidence level).
Work Step by Step
Here we have: $H_{o}$: p = 0.05, $H_{a}: p > 0.05$, p̂ = 0.068
Using the p-value approach:
$z = \frac{0.05 - 0.068}{\sqrt \frac{0.05 (1-0.05)}{250}} = \frac{0.18}{0.0138} = -1.31$
P(z > -1.31) = 1 - P(z < 1.31) = 0.0951
Since 0.0951 > 0.05, we fail to reject the null hypothesis.
