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.

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Answer 1

Question

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.

Expert Solution & Answer

Answer to Problem 29E

Solution: The confidence interval is (0.474,0.506).

Explanation of Solution

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 70% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.04(0.49(1−0.49)1004)]=(0.49±0.016)=(0.474,0.506)

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 (0.470,0.510).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 80% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.24(0.49(1−0.49)1004)]=(0.49±0.02)=(0.47,0.51)

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 (0.464,516).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 90% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.64(0.49(1−0.49)1004)]=(0.49±0.026)=(0.464,0.516)

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 (0.449,0.531).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 99% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±2.58(0.49(1−0.49)1004)]=(0.49±0.041)=(0.449,0.531)

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:

70% width=Upper limit−Lower limit=0.506−0.474=0.032

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

80% width=Upper limit−Lower limit=0.510−0.470=0.04

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

90% width=Upper limit−Lower limit=0.516−0.464=0.052

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

99% width=Upper limit−Lower limit=0.531−0.449=0.082

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.

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Answer 2

Question

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.

Expert Solution & Answer

Answer to Problem 29E

Solution: The confidence interval is (0.474,0.506).

Explanation of Solution

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 70% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.04(0.49(1−0.49)1004)]=(0.49±0.016)=(0.474,0.506)

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 (0.470,0.510).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 80% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.24(0.49(1−0.49)1004)]=(0.49±0.02)=(0.47,0.51)

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 (0.464,516).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 90% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.64(0.49(1−0.49)1004)]=(0.49±0.026)=(0.464,0.516)

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 (0.449,0.531).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 99% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±2.58(0.49(1−0.49)1004)]=(0.49±0.041)=(0.449,0.531)

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:

70% width=Upper limit−Lower limit=0.506−0.474=0.032

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

80% width=Upper limit−Lower limit=0.510−0.470=0.04

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

90% width=Upper limit−Lower limit=0.516−0.464=0.052

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

99% width=Upper limit−Lower limit=0.531−0.449=0.082

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.

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Answer 3

Question

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.

Expert Solution & Answer

Answer to Problem 29E

Solution: The confidence interval is (0.474,0.506).

Explanation of Solution

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 70% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.04(0.49(1−0.49)1004)]=(0.49±0.016)=(0.474,0.506)

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 (0.470,0.510).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 80% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.24(0.49(1−0.49)1004)]=(0.49±0.02)=(0.47,0.51)

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 (0.464,516).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 90% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.64(0.49(1−0.49)1004)]=(0.49±0.026)=(0.464,0.516)

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 (0.449,0.531).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 99% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±2.58(0.49(1−0.49)1004)]=(0.49±0.041)=(0.449,0.531)

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:

70% width=Upper limit−Lower limit=0.506−0.474=0.032

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

80% width=Upper limit−Lower limit=0.510−0.470=0.04

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

90% width=Upper limit−Lower limit=0.516−0.464=0.052

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

99% width=Upper limit−Lower limit=0.531−0.449=0.082

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.

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Answer 4

Question

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.

Expert Solution & Answer

Answer to Problem 29E

Solution: The confidence interval is (0.474,0.506).

Explanation of Solution

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 70% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.04(0.49(1−0.49)1004)]=(0.49±0.016)=(0.474,0.506)

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 (0.470,0.510).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 80% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.24(0.49(1−0.49)1004)]=(0.49±0.02)=(0.47,0.51)

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 (0.464,516).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 90% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.64(0.49(1−0.49)1004)]=(0.49±0.026)=(0.464,0.516)

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 (0.449,0.531).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 99% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±2.58(0.49(1−0.49)1004)]=(0.49±0.041)=(0.449,0.531)

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:

70% width=Upper limit−Lower limit=0.506−0.474=0.032

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

80% width=Upper limit−Lower limit=0.510−0.470=0.04

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

90% width=Upper limit−Lower limit=0.516−0.464=0.052

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

99% width=Upper limit−Lower limit=0.531−0.449=0.082

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.

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Answer 5

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.

Expert Solution & Answer

Answer to Problem 29E

Solution: The confidence interval is (0.474,0.506).

Explanation of Solution

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 70% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.04(0.49(1−0.49)1004)]=(0.49±0.016)=(0.474,0.506)

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 (0.470,0.510).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 80% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.24(0.49(1−0.49)1004)]=(0.49±0.02)=(0.47,0.51)

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 (0.464,516).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 90% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.64(0.49(1−0.49)1004)]=(0.49±0.026)=(0.464,0.516)

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 (0.449,0.531).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 99% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±2.58(0.49(1−0.49)1004)]=(0.49±0.041)=(0.449,0.531)

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:

70% width=Upper limit−Lower limit=0.506−0.474=0.032

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

80% width=Upper limit−Lower limit=0.510−0.470=0.04

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

90% width=Upper limit−Lower limit=0.516−0.464=0.052

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

99% width=Upper limit−Lower limit=0.531−0.449=0.082

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.

Answer 6

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.

Expert Solution & Answer

Answer to Problem 29E

Solution: The confidence interval is (0.474,0.506).

Explanation of Solution

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 70% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.04(0.49(1−0.49)1004)]=(0.49±0.016)=(0.474,0.506)

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 (0.470,0.510).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 80% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.24(0.49(1−0.49)1004)]=(0.49±0.02)=(0.47,0.51)

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 (0.464,516).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 90% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.64(0.49(1−0.49)1004)]=(0.49±0.026)=(0.464,0.516)

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 (0.449,0.531).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 99% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±2.58(0.49(1−0.49)1004)]=(0.49±0.041)=(0.449,0.531)

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:

70% width=Upper limit−Lower limit=0.506−0.474=0.032

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

80% width=Upper limit−Lower limit=0.510−0.470=0.04

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

90% width=Upper limit−Lower limit=0.516−0.464=0.052

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

99% width=Upper limit−Lower limit=0.531−0.449=0.082

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.

Answer 7

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 8

Expert Solution & Answer

Answer to Problem 29E

Solution: The confidence interval is (0.474,0.506).

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 70% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.04(0.49(1−0.49)1004)]=(0.49±0.016)=(0.474,0.506)

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 (0.470,0.510).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 80% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.24(0.49(1−0.49)1004)]=(0.49±0.02)=(0.47,0.51)

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 (0.464,516).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 90% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±1.64(0.49(1−0.49)1004)]=(0.49±0.026)=(0.464,0.516)

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 (0.449,0.531).

Explanation:

Calculation:

Compute the confidence interval using formula as follows:

Confidence interval=[p^±z(p^(1−p^)n)]

where

p^ is the sample proportionz is the tabulated value for desired confidence intervaln is the sample size

Now, compute the 99% confidence interval as follows:

Confidence interval=[p^±z(p^(1−p^)n)]=[0.49±2.58(0.49(1−0.49)1004)]=(0.49±0.041)=(0.449,0.531)

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:

70% width=Upper limit−Lower limit=0.506−0.474=0.032

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

80% width=Upper limit−Lower limit=0.510−0.470=0.04

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

90% width=Upper limit−Lower limit=0.516−0.464=0.052

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

99% width=Upper limit−Lower limit=0.531−0.449=0.082

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.