I have attempted to work this out twice use 2 different formulas (1.interval estimation of a population mean (small-sample case) 2. interval estimation of a population proportion) and still not getting the correct answer.....

QUESTION:
The monthly incomes from a random sample of faculty at a university are shown as:

monthly income
(in $1,000)
3.0
4.0
6.0
3.0
5.0
5.0
6.0
8.0

Compute a 90% confidence interval for the mean of the population. Give your answer in dollars.


Here's what i did:
1. used the population standard deviation formula to calculate and got S.D.=1581.13883

2. I plug that into the formula: both mentioned above.........PLEASE let me know which one is the correct formula to use???? I'm not sure because I got the wrong answer both times....


Can you please make any suggestions... and help me along the right direction, thanks!:)

what i know so far....

40,000/8=$5,000 mean
8 random selected
standard deviation=1581.13883 (if i calculated correctly)
90% confidence = 1.645 margin of error

This is a small-sample case, so you need to compute the sample mean, the sample standard deviation, and then use the t-distribution (t-table in the back of your book) to find the 90% confidence interval for the mean.

use the t-table (two-sided) with degrees of freedom n-1 = 7

let us know if you have any questions.....I'm sure your text has an example just like this:)