Question

a national air traffic control system handled an average of 47,708 flights during 29 randomly selected days in a recent year. the standard deviation for this sample is 6,486 flights per day. complete parts a through c below.
a. construct a 99% confidence interval to estimate the average number of flights per day handled by the system.
the 99% confidence interval to estimate the average number of flights per day handled by the system is from a lower limit of to an upper limit of . (round to the nearest whole numbers.)

Explanation:

Step1: Identify the formula

For a confidence interval when the population standard - deviation is unknown (we use the sample standard - deviation $s$), the formula is $\bar{x}\pm t_{\alpha/2}\frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $t_{\alpha/2}$ is the critical value, $s$ is the sample standard - deviation, and $n$ is the sample size.

Step2: Determine the values of $\bar{x}$, $s$, and $n$

We are given that $\bar{x} = 47708$, $s = 6486$, and $n=29$.

Step3: Find the degrees of freedom and the critical value $t_{\alpha/2}$

The degrees of freedom $df=n - 1=29 - 1 = 28$. For a 99% confidence interval, $\alpha=1 - 0.99 = 0.01$, and $\alpha/2=0.005$. Looking up in the $t$ - distribution table, $t_{0.005,28}\approx 2.763$.

Step4: Calculate the margin of error $E$

$E=t_{\alpha/2}\frac{s}{\sqrt{n}}=2.763\times\frac{6486}{\sqrt{29}}\approx2.763\times\frac{6486}{5.385}\approx2.763\times1204.457\approx3328.9$.

Step5: Calculate the lower and upper limits of the confidence interval

The lower limit is $\bar{x}-E=47708 - 3328.9\approx44379$.
The upper limit is $\bar{x}+E=47708 + 3328.9\approx51037$.

Answer:

The 99% confidence interval to estimate the average number of flights per day handled by the system is from a lower limit of 44379 to an upper limit of 51037.