Suppose you survey a random sample from a large population to understand the population proportion. How large must your sample be if you want the Margin of Error to be 6% = 0.06 with a confidence level of 98%? (Be sure to show your calculations.)
Suppose you survey a random sample from a large population to understand the population proportion. How large must your sample be if you want the Margin of Error to be 6% = 0.06 with a confidence level of 98%? (Be sure to show your calculations.)
MATLAB: An Introduction with Applications
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Author:Amos Gilat
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![### Sample Size Calculation for Population Proportion
**Question:**
Suppose you survey a random sample from a large population to understand the population proportion. How large must your sample be if you want the Margin of Error to be 6% (0.06) with a confidence level of 98%? (Be sure to show your calculations.)
**Explanation and Solution:**
To determine the required sample size for estimating a population proportion within a certain margin of error at a given confidence level, you can use the formula:
\[ n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right) \]
where:
- \( n \) = required sample size
- \( Z \) = Z-score (standard score) corresponding to the desired confidence level
- \( p \) = estimated population proportion (if unknown, use 0.5 for maximum variability)
- \( E \) = margin of error
1. **Determine the Z-score for a 98% confidence level:**
The Z-score that corresponds to a 98% confidence level can be found using a Z-table or standard normal distribution table. For a 98% confidence level, the Z-score is approximately 2.33.
2. **Assume the population proportion \( p \):**
If the population proportion is unknown, we use \( p \) = 0.5 to ensure maximum variability.
3. **Use the margin of error \( E \):**
The margin of error given is 6% (0.06).
Plug the values into the formula:
\[
n = \left( \frac{2.33^2 \cdot 0.5 \cdot (1 - 0.5)}{0.06^2} \right)
= \left( \frac{2.33^2 \cdot 0.25}{0.0036} \right)
= \left( \frac{5.4289 \cdot 0.25}{0.0036} \right)
= \left( \frac{1.357225}{0.0036} \right)
\approx 377.00695
\]
4. **Round up to ensure adequacy:**
Since the sample size must be a whole number, round up to the next whole number.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2597422-a53b-4ae1-a6c2-4e1890f10108%2F62179426-9ee7-4304-b062-ae45a3b251ff%2Fpwtys5.png&w=3840&q=75)
Transcribed Image Text:### Sample Size Calculation for Population Proportion
**Question:**
Suppose you survey a random sample from a large population to understand the population proportion. How large must your sample be if you want the Margin of Error to be 6% (0.06) with a confidence level of 98%? (Be sure to show your calculations.)
**Explanation and Solution:**
To determine the required sample size for estimating a population proportion within a certain margin of error at a given confidence level, you can use the formula:
\[ n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right) \]
where:
- \( n \) = required sample size
- \( Z \) = Z-score (standard score) corresponding to the desired confidence level
- \( p \) = estimated population proportion (if unknown, use 0.5 for maximum variability)
- \( E \) = margin of error
1. **Determine the Z-score for a 98% confidence level:**
The Z-score that corresponds to a 98% confidence level can be found using a Z-table or standard normal distribution table. For a 98% confidence level, the Z-score is approximately 2.33.
2. **Assume the population proportion \( p \):**
If the population proportion is unknown, we use \( p \) = 0.5 to ensure maximum variability.
3. **Use the margin of error \( E \):**
The margin of error given is 6% (0.06).
Plug the values into the formula:
\[
n = \left( \frac{2.33^2 \cdot 0.5 \cdot (1 - 0.5)}{0.06^2} \right)
= \left( \frac{2.33^2 \cdot 0.25}{0.0036} \right)
= \left( \frac{5.4289 \cdot 0.25}{0.0036} \right)
= \left( \frac{1.357225}{0.0036} \right)
\approx 377.00695
\]
4. **Round up to ensure adequacy:**
Since the sample size must be a whole number, round up to the next whole number.
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