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{(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=(d)/(n)=(141)/(523) \\ \\ 1-p=(382)/(523) \end{gathered}`

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

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