Question

A police officer would like to estimate the proportion of petty theft cases that have gone unsolved in a large city. He takes a random sample of 72 petty theft cases and finds 45 of these have gone unsolved.
Set up the equation for the 90% confidence interval for p.
Format: Estimate ± Crit. Value times x Standard Error

Answer 1

Answer:

$0.625 \pm 1.645*0.0571$

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

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

In which

$\pi$ is the estimate

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$, also called the critical value.

The standard error is:

$s = \sqrt{\frac{\pi(1-\pi)}{n}}$

90% confidence level critical value

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.955$, so $z = 1.645$.

Estimate:

72 petty theft cases and finds 45 of these have gone unsolved.

So $\pi = \frac{45}{72} = 0.625$

Standard error:

$s = \sqrt{\frac{\pi(1-\pi)}{n}} = \sqrt{\frac{0.625*0.375}{72}} = 0.0571$

Answer:

$0.625 \pm 1.645*0.0571$