# The monthly incomes for 12 randomly selected people, each with a bachelor's degree in economics, are shown on the right. Complete parts 1–3 below.

Assume the population is normally distributed.

• 4450.35
• 4596.21
• 4366.32
• 4455.53
• 4151.38
• 3727.04
• 4283.95
• 4527.64
• 4407.75
• 3946.26
• 4023.07
• 4221.97
• Find the sample mean.x overbarxequals= (Round to one decimal place as needed.)
• Find the sample standard deviation.sequals= (Round to one decimal place as needed.)
• Construct a 99% confidence interval for the population mean muμ. A 99% confidence interval for the population mean is ( , ).
• The 99% confidence interval for the data in question is 4029.7 < pop. mean < 4496.3 or 4263.1 plus/minus 233.3.

We are given a list of twelve monthly salaries and are asked to find the 99% confidence interval about the population mean.

Adding the twelve pieces of data, we get `sum x=51157.47` so `bar(x)=51157.47/12~~4263.12` .

We are not given the population standard deviation, so we calculate the sample standard deviation `s~~260.17` .

Since we are using the sample standard deviation, we use a t-table to find the element of the standard error that takes into account our confidence level. (The standard error is `t_(alpha/2) s/sqrt(n)` .) From a student's t-table with degrees of freedom 11 (12-1) and `alpha=.01`, we get `t_(alpha/2)~~3.106`.

So, we can substitute the known values into the formula for the confidence interval:

`bar(x)-t_(alpha/2)(s/sqrt(n))<mu<bar(x)+t_(alpha/2)(s/sqrt(n))`

`4263.12-3.106(260.17/sqrt(12))<mu<4263.12+3.106(260.17/sqrt(12))`

`4029.73<mu<4496.27`

My calculator (rounding further out on the values for t, s, and the sample mean) gives `4029.9<mu<4496.4`.