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.

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Expert Answers

| Certified Educator

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`.

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