Chapter 25, Problem 24E

(a)

To determine

To write a $95\%$ confidence interval for the mean dry pavement stopping distance and explain what your interval means.

Answer to Problem 24E

The mean dry pavements stopping distance is between $(133.6064,145.1936)$ feet.

Explanation of Solution

The table of another set of cars whose tires are tested for dry and wet pavements are given in the question. Thus, now let us first check the conditions and assumptions for inference are as follows:

Random condition: It is satisfied because we are assuming that the cars were randomly selected.

Independent condition: It is satisfied because the cars are different.

Normal condition: It is satisfied because the histogram of the data is roughly symmetric and uni-modal.

Thus, all the conditions for the data are satisfied.

Now, let us calculate the mean and standard deviation of the dry pavements for the cars as:

  $\begin{array}{l}\overline{x}=\frac{145+152+\mathrm{....}+135+133}{10}\\ =139.4\\ s=\sqrt{\frac{{(145-139.4)}^2+{(152-139.4)}^2+\mathrm{....}+{(135-139.4)}^2+{(133-139.4)}^2}{10-1}}\\ =8.0994\end{array}$

And the degree of freedom will be as:

  $\begin{array}{l}df=n-1\\ =10-1\\ =9\end{array}$

Now let us find out the t -value by looking in the row starting with degree of freedom and in table T of appendix F is as:

  $t=2.262$

Now the confidence interval will be as:

  $\begin{array}{l}\overline{x}-t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =139.4-2.262\times \frac{8.0994}{\sqrt{10}}\\ =133.6064\\ \overline{x}+t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =139.4+2.262\times \frac{8.0994}{\sqrt{10}}\\ =145.1936\end{array}$

Thus, we conclude that we are $95\%$ confident that the mean dry pavements stopping distance is between $(133.6064,145.1936)$ feet.

(b)

To determine

To write a $95\%$ confidence interval for the mean increase in stopping distance on wet pavement and also explain what your interval means.

Answer to Problem 24E

We are $95\%$ confident that the mean increase in stopping distance is between $(-75.5853,-50.4147)$ feet.

Explanation of Solution

The table of the cars whose tires are tested for dry and wet pavements are given in the question. Thus, now let us first check the conditions and assumptions for inference are as follows:

Random condition: It is satisfied because we are assuming that the cars were randomly selected.

Independent condition: It is satisfied because the cars are different.

Normal condition: It is satisfied because the histogram of the data is roughly symmetric and uni-modal.

Thus, all the conditions for the data are satisfied.

It is also given that,

  $n=10$

And we will find the difference between the two samples as:

  

Now, the sample mean and the standard deviation of the difference are as follows:

  $\begin{array}{l}\overline{d}=\frac{-66-39-\mathrm{....}-48-90}{10}\\ =-63\\ s_d=\sqrt{\frac{{(-66-(-63))}^2+{(-39-(-63))}^2+\mathrm{....}+{(-48-(-63))}^2+{(-90-(-63))}^2}{10-1}}\\ =17.5942\end{array}$

Now the degree of freedom is then as:

  $\begin{array}{l}df=n-1\\ =10-1\\ =9\end{array}$

Now let us find out the t -value by looking in the row starting with degree of freedom and in table T of appendix F is as:

  $t=2.262$

Now the confidence interval will be as:

  $\begin{array}{l}\overline{d}-t_{\alpha /2}\times \frac{s_d}{\sqrt{n}}\\ =-63-2.262\times \frac{17.5942}{\sqrt{10}}\\ =-75.5853\\ \overline{d}+t_{\alpha /2}\times \frac{s_d}{\sqrt{n}}\\ =-63+2.262\times \frac{17.5942}{\sqrt{10}}\\ =-50.4147\end{array}$

Thus, we conclude that we are $95\%$ confident that the mean increase in stopping distance is between $(-75.5853,-50.4147)$ feet.