need the lower and upper endpoint

Question

An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California.  Suppose that the mean income is found to be  $24.4 for a random sample of  748 people.  Assume the population standard deviation is known to be  $10.6.  Construct the  85% confidence interval for the mean per capita income in thousands of dollars. 

Answer 1

Chels,

 

Since we know the population std deviation (10.6) and the sample size is >30, we'll use the normal z-distribution to help answer this question (we use the z-distribution since it's characteristics [mean, std deviation, etc] are all known; we don't know all of these for our real population).

 

The confidence interval represents that range of z-values we'd expect our population mean to fall within.  The 85% level represents the amount of risk we want to take that the population value of the mean lies OUTSIDE the range.

 

Now, the general expression for the CI in this case is:

 

sample mean ± z_critical * σ/√n

 

where z_critical is the boundary value of the z-distribution given by our 85% level of risk. σ is the population standard deviation and n is the sample size.

 

In this case then, we have:

 

24.4 ± z_critical * 10.6/√748

 

To find z_critical , look up 85% on a z-table (in your text or online).  When you do, you should find a value of 1.036. 

 

So, we now have;

 

24.4 ± 1.036 * 10.6/√748 = 24.4 ± .40

 

So the CI for 85% is:

 

$24,000 - $24800; the population mean will lie in this range with 85% probability.