Richard C. answered • 07/24/16

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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/√nwhere 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/√748To 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.