On-Time Arrivals According to the Bureau of Transportation, American Airlines had an on-time arrival rate of 80.3% for a given year. Assume that this statistic is based on a sample of 1000 randomly selected American Airlines flights. Find the 99% confidence interval estimate of the percentage of all American Airlines flights that arrive on time.

The proportion of flights that have an on-time arrival is equal to 80.3%. The sample size is equal to 1000. The confidence level is given to be equal to 99%.

The formula for computing the confidence interval estimate of the population proportion of flights that arrive on time is given below:

$CI=\left(\hat{p}-E,\hat{p}+E\right)$

Here, $\hat{p}$is the sample proportion, and E is the margin of error.

$E=z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}\hat{q}}{n}}$

The sample proportion of flights that arrive on time is calculated in the following manner:

$$\begin{gathered} \hat{p}=80.3\% \\ =\frac{80.3}{100} \\ =0.803 \end{gathered}$$

The sample proportion of flights that do not arrive on time is is calculated in the following manner:

$$\begin{gathered} \hat{q}=1-\hat{p} \\ =1-0.803 \\ =0.197 \end{gathered}$$

The sample size is given as n=1000.

The confidence level is given to be equal to 99%. Thus, the level of significance is equal to 0.01.

The corresponding value of $z_{\frac{\alpha }{2}}$is equal to 2.5758.

The margin of error is calculated in the following manner:

$$\begin{gathered} E=z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}\hat{q}}{n}} \\ =2.5758\sqrt{\frac{\left(0.803\right)\left(0.197\right)}{1000}} \\ =0.0324 \end{gathered}$$

The 99% confidence interval is as follows:

$$\begin{gathered} CI=\left(\hat{p}-E,\hat{p}+E\right) \\ =\left(0.803-0.0324,0.803+0.0324\right) \\ =\left(0.7706,0.8354\right) \\ =\left(77.06\%,83.54\%\right) \end{gathered}$$

$=\left(77.1\%,83.5\%\right)$

Therefore, the 99% confidence interval estimate of the population percentage of flights that arrive on time is equal to (77.1%,83.5%).