Sample Size Find the sample size necessary to estimate the mean arrival delay time for all American Airlines flights from JFK to LAX. Assume that we want 95% confidence that the sample mean is in error by no more than 5 min. Based on a larger sample than the one given for Exercises 1–4, assume that all arrival delay times have a standard deviation of 30.4 min.

It is given that a 95% confidence level and a margin of error equal to five minutes is used to determine the sample size for estimating the mean arrival delay time.

The population standard deviation is equal to 30.4 minutes.

The formula used to determine the sample size for estimating the mean arrival delay time is given as follows:

\(n = {\left( {\frac{{{z_{\frac{\alpha }{2}}}\sigma }}{E}} \right)^2}\)

Here, \(\sigma \) is given to be equal to 30.4 minutes.

Since the confidence level is equal to 95%, the value of the level of significance will be equal to 0.05.

Thus, the corresponding value of \({z_{\frac{\alpha }{2}}}\) is equal to 1.96.

The margin of error (E) is equal to five minutes.

Now, the value of the sample size is computed below:

\(\begin{array}{c}n = {\left( {\frac{{{z_{\frac{\alpha }{2}}}\sigma }}{E}} \right)^2}\\ = {\left( {\frac{{\left( {1.96} \right)\left( {30.4} \right)}}{5}} \right)^2}\\ = 142.0101\\ \approx 143\end{array}\)

Thus, the sample size for estimating themean arrival delay time with a 95% confidence level and margin of error equal to five minutes is equal to 143 American Airlines flights.