Overbooking policies for major airlines.Airlines overbook flights to reduce the odds of flying with unused seats. An article in Transportation Research (Vol. 38, 2002) investigated the optimal overbooking policies for major airlines. One of the variables measured for each airline was the compensation (in dollars) per bumped passenger required to maximize future revenue. Consider the threshold compensation levels for a random sample of $\mathbf{10}$ major airlines shown in the next table. Estimate the true mean threshold compensation level for all major worldwide airlines using a$\mathbf{90}\mathbf{\%}$ confidence interval. Interpret the result practically.

825	850	1,210	1,370	1,415
1,500	1,560	1,625	2,155	2,220

Consider threshold levels of compensation for a random sample of $10$major airline samples shown in the table.

First, in this problem, we find the mean from the given table. Sample size $\mathrm{n}=10$.Then, the mean will be given by:

$\begin{array}{c}\overline{X}=\frac{825+850+1210+1370+1500+1560+1625+2155+2220}{10}\\ =\frac{13320}{10}\\ =1332\end{array}$

Then the sample variance will be given by:

$\begin{array}{c}{s}^2=\frac{{\displaystyle \sum _{i=1}^n{(X_i-\overline{X})}^2}}{n-1}\\ =\frac{{\displaystyle \sum _{i=1}^{10}{(X_i-\overline{X})}^2}}{10-1}\\ =\frac{{\displaystyle \sum _{i=1}^{10}{(X_i-\overline{X})}^2}}{9}\end{array}$

Then

$\begin{array}{c}{s}^2=\frac{\begin{array}{c}{\left(825-1332\right)}^2+{\left(850-1332\right)}^2+{\left(1210-1332\right)}^2+{\left(1370-1332\right)}^2+{(1415-1332)}^2\\ +{(1500-1332)}^2+{(1560-1332)}^2+{(1625-1332)}^2+{(2155-1332)}^2+{(2220-1332)}^2\end{array}}{9}\\ =\frac{214520}{9}\\ =238280\end{array}$

We know for interval estimate, the confidence interval will be given by

$\mathrm{CI}=\overline{\mathrm{X}}\pm \mathrm{z}\times \frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}$

This is the point estimate for $m$.

We know that for $90\%$confidence interval, the estimate will be written as

$\mathrm{CI}=\overline{\mathrm{X}}\pm \mathrm{z}\times \frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}$

For$90\%$confidence interval, the$z$score will be given by

$\mathrm{z}=1.645$

Therefore the confidence interval is written as

$\begin{array}{c}\mathrm{CI}=\overline{\mathrm{X}}\pm \mathrm{z}\times \frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}\\ =\overline{\mathrm{X}}\pm 1.645\times \frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}\end{array}$

Here we calculate the variance from the given table

$\mathrm{var}\left(\mathrm{x}\right)=\mathrm{E}\left({\mathrm{x}}^2\right)-{\mathrm{E}}^2\left(\mathrm{x}\right)$

Then

$\begin{array}{c}\mathrm{E}\left({\mathrm{x}}^2\right)={\left(825\right)}^2+{\left(850\right)}^2+{\left(1210\right)}^2+{\left(1370\right)}^2+{\left(1415\right)}^2+{\left(1500\right)}^2+{\left(1560\right)}^2+{\left(1625\right)}^2+{\left(2155\right)}^2\\ +{\left(2220\right)}^2\\ =23643000\end{array}$

Then the variance will be given by

$\begin{array}{c}\mathrm{var}\left(\mathrm{x}\right)=\mathrm{E}\left({\mathrm{x}}^2\right)-{\mathrm{E}}^2\left(\mathrm{x}\right)\\ =23643000-{\left(1332\right)}^2\\ =21868776\end{array}$

Mean is $\overline{\mathrm{X}}=1332$

Therefore the confidence interval will be given by

$\begin{array}{c}\mathrm{CI}=\overline{\mathrm{X}}\pm \mathrm{z}\times \frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}\\ =\overline{\mathrm{X}}\pm 1.645\times \frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}\\ =1332\pm 1.645\times \frac{21868776}{\sqrt{10}}\\ =113695.32\end{array}$

And

$\begin{array}{c}\mathrm{CI}=1332-11368603.32\\ =-11367271.32\end{array}$