Question

Out of 200 people sampled, 170 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids.Give your answers as decimals, to three places

Answer 1

The sample proportion is calculated as:

$p=\frac{170}{200}=0.85$

The critical value for significance level = 1 - 0.99 = 0.01, from the z table is z = 2.3263

Therefore, the 99%confidence interval is computed as follows:

$CI=(p-z_c\sqrt{\frac{p(1-p)}{n}},p+z_c\sqrt{\frac{p(1-p)}{n}})$

Where:

p = 0.85

zc = 2.3263

n = 200

Substitute the values we have:

$CI=(0.85-2.3263\sqrt{\frac{0.85(1-0.85)}{200}},0.85+2.3263\sqrt{\frac{0.85(1-0.85)}{200}})$

Simplify:

$\begin{gathered} CI=(0.85-0.0587,0.85+0.0587) \\ CI=(0.7913,0.9087) \end{gathered}$

For 3 decimal places is (0.791, 0.909).

Answer: (0.791, 0.909)