Question

Assume you are provided the following 3 types of information and for each case you are assigned to calculate a 90% confidence interval around the sample mean. Answer the questions below based on the provided information.

An airline wants to estimate the number of tickets sold for a daily flight between Los Angeles and New York. Suppose a sample of 10 flights gives a sample mean of 125.5  (Assume that the population standard deviation is σ = 20.5 and the population follows a normal distribution).

Answer 1

Solution:
Given that, μ= 125.5, σ=20.5 , n=10
(1–α)%=90%
α =0.10
α/2 =0.05
Zα/2 =1.645 ....from standard normal table.
Margin of error=E=Zα/2 ×(σ/✓n)
=1.645 ×(20.5/✓10)
=1.645 × 6.4827
=10.6640
Margin of error=E=10.6640
90% confidence interval for true population mean given as,
x̄ ± Margin of error

=(125.5 -10.6640,125.5+10.6640)

=(114.836,136.164)
Lower limit =114.836
Upper limit=136.164
90% confidence interval for true population mean is (114.836,136.164)