A survey asked, "Do you think that homeless people are responsible for the situation they are in?" Of the 2,500 Americans surveyed, 1,408 said no.
(a) Determine the sample proportion of Americans who think that homeless people are not responsible for the situation they are in. (Round your answer to the nearest whole number.)
Sample proportion is
%

(b) With an 80% level of confidence, find the margin of error (MOE) associated with the sample proportion. (Give the answer to one decimal place, if needed.)
MOE =
%

(c) With a 95% level of confidence, find the margin of error (MOE) associated with the sample proportion. (Give the answer to one decimal place, if needed.)
MOE =
%

(d) The answer in part (c) is
---Select---
the answer in part (b). Why?
In order to guarantee 95% accuracy, we must
---Select---
the error.
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### (a) Sample Proportion
The sample proportion (\( \hat{p} \)) is calculated as the number of people who said "no" divided by the total number of people surveyed.

\[
\hat{p} = \frac{1408}{2500}
\]

Now, let's compute this and convert it into a percentage:

\[
\hat{p} = \frac{1408}{2500} \times 100 = 56.32\%
\]

Rounding to the nearest whole number:

\[
\hat{p} = 56\%
\]

So, the sample proportion is **56%**.

### (b) Margin of Error with 80% Confidence Level
The formula for the margin of error (MOE) is given by:

\[
MOE = z \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]

Where:
- \( z \) is the z-value corresponding to the desired confidence level,
- \( \hat{p} \) is the sample proportion (in decimal form),
- \( n \) is the sample size.

For an 80% confidence level, the z-value is approximately **1.28**.

We know:
- \( \hat{p} = 0.5632 \),
- \( n = 2500 \).

Let’s compute the margin of error for 80% confidence.

\[
MOE = 1.28 \times \sqrt{\frac{0.5632 \times (1 - 0.5632)}{2500}}
\]

### (c) Margin of Error with 95% Confidence Level
For a 95% confidence level, the z-value is approximately **1.96**.

We will use the same formula but with the z-value for 95% confidence:

\[
MOE = 1.96 \times \sqrt{\frac{0.5632 \times (1 - 0.5632)}{2500}}
\]

I will now compute the MOE for both (b) and (c).### (b) Margin of Error with 80% Confidence
The margin of error for an 80% confidence level is approximately **1.3%**.

### (c) Margin of Error with 95% Confidence
The margin of error for a 95% confidence level is approximately **1.9%**.

### (d) Interpretation
The answer in part (c) is **greater than** the answer in part (b). This is because a higher confidence level (95% vs. 80%) requires a **larger margin of error** to ensure a higher degree of accuracy.

To guarantee 95% accuracy, we must **increase** the error (margin of error) to account for more uncertainty in the data.

---

Would you like more details or have any questions? Here are 5 related questions for further exploration:
1. How does the sample size affect the margin of error?
2. What is the impact of the confidence level on the margin of error?
3. How would the MOE change if the sample proportion were closer to 50%?
4. Can you explain why the z-value differs for each confidence level?
5. What happens to the margin of error if the sample size increases?

**Tip:** A larger sample size reduces the margin of error, leading to more precise estimates.

A survey asked, 'Do you think that homeless people are responsible for the situation they are in?' Of the 2,500 Americans surveyed, 1,408 said no. (a) Determine the sample proportion of Americans who think that homeless people are not responsible for the situation they are in. (Round your answer to the nearest whole number.) Sample proportion is % (b) With an 80% level of confidence, find the margin of error (MOE) associated with the sample proportion. (Give the answer to one decimal place, if needed.) MOE = % (c) With a 95% level of confidence, find the margin of error (MOE) associated with the sample proportion. (Give the answer to one decimal place, if needed.) MOE = % (d) The answer in part (c) is ---Select--- the answer in part (b). Why? In order to guarantee 95% accuracy, we must ---Select--- the error.

Learn how to calculate the sample proportion and margin of error in a survey about homelessness. This problem demonstrates statistical concepts such as proportion, margin of error, and confidence intervals, using both 80% and 95% confidence levels. Suitable for students in Grades 10-12.

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