Answer
For 95% confidence interval: $z_{0.025} = 1.96$
For 99% confidence interval: $z_{0.005} = 2.575$
$E = z.\sqrt \frac{p(1-p)}{n}$
Assuming p is the same for both, we have:
$\frac{2.575}{\sqrt 400} < \frac{1.96}{\sqrt 200} $
Hence Matthew's estimate will have a smaller margin of error.
Work Step by Step
Given above.
