Question

A random sample of 50 airline pilots had an average yearly income of $103,450 with a standard deviation of $10,000.
1. if we want to determine a 90% co fidence interval for the average yearly income, what is the value of t?
2. Develop a 90% confidence interval for the average yearly income of all pilots.

Answer 1

Given : = $103,450, s = $10,000, n = 50

(1) The tcritical (2 tail) for = 0.10, for df = n -1 = 49, is 1.677

(2) The Confidence Interval is given by ME, where

The Lower Limit = 103,450 - 2371.436 = 101,078.364

The Upper Limit = 103,450 + 2371.436 = 105,821.636

The 90% Confidence Interval is (101078.364 , 105821.636)

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If rounded to 2 decimals the CI will be (101078.36, 105821.64)

If Rounded to 1 decimal the CI will be (101078.4, 105821.6)

If Rounded to the nearest integer the CI will be (101078, 105822)

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