Question

A 2003 survey showed that 14 out of 250 Americans surveyed had suffered some kind of identity theft in the past 12 months. What is the lower confidence limit of the 95% confidence i terval for the population proportion of Americans who were victims of identity theft

Answer 1

Answer:

The lower confidence limit of the 95% confidence interval for the population proportion of Americans who were victims of identity theft is 0.0275.

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

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

A 2003 survey showed that 14 out of 250 Americans surveyed had suffered some kind of identity theft in the past 12 months.

This means that $n = 250, \pi = \frac{14}{250} = 0.056$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.056 - 1.96\sqrt{\frac{0.056*0.944}{250}} = 0.0275$

The lower confidence limit of the 95% confidence interval for the population proportion of Americans who were victims of identity theft is 0.0275.