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united airlines flight 179 is a nonstop flight from boston to san francisco. during 2016 the mean length of time for all flights was 357.9 minutes and population standard deviation was 20.18 minutes. you want to construct a 95% confidence interval for the mean length of time of the flight in 2018 based on a sample of flights in 2018. what is the minimum size of the sample needed in order to have a margin of error of two minutes or less?



Answer :

Using the concepts of statistics, we got that at least 79 is the minimum size of the sample needed in order to have a margin of error of two minutes or less, when we need to construct a 95% confidence interval for the mean length of time.

We have mean as  357.9minutes and standard deviation as 20.18 minutes.

To have the margin error 2 minutes or less than it we need to take

Z-score of the given sample distribution which is given by:

                Z=(X-μ)/σ

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation .

Here ,μ=357.9,σ=20.18,X=?,Z=95/100

=>95/100=(X-357.9)/20.18

=> -0.95×20.18=X-357.9

=>X=-357.9+0.95×20.18

=>X=-357.9+19.171

>X=-338.729

When X=-338.729,the p score is around 79.

Hence, when I want to construct a 95% confidence interval for the mean length of time of the flight in 2018 based on a sample of flights in 2018., the minimum size of the sample needed in order to have a margin of error of two minutes or less is 79.

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