To estimate the sample size needed to achieve a certain margin of error for a proportion with a specified confidence level, we can use the formula for the sample size of proportions:
\[ n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right) \]
Where:
- \( n \) is the sample size.
- \( Z \) is the z-score corresponding to the desired confidence level.
- \( p \) is the estimated proportion (if unknown, we typically use 0.5 for maximum sample size).
- \( E \) is the margin of error.
Given:
- Confidence level = 90%
- Margin of error \( E = 0.05 \)
Step 1: Find the Z-Score
For a 90% confidence level, the critical value (z-score) can be found using a standard normal distribution table or calculator:
- The z-score for a 90% confidence level is approximately \( Z = 1.645 \).
Step 2: Estimating Proportion \( p \)
If we don't have an estimate for the proportion of office workers who will respond to e-mails, we can use \( p = 0.5 \) since it provides the maximum variability and thus the largest required sample size.
Step 3: Plug Values into the Formula
Now we can substitute the values into the formula:
\[ n = \left( \frac{(1.645)^2 \cdot 0.5 \cdot (1 - 0.5)}{(0.05)^2} \right) \]
Calculating this step-by-step:
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Calculate \( (1.645)^2 \): \[ (1.645)^2 \approx 2.706 \]
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Calculate \( 0.5 \cdot (1 - 0.5) \): \[ 0.5 \cdot 0.5 = 0.25 \]
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Calculate the margin of error squared \( (0.05)^2 \): \[ (0.05)^2 = 0.0025 \]
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Substitute back into the formula: \[ n = \frac{2.706 \cdot 0.25}{0.0025} = \frac{0.6765}{0.0025} \approx 270.6 \]
Conclusion:
Since the sample size must be a whole number, we round up to the next whole number:
\[ n \approx 271 \]
Thus, you would need a sample size of 271 office workers to be 90% confident that the proportion of those who will respond to emails in an hour is within 0.05 of the true proportion.
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