Question

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.1 seats.

a. i. $\overline{x}=$__________

ii. $s_x=$__________

iii. $n=$__________

iv. $n-1=$__________

b. Define the random variables $X$and $\overline{X}$in words.

c. Which distribution should you use for this problem? Explain your choice.

d. Construct a 92% confidence interval for the population mean number of unoccupied seats per flight.

i. State the confidence interval.

ii. Sketch the graph.

iii. Calculate the error bound.

Short answer

a. i. $\overline{x}=6$

ii. $s_x=3$

iii. $n=14$

iv. $n-1=13$

b. $X$is the number of unoccupied seats on a single flight. $\overline{X}$is the mean amount of unoccupied seats from a sample of $225$flights.

c. $t_{n-1}$

d. i. localid="1652100608787" $CI=(11.119,12.081)$

ii. The graph is drawn

iii. localid="1652100644953" $EBM=0.48$

Step-by-step solution

Given Information

number of flights= $225$

sample mean =$11.6$

standard deviation =$4.1$

Explanation (Part a)

i. The sample mean is given by $\overline{x}=11.6$

ii. The standard deviation is given by $s_x=4.1$

iii. Number of flights selected$n=225$

iv. $n-1=224$

Explanation (Part b)

According to the given information, the random variables of$X$and$\overline{X}$are as follows:

$X$is the number of unoccupied seats on a single flight.

$\overline{X}$is the mean amount of unoccupied seats from a sample of$225$flights.

Explanation (Part c)

Since the distribution is $t$We will use $t_{n-1}$as a parameter.

Explanation (Part d)

i. Given,$\stackrel{-}{x}=11\cdot 6$and the standard deviation is $4.1$

sample size n = $225$

Confidence level z =$92\%$

$$\begin{gathered} CI=\overline{x}\pm z\frac{s}{\sqrt{n}} \\ =11.6\pm 92\times \frac{4.1}{\sqrt{225}} \end{gathered}$$

$CI=(11.119,12.081)$

ii. The graph is,

iii. Error bound margin is calculated,

$EBM=t_{n-1}\left(\frac{\alpha }{2}\right)\frac{s}{\sqrt{n}}$

$EBM=t_{225-1}\left(\frac{0.08}{2}\right)\frac{4.1}{\sqrt{225}}=0.48$