A survey asked, "How important is it to you to buy products that are made in America?" Of the 600 Americans surveyed, 450 responded, "It is important." For each of the following levels of confidence, find the sample proportion and the margin of error associated with the poll. (Round the margin of error to one decimal place.)
(a) an 80% level of confidence
sample proportion    		
.75
 
Correct: Your answer is correct.
margin of error	±	
 
Incorrect: Your answer is incorrect.
 %

(b) a 99% level of confidence
sample proportion    		
.75
 
Correct: Your answer is correct.
margin of error	±	
 
Incorrect: Your answer is incorrect.
 %


To find the margin of error (ME) for each level of confidence, we need to follow these steps:

### Step 1: Compute the Sample Proportion
The sample proportion is given by:

\[
p = \frac{450}{600} = 0.75
\]

### Step 2: Identify the Critical Value \(z^*\)
For different confidence levels, the critical value \(z^*\) corresponds to the z-score that captures the confidence interval. Here are the standard \(z^*\) values:

- For an 80% confidence level, \( z^* \approx 1.28 \)
- For a 99% confidence level, \( z^* \approx 2.576 \)

### Step 3: Calculate the Margin of Error (ME)
The formula for margin of error is:

\[
ME = z^* \times \sqrt{\frac{p(1 - p)}{n}}
\]

Where:
- \( p \) is the sample proportion.
- \( n = 600 \) is the sample size.
- \( z^* \) is the critical value for the given confidence level.

### (a) For 80% confidence level:

1. **Sample Proportion:**
   \[
   p = 0.75
   \]
   
2. **Critical value:**
   \[
   z^* = 1.28
   \]
   
3. **Margin of Error:**
   \[
   ME = 1.28 \times \sqrt{\frac{0.75 \times 0.25}{600}}
   \]
   Let’s calculate this:

\[
ME \approx 1.28 \times \sqrt{\frac{0.1875}{600}} 
\approx 1.28 \times \sqrt{0.0003125} 
\approx 1.28 \times 0.01768 
\approx 0.0226 \text{ or } 2.3\%
\]

Thus, the margin of error for 80% confidence is approximately \( \pm 2.3\% \).

### (b) For 99% confidence level:

1. **Sample Proportion:**
   \[
   p = 0.75
   \]
   
2. **Critical value:**
   \[
   z^* = 2.576
   \]
   
3. **Margin of Error:**
   \[
   ME = 2.576 \times \sqrt{\frac{0.75 \times 0.25}{600}}
   \]
   Let’s calculate this:

\[
ME \approx 2.576 \times \sqrt{\frac{0.1875}{600}} 
\approx 2.576 \times \sqrt{0.0003125} 
\approx 2.576 \times 0.01768 
\approx 0.0455 \text{ or } 4.6\%
\]

Thus, the margin of error for 99% confidence is approximately \( \pm 4.6\% \).

### Final Answers:
- **(a) 80% confidence level:**
   - Margin of error: \( \pm 2.3\% \)

- **(b) 99% confidence level:**
   - Margin of error: \( \pm 4.6\% \)

---

Would you like further details on the calculations or an explanation of confidence intervals?

Here are 5 related questions:
1. How do critical values differ for various confidence levels?
2. What is the relationship between sample size and margin of error?
3. How can we reduce the margin of error in surveys?
4. What happens to the margin of error as the sample proportion moves closer to 0 or 1?
5. How is the margin of error affected if we increase the confidence level?

**Tip:** Larger sample sizes generally reduce the margin of error, making the results more precise.

A survey asked, 'How important is it to you to buy products that are made in America?' Of the 600 Americans surveyed, 450 responded, 'It is important.' For each of the following levels of confidence, find the sample proportion and the margin of error associated with the poll. (a) an 80% level of confidence, (b) a 99% level of confidence.

Learn how to calculate the margin of error for different confidence levels using a sample proportion. This problem explores margin of error calculations for an 80% and 99% confidence level, using a survey where 450 out of 600 Americans said buying American-made products is important.

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