Chapter 23, Problem 20E

(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 reported the percentage of flights that were late each month from $\text{1995 through 2006}$ . The histogram is given in the question. We consider data to be a representative sample of all months. The correlation of late arrival% with time is $r=-0.07$ .

Now, let us assume each flight is independent of one another and the histogram of late 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 $\text{99\%}$ confidence interval for the true percentage of flights that arrive late.

Answer to Problem 20E

The $\text{99\%}$ confidence interval for the true percentage of flights that arrive lateis $(19.1865,20.9649)$ .

Explanation of Solution

It is given that the U.S. Bureau of Transportation Statistics of the Department of Transportation reported the percentage of flights that were late each month from $\text{1995 through 2006}$ . The histogram is given in the question. We consider data to be a representative sample of all months. The correlation of late arrival% with time is $r=-0.07$ .It is given also,

  $\begin{array}{l}n=144\\ \overline{y}=20.0757\\ s=4.08837\end{array}$

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 $\text{99\%}$ confidence interval be:

  $t_{\alpha /2}=2.610$

The confidence interval is calculated as:

  $\begin{array}{l}\overline{y}-t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =20.0757-2.610\times \frac{4.08837}{\sqrt{144}}\\ =19.1865\\ \overline{y}+t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =20.0757+2.610\times \frac{4.08837}{\sqrt{144}}\\ =20.9649\end{array}$

Thus, the $\text{99\%}$ confidence interval for the true percentage of flights that arrive lateis $(19.1865,20.9649)$ .

(c)

To determine

To interpret this interval for a traveller planning to fly.

Answer to Problem 20E

We are $\text{99\%}$ confident that the true percentage of flights that arrive late is between $19.1865\%\text{ and }20.9649\%$ .

Explanation of Solution

It is given that the U.S. Bureau of Transportation Statistics of the Department of Transportation reported the percentage of flights that were late each month from $\text{1995 through 2006}$ . The histogram is given in the question. We consider data to be a representative sample of all months. The correlation of late arrival% with time is $r=-0.07$ .It is given also,

  $\begin{array}{l}n=144\\ \overline{y}=20.0757\\ s=4.08837\end{array}$

Thus, the $\text{99\%}$ confidence interval for the true percentage of flights that arrive lateis $(19.1865,20.9649)$ . Therefore, it means that we are $\text{99\%}$ confident that the true percentage of flights that arrive late is between $19.1865\%\text{ and }20.9649\%$ .