To determine the required sample size for a given margin of error and confidence level, we can use the formula for sample size in estimating proportions:
\[ n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right) \]
Where:
- \( n \) = required sample size
- \( Z \) = Z-score corresponding to the desired confidence level
- \( p \) = estimated proportion of the population supporting the candidate (typically 0.5 when no prior estimate is available as it provides the maximum sample size)
- \( E \) = margin of error (as a decimal)
Step 1: Determine the Z-score for a 90% confidence level
For a 90% confidence level, the Z-score is approximately 1.645.
Step 2: Estimate the population proportion
If we do not have prior information about the proportion of people who support the candidate, we will use \( p = 0.5 \). This maximizes the required sample size.
Step 3: Set the margin of error
The margin of error is given as 1%, which is \( E = 0.01 \).
Step 4: Plug values into the formula
Now we can calculate \( n \):
\[ n = \left( \frac{(1.645)^2 \cdot 0.5 \cdot (1 - 0.5)}{(0.01)^2} \right) \]
Calculating:
\[ n = \left( \frac{(1.645)^2 \cdot 0.5 \cdot 0.5}{0.0001} \right) \]
\[ n = \left( \frac{2.706025 \cdot 0.25}{0.0001} \right) \]
\[ n = \left( \frac{0.67650625}{0.0001} \right) = 6765.0625 \]
Step 5: Round to the nearest whole person
Since we cannot have a fraction of a person, we round up:
\[ n = 6766 \]
Thus, the required sample size needed to achieve a 1% margin of error at a 90% confidence level is 6766 people.
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