Originally Posted by

**TKHunny** You should not be struggling with this. Go ahead and construct the confidence interval.

Find the Sample Mean. *

Find the Sample Standard Deviation. *

Find the Standard Deviation of the Sample Mean. <== This is the important part that will allow you to answer the second question. ? Mr F says: This is equal to $\displaystyle \, \frac{\sigma}{\sqrt{n}} \,$ where $\displaystyle \, \sigma \,$ is the population standard deviation. If $\displaystyle \, \sigma \,$ is unknown, it needs to be estimated by $\displaystyle \, s \,$. If $\displaystyle \, n \,$ is large ($\displaystyle \, n \geq 50 \, $, although some books will say $\displaystyle \, n \geq 30 \, $) then $\displaystyle \, s \approx \sigma$ and you can use the z-test. Otherwise you must use the t-test (assuming a roughly normal distribution).

Decide if you can use a z-score or a t-score. ?

Find the right value for a 90% confidence interval. Is it one-tail or two-tail? ?

That's a good start. Show us what you get and we can move on from there.