Confidence Interval Question

A social service agency plans to conduct a survey to determine the mean income of its clients. The director of the agency prefers that you measure the mean income very accurately, to within +- $500. From a sample taken 2 years ago, you estimate that the standard deviation of income for this population is about $5000. Your job is to figure out the necessary sample size to reduce sampling error to +- $500.

a. Do you need to have an estimate of the current mean income to answer this question? Why or why not.

b. what sample size should be drawn to meet the director's requirement at the 95 percent level of confidence?

c. what sample size should be drawn to meet the director's requirement at the 99 percent level of confidence?

I just copied the entire question, can someone point me in the right direction. The confidence interval formula does not take into account the size of the sample directly, so I do not understand how to figure the sample size needed.

Re: Confidence Interval Question

Quote:

Originally Posted by

**claw24** The confidence interval formula does not take into account the size of the sample directly...

Yes it does. It usually looks like this: $\displaystyle \frac{\sigma}{\sqrt{n}}$. See that 'n'? There's your sample size.

Let's see what you get.

Re: Confidence Interval Question

Quote:

Originally Posted by

**TKHunny** Yes it does. It usually looks like this: $\displaystyle \frac{\sigma}{\sqrt{n}}$. See that 'n'? There's your sample size.

Let's see what you get.

So it would be 5000 / sqrt(N)? I don't understand how to figure out N because I do not have the sampling error that would be included in the confidence interval formula.

Re: Confidence Interval Question

Look carefully at your confidence interval. How do you calculate the upper bound?

Something like: $\displaystyle \;\;\mu\;+\;z_{\alpha}\frac{\sigma}{\sqrt{n}}$ right?

How can we use this information, compared to the maximum acceptable error, to back into the Sample Size?