91影视

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:

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

Then the sample variance will be given by:

$\begin{array}{c}{s}^2=\frac{{\displaystyle \underset{i=1}{\overset{n}{鈭�}}{(X_i鈭�\stackrel{炉}{X})}^2}}{n鈭�1}\\ =\frac{{\displaystyle \underset{i=1}{\overset{10}{鈭�}}{(X_i鈭�\stackrel{炉}{X})}^2}}{10鈭�1}\\ =\frac{{\displaystyle \underset{i=1}{\overset{10}{鈭�}}{(X_i鈭�\stackrel{炉}{X})}^2}}{9}\end{array}$

Then

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

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

$\mathrm{CI}=\stackrel{炉}{\mathrm{X}}卤\mathrm{z}脳\frac{\mathrm{蟽}}{\sqrt{\mathrm{n}}}$

This is the point estimate for $m$.

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

$\mathrm{CI}=\stackrel{炉}{\mathrm{X}}卤\mathrm{z}脳\frac{\mathrm{蟽}}{\sqrt{\mathrm{n}}}$

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

$\mathrm{z}=1.645$

Therefore the confidence interval is written as

$\begin{array}{c}\mathrm{CI}=\stackrel{炉}{\mathrm{X}}卤\mathrm{z}脳\frac{\mathrm{蟽}}{\sqrt{\mathrm{n}}}\\ =\stackrel{炉}{\mathrm{X}}卤1.645脳\frac{\mathrm{蟽}}{\sqrt{\mathrm{n}}}\end{array}$

Here we calculate the variance from the given table

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

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

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

Mean is $\stackrel{炉}{\mathrm{X}}=1332$

Therefore the confidence interval will be given by

$\begin{array}{c}\mathrm{CI}=\stackrel{炉}{\mathrm{X}}卤\mathrm{z}脳\frac{\mathrm{蟽}}{\sqrt{\mathrm{n}}}\\ =\stackrel{炉}{\mathrm{X}}卤1.645脳\frac{\mathrm{蟽}}{\sqrt{\mathrm{n}}}\\ =1332卤1.645脳\frac{21868776}{\sqrt{10}}\\ =113695.32\end{array}$

And

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