To derive the formula for the sample size \( n \) needed for a given margin of error \( m \), we start from the margin of error formula that relates \( m \), the critical value \( z_{\alpha/2} \), the standard deviation \( \sigma \), and the sample size \( n \). The margin of error \( m \) can be expressed as:
\[ m = z_{\alpha/2} \left( \frac{\sigma}{\sqrt{n}} \right) \]
where:
- \( m \) is the margin of error,
- \( z_{\alpha/2} \) is the z-score corresponding to the desired level of confidence,
- \( \sigma \) is the population standard deviation,
- \( n \) is the sample size.
Steps to Solve for \( n \):
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Start with the margin of error formula: \[ m = z_{\alpha/2} \left( \frac{\sigma}{\sqrt{n}} \right) \]
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Multiply both sides of the equation by \( \sqrt{n} \) to eliminate the fraction: \[ m \sqrt{n} = z_{\alpha/2} \sigma \]
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Now, isolate \( \sqrt{n} \) by dividing both sides by \( m \): \[ \sqrt{n} = \frac{z_{\alpha/2} \sigma}{m} \]
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Square both sides to solve for \( n \): \[ n = \left( \frac{z_{\alpha/2} \sigma}{m} \right)^2 \]
Result:
Thus, the formula for the required sample size \( n \) needed to achieve a specified margin of error \( m \) is:
\[ n = \left( \frac{z_{\alpha/2} \sigma}{m} \right)^2 \]
This formula indicates that to achieve a smaller margin of error \( m \), you would need to either increase the sample size \( n \) or adjust the level of confidence (which changes \( z_{\alpha/2} \)).