- The margin of error formula for a proportion is \(E = z^* \sqrt{\frac{p^*(1-p^*)}{n}}\).
- Rearranging for the sample size \(n\) gives \(n = \frac{(z^*)^2 p^*(1-p^*)}{E^2}\).
- Substituting \(z^* = 1.96\), \(p^* = 0.26\), and \(E = 0.04\), we find \(n \approx 461.95\).
- Rounding up to the nearest whole number, we get the required sample size: \(\boxed{462}\).
Explanation
- Identify the problem parameters and formula
To estimate the population proportion, we need a sample size that ensures our estimate is within the desired margin of error with a specific level of confidence. We are given the following:
- The preliminary estimate of the population proportion: \(p^* = 0.26\).
- The confidence level: \(95\%\).
- The margin of error: \(E = 0.04\).
The formula for the margin of error in estimating a population proportion is given by:
\[E = z^* \sqrt{\frac{p^*(1-p^*)}{n}}\]
where \(z^*\) is the critical value corresponding to the chosen confidence level. For a \(95\%\) confidence level, the standard critical value is \(z^* \approx 1.96\).
- Calculate the sample size n
We need to solve for the sample size \(n\). Rearranging the margin of error formula to isolate \(n\):
\[E^2 = (z^*)^2 \frac{p^*(1-p^*)}{n}\]
\[n = \frac{(z^*)^2 p^*(1-p^*)}{E^2}\]
Now, we substitute the given values:
- \(z^* = 1.96\)
- \(p^* = 0.26\)
- \(1 - p^* = 1 - 0.26 = 0.74\)
- \(E = 0.04\)
Performing the calculation:
\[n = \frac{(1.96)^2 \times 0.26 \times 0.74}{(0.04)^2}\]
\[n = \frac{3.8416 \times 0.1924}{0.0016}\]
\[n = \frac{0.73912384}{0.0016} \approx 461.95\]
- Determine the final sample size
Since the sample size \(n\) must be a whole number, and we need to ensure the margin of error is at most \(4\%\), we always round up to the next whole number.
Rounding \(461.95\) up gives us \(462\).
Therefore, the required sample size is \(462\).
Examples
This calculation is vital in fields like public health and marketing research. For example, if a researcher wants to estimate what proportion of a city's population uses a specific type of public transportation, they can use this formula to determine how many people to survey to be confident that their result is accurate within a desired margin of error, helping them allocate resources effectively and minimize survey costs. If \(p^* = 0.26\), \(E = 0.04\), and \(z^* = 1.96\), then \(n = \frac{1.96^2 \cdot 0.26 \cdot 0.74}{0.04^2} \approx 462\).