Chapter 19, Problem 17E

(a)

To determine

To create a $95\%$ confidence interval for the percentage of all auto accidents that involve teenage drivers.

Answer to Problem 17E

  $(0.12685,0.18586)$ .

Explanation of Solution

It is given in the question that an insurance company checks police records on accidents selected at random and notes that teenagers were at the wheel in $91$ of them. Thus, we have,

Let us first check the conditions for the following as:

Random condition: It is satisfied as it is assumed that the sample is randomly selected.

Normal condition: It is satisfied as,

  $\begin{array}{l}np=91>10\\ n(1-p)=491>10\end{array}$

$10\%$ condition: It is satisfied as the sample size is less than $10\%$ of the population size.

Thus, all the conditions are met. Therefore, now, to find the confidence interval we will use the calculator $\text{TI}-89$ . So, we have to first go to the $\text{Ints}$ menu and select $5:1-\text{PropZInt}$ . Then we have to enter the number of successes observed and the sample size. Then specify a confidence level. Then the interval will be calculated. Thus, we have,

  $\begin{array}{l}x=91\\ n=582\\ c=0.95\end{array}$

Thus, the confidence interval will be as:

  $(0.12685,0.18586)$

(b)

To determine

To explain what your interval means.

Explanation of Solution

It is given in the question that an insurance company checks police records on accidents selected at random and notes that teenagers were at the wheel in $91$ of them. Thus, the confidence interval will be as:

  $(0.12685,0.18586)$

Thus we conclude that we are $95\%$ confident that the interval $(0.12685,0.18586)$ contains the true population proportion of auto accidents that involve teenagers.

(c)

To determine

To explain what “ $95\%$ confidence” means.

Explanation of Solution

It is given in the question that an insurance company checks police records on accidents selected at random and notes that teenagers were at the wheel in $91$ of them. Thus, the confidence interval will be as:

  $(0.12685,0.18586)$

Thus we conclude that we are $95\%$ confident that the interval $(0.12685,0.18586)$ contains the true population proportion of auto accidents that involve teenagers. Thus, “ $95\%$ confidence” means that we will guess that the true population proportion correctly $95\%$ of the time.

(d)

To determine

To explain does your confidence interval support or contradict this statement.

Answer to Problem 17E

Our confidence interval contradict this statement.

Explanation of Solution

It is given in the question that an insurance company checks police records on accidents selected at random and notes that teenagers were at the wheel in $91$ of them. Thus, the confidence interval will be as:

  $(0.12685,0.18586)$

Thus we conclude that we are $95\%$ confident that the interval $(0.12685,0.18586)$ contains the true population proportion of auto accidents that involve teenagers. Thus, we can say that our confidence interval contradict this statement because the police officer is saying that $0.2$ accidents involve a teenager, but our confidence interval shows values less than $0.2$ .