Question

Suppose a sample of 523 suspected criminals is drawn. Of these people, 141 were captured. Using the data, construct a 95% confidence interval for the population proportion of people who are captured after appearing on the 10 most wanted list. Round your answers to three decimal places.

Answer 1

To solve this problem we use the following formula:

$p\pm z\cdot\sqrt{\frac{p(1-p)}{n}}$

Where:

p: proportion of the population

n: sample

z: This is the value given by the % of confidence and the normal distribution

First, we identify the variables of the problem:

n = 523

d = 141

alpha = (1-%confidence)/2

Second, we find the values of p and z:

$\begin{gathered} p=\frac{d}{n}=\frac{141}{523} \\ \\ 1-p=\frac{382}{523} \end{gathered}$

And for Z, we find the value of B such that:

$P(ZUsing excel, we can see that B = 1.96

Finally, we replace these values with the formula:

[tex]\frac{141}{523}\pm1.96\sqrt{\frac{141}{523}\cdot\frac{382}{523}\frac{1}{523}}$

As result, we get that:

$0.270\pm0.038$

And the inferior limit is equal to 0.232 and the superior limit is equal to 0.308