91影视

The values for different arriving times of flights are given.

Sample size $n=12$.

Let X represent the arrival times of different flights.

The sample mean is given as follows.

$$\begin{gathered} \stackrel{炉}{x}=\frac{{鈭�}_{i=1}^n{x}_i}{n} \\ =\frac{\left(-5\right)+\left(-32\right)+...+\left(11\right)}{12} \\ =-9.1667 \end{gathered}$$

Therefore, the mean value is -9.1667 min.

The standard deviation is given as follows.

$$\begin{gathered} s=\sqrt{\frac{{鈭�}_{i=1}^n{(x_i-\stackrel{炉}{x})}^2}{n-1}} \\ =\sqrt{\frac{{\left(-5-\left(-9.1667\right)\right)}^2+{\left(-32-\left(-9.1667\right)\right)}^2+...+\left(11-\left(-9.1667\right)\right)}{12-1}} \\ =23.8626 \end{gathered}$$

Therefore, the standard deviation is 23.8626 min.

The degree of freedom is computed as follows.

$$\begin{gathered} df=n-1 \\ =12-1 \\ =11 \end{gathered}$$

The critical value is computed by using the significance level of 0.05 and the degree of freedom 11 from the standard normal table.

The level of significance is given as follows.

$$\begin{gathered} 伪=\left(100-95\right)\% \\ =0.05 \end{gathered}$$

From the t-table, the critical value is $t_{\frac{伪}{2}}=2.201$.

The margin of error is as follows.

$$\begin{gathered} E=t_{\frac{伪}{2}}脳\frac{s}{\sqrt{n}} \\ =2.201脳\frac{23.8626}{\sqrt{12}} \\ =15.1616 \end{gathered}$$

Therefore, the margin of error is 15.1616 min.

The confidence interval for the mean is given as

$$\begin{gathered} \stackrel{炉}{x}-E<渭<\stackrel{炉}{x}+E \\ -9.1667-15.1616<渭<-9.1667+15.1616 \\ -24.328<渭<5.995 \end{gathered}$$

The lower limit of the 95% confidence interval is -24.3 min.

The upper limit of the 95% confidence interval is 6.00 min.

As the confidence interval includes more negative measures, it implies that most flights arrive before the scheduled time. It indicates a good on-time schedule.