Chapter 21, Problem 29E

Section 1

To determine

To find: The 70% confidence interval for the proportion of American adults who reported looking for information on food labels about whether the food they are purchasing is organic.

Answer to Problem 29E

Solution: The confidence interval is $\left(0.474,0.506\right)$.

Explanation of Solution

Calculation:

Compute the confidence interval using formula as follows:

$\text{Confidence interval}=\left[\widehat{p}\pm z\left(\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\right)\right]$

where

$\begin{array}{l}\widehat{p}\text{ is the sample proportion}\\ z\text{ is the tabulated value for desired confidence interval}\\ n\text{ is the sample size}\end{array}$

Now, compute the 70% confidence interval as follows:

$\begin{array}{c}\text{Confidence interval}=\left[\widehat{p}\pm z\left(\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\right)\right]\\ =\left[0.49\pm 1.04\left(\sqrt{\frac{0.49\left(1-0.49\right)}{1004}}\right)\right]\\ =\left(0.49\pm 0.016\right)\\ =\left(0.474,0.506\right)\end{array}$

Interpretation: It can be concluded with 70% confidence that proportion of American adults who reported looking for information on food labels about whether the food they are purchasing is organic lies between 47.4% and 50.6%.

Section 2

To find: The 80% confidence interval for the proportion of American adults who reported looking for information on food labels about whether the food they are purchasing is organic.

Solution: The confidence interval is $\left(0.470,0.510\right)$.

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

$\text{Confidence interval}=\left[\widehat{p}\pm z\left(\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\right)\right]$

where

$\begin{array}{l}\widehat{p}\text{ is the sample proportion}\\ z\text{ is the tabulated value for desired confidence interval}\\ n\text{ is the sample size}\end{array}$

Now, compute the 80% confidence interval as follows:

$\begin{array}{c}\text{Confidence interval}=\left[\widehat{p}\pm z\left(\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\right)\right]\\ =\left[0.49\pm 1.24\left(\sqrt{\frac{0.49\left(1-0.49\right)}{1004}}\right)\right]\\ =\left(0.49\pm 0.02\right)\\ =\left(0.47,0.51\right)\end{array}$

Interpretation: It can be concluded with 80% confidence that proportion of American adults who reported looking for information on food labels about whether the food they are purchasing is organic lies between 47% and 51%.

Section 3

To find: The 90% confidence interval for the proportion of American adults who reported looking for information on food labels about whether the food they are purchasing is organic.

Solution: The confidence interval is $\left(0.464,516\right)$.

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

$\text{Confidence interval}=\left[\widehat{p}\pm z\left(\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\right)\right]$

where

$\begin{array}{l}\widehat{p}\text{ is the sample proportion}\\ z\text{ is the tabulated value for desired confidence interval}\\ n\text{ is the sample size}\end{array}$

Now, compute the 90% confidence interval as follows:

$\begin{array}{c}\text{Confidence interval}=\left[\widehat{p}\pm z\left(\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\right)\right]\\ =\left[0.49\pm 1.64\left(\sqrt{\frac{0.49\left(1-0.49\right)}{1004}}\right)\right]\\ =\left(0.49\pm 0.026\right)\\ =\left(0.464,0.516\right)\end{array}$

Interpretation: It can be concluded with 90% confidence that proportion of American adults who reported looking for information on food labels about whether the food they are purchasing is organic lies between 46% and 52%.

Section 4

To find: The 99% confidence interval for the proportion of American adults who reported looking for information on food labels about whether the food they are purchasing is organic.

Solution: The confidence interval is $\left(0.449,0.531\right)$.

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

$\text{Confidence interval}=\left[\widehat{p}\pm z\left(\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\right)\right]$

where

$\begin{array}{l}\widehat{p}\text{ is the sample proportion}\\ z\text{ is the tabulated value for desired confidence interval}\\ n\text{ is the sample size}\end{array}$

Now, compute the 99% confidence interval as follows:

$\begin{array}{c}\text{Confidence interval}=\left[\widehat{p}\pm z\left(\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\right)\right]\\ =\left[0.49\pm 2.58\left(\sqrt{\frac{0.49\left(1-0.49\right)}{1004}}\right)\right]\\ =\left(0.49\pm 0.041\right)\\ =\left(0.449,0.531\right)\end{array}$

Interpretation: It can be concluded with 99% confidence that proportion of American adults who reported looking for information on food labels about whether the food they are purchasing is organic lies between 45% and 53%.

Section 5

To find: The comparison of confidence intervals obtained in sections 1, 2, 3, and 4.

Solution: The width of confidence interval widens with increase in confidence level.

Explanation:

Calculation:

Now, compute the width of the 70%, 80%, 90%, and 99% confidence intervals.

The width of 70% confidence interval is calculated as shown below:

$\begin{array}{c}70\%\text{ width}=\text{Upper limit}-\text{Lower limit}\\ =0.506-0.474\\ =0.032\end{array}$

The width of 80% confidence interval is calculated as shown below:

$\begin{array}{c}80\%\text{ width}=\text{Upper limit}-\text{Lower limit}\\ =0.510-0.470\\ =0.04\end{array}$

The width of 90% confidence interval is calculated as shown below:

$\begin{array}{c}90\%\text{ width}=\text{Upper limit}-\text{Lower limit}\\ =0.516-0.464\\ =0.052\end{array}$

The width of 99% confidence interval is calculated as shown below:

$\begin{array}{c}99\%\text{ width}=\text{Upper limit}-\text{Lower limit}\\ =0.531-0.449\\ =0.082\end{array}$

From the above calculations, it is clear that if there is an increase in the confidence level, then there is an increase in the width of the confidence interval.