Chapter 8.2, Problem 45E

(a)

To determine

To construct and interpret a $99\%$ confidence interval for the proportion of all U.S. adults who think that organic produce is better for health than conventionally grown produce.

Answer to Problem 45E

We are $99\%$ confident that the true proportion of all U.S. adults think that the organic produce is better for health is between $0.5167\text{ and 0}\text{.5833}$ .

Explanation of Solution

It is given that,

  $\begin{array}{l}c=99\%=0.99\\ \widehat{p}=0.55\\ n=1480\end{array}$

Conditions to be met are:

The three conditions for calculating a confidence interval for the population proportion p are:

Random: It is satisfied because the U.S. adults are randomly selected.

Independent: It is satisfied because the sample of $1480$ U.S. adults is less than $10\%$ of the population of all U.S. adults.

Normal: It is satisfied because there are $1480(0.55)=814$ successes and $1480(1-0.55)=666$ failures, which are bout at least ten.

Thus, all the three conditions are satisfied.

The calculation is as:

For the confidence interval $1-\alpha =0.99$ determine $z_{\alpha /2}=z_{0.005}$ using the normal probability table in the appendix as:

  $z_{\alpha /2}=2.575$

Thus, the margin of error is then:

  $\begin{array}{l}E=z_{\alpha /2}\times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\\ =2.575\times \sqrt{\frac{0.55(1-0.55)}{1480}}\\ =0.0333\end{array}$

Thus, the confidence interval is as:

  $\begin{array}{l}\widehat{p}-E=0.55-0.0333\\ =0.5167\\ \widehat{p}+E=0.55+0.0333\\ =0.5833\end{array}$

So, we are $99\%$ confident that the true proportion of all U.S. adults think that the organic produce is better for health is between $0.5167\text{ and 0}\text{.5833}$ .

(b)

To determine

To explain does the interval from part (a) provide convincing evidence that a majority of all U.S. adults think that organic produce is better for health.

Answer to Problem 45E

Yes, there is convincing evidence that a majority of all U.S. adults think that organic produce is better.

Explanation of Solution

It is given that,

  $\begin{array}{l}c=99\%=0.99\\ \widehat{p}=0.55\\ n=1480\end{array}$

For the confidence interval $1-\alpha =0.99$ determine $z_{\alpha /2}=z_{0.005}$ using the normal probability table in the appendix as:

  $z_{\alpha /2}=2.575$

Thus, the margin of error is then:

  $\begin{array}{l}E=z_{\alpha /2}\times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\\ =2.575\times \sqrt{\frac{0.55(1-0.55)}{1480}}\\ =0.0333\end{array}$

Thus, the confidence interval is as:

  $\begin{array}{l}\widehat{p}-E=0.55-0.0333\\ =0.5167\\ \widehat{p}+E=0.55+0.0333\\ =0.5833\end{array}$

Thus, the confidence interval $(0.5167,0.5833)$ does not contain $0.5$ and contains only values above $0.5$ which indicates that more than $50\%$ of the U.S. adults think that organic produce is better for health and thus there is convincing evidence that a majority of all U.S. adults think that organic produce is better.