Chapter 23, Problem 19E

(a)

To determine

To check the assumptions and conditions for inference.

Explanation of Solution

It is given that the U.S. Bureau of Transportation Statistics of the Department of Transportation publishes information about airline performance. The histogram is given in the question. There is no evidence of a trend over time. The correlation of On Time Departure% with time is $r=-0.016$ .

Now, let us assume each flight is independent of one another and the histogram of on-time departure percentage is approximately uni-modal and symmetric. Also, assume data are representative of the population $\text{144 is not less than 10\%}$ of all months from $1995-2006$ .

(b)

To determine

To find a $90\%$ confidence interval for the true percentage of flights that depart on time.

Answer to Problem 19E

The $90\%$ confidence interval for the true percentage of flights that depart on time is $\left(80.5653,81.8023\right)$ .

Explanation of Solution

It is given that the U.S. Bureau of Transportation Statistics of the Department of Transportation publishes information about airline performance. The histogram is given in the question. There is no evidence of a trend over time. The correlation of On Time Departure% with time is $r=-0.016$ . It is given also,

  $\begin{array}{l}n=144\\ \overline{y}=81.1838\\ s=4.47094\end{array}$

Now we will find the t -value by looking in the row starting with degrees of freedom. The degree of freedom is as:

  $\begin{array}{l}df=n-1\\ =144-1\\ =143\end{array}$

Thus, the t -value with $90\%$ confidence interval be:

  $t_{\alpha /2}=1.660$

The confidence interval is calculated as:

  $\begin{array}{l}\overline{x}-t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =81.1838-1.660\times \frac{4.47094}{\sqrt{144}}\\ =80.5653\\ \overline{x}+t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =81.1838+1.660\times \frac{4.47094}{\sqrt{144}}\\ =81.8023\end{array}$

Thus, the $90\%$ confidence interval for the true percentage of flights that depart on timeis $\left(80.5653,81.8023\right)$ .

(c)

To determine

To interpret this interval for a traveler planning to fly.

Answer to Problem 19E

We are $90\%$ confident that the true percentage of flights that depart on time is between $\text{80}\text{.5653\% and 81}\text{.8023\%}$ .

Explanation of Solution

It is given that the U.S. Bureau of Transportation Statistics of the Department of Transportation publishes information about airline performance. The histogram is given in the question. There is no evidence of a trend over time. The correlation of On Time Departure% with time is $r=-0.016$ . It is given also,

  $\begin{array}{l}n=144\\ \overline{y}=81.1838\\ s=4.47094\end{array}$

Thus, the $90\%$ confidence interval for the true percentage of flights that depart on time is $\left(80.5653,81.8023\right)$ . Therefore, it means that we are $90\%$ confident that the true percentage of flights that depart on time is between $\text{80}\text{.5653\% and 81}\text{.8023\%}$ .