Chapter 19, Problem 18E

(a)

To determine

To create $90\%$ confidence interval for the percentage of people the company contacts who may buy something.

Answer to Problem 18E

  $(0.10592,0.14008)$ .

Explanation of Solution

It is given in the question that direct mail advertise send solicitations to thousands of potential customers in the hope that some will buy the company’s product. The acceptance rate is usually low. 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,

  $\begin{array}{l}np=123>10\\ n(1-p)=877>10\end{array}$

$10\%$ 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,

  $\begin{array}{l}x=123\\ n=1000\\ c=0.90\end{array}$

Thus, the confidence interval will be as:

  $(0.10592,0.14008)$

(b)

To determine

To explain what this interval means.

Explanation of Solution

It is given in the question that direct mail advertise send solicitations to thousands of potential customers in the hope that some will buy the company’s product. The acceptance rate is usually low. Thus, the confidence interval will be as:

  $(0.10592,0.14008)$

Thus, we conclude from the above that we are $95\%$ confident that the interval $(0.10592,0.14008)$ contains the true population proportion of orders.

(c)

To determine

To explain what “ $90\%$ confidence” means.

Explanation of Solution

It is given in the question that direct mail advertise send solicitations to thousands of potential customers in the hope that some will buy the company’s product. The acceptance rate is usually low. Thus, the confidence interval will be as:

  $(0.10592,0.14008)$

Thus, we conclude from the above that we are $95\%$ confident that the interval $(0.10592,0.14008)$ contains the true population proportion of orders. Thus, “ $90\%$ confidence” means that $90\%$ confidence means that we will guess that true population proportion correctly $90\%$ of the time.

(d)

To determine

To explain what does your confidence interval suggest.

Explanation of Solution

It is given in the question that direct mail advertise send solicitations to thousands of potential customers in the hope that some will buy the company’s product. The acceptance rate is usually low. Thus, the confidence interval will be as:

  $(0.10592,0.14008)$

Thus, we conclude from the above that we are $95\%$ confident that the interval $(0.10592,0.14008)$ contains the true population proportion of orders. Thus, the confidence interval contains values well about $5\%$ suggesting that the mailing would be very successful. So, the company should do a mass mailing.