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.

    Expert Answers

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    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:




    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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