To determine the total sample size necessary for a specified confidence interval width, we can use the formula for the margin of error in confidence intervals. The margin of error (E) can be expressed as:
\[ E = z \times \frac{\sigma}{\sqrt{n}} \]
Where:
- \(E\) is the margin of error
- \(z\) is the z-score that corresponds to the confidence level
- \(\sigma\) is the population standard deviation
- \(n\) is the sample size
For a 95% confidence interval, the z-score (based on the standard normal distribution) is approximately 1.96.
Given:
- Population standard deviation (\(\sigma\)) = R1000
- Desired margin of error (E) = R100 (half-width of the confidence interval)
We can rearrange the formula to solve for the required sample size \(n\):
\[ E = z \times \frac{\sigma}{\sqrt{n}} \implies \sqrt{n} = z \times \frac{\sigma}{E} \implies n = \left(\frac{z \times \sigma}{E}\right)^2 \]
Substituting the values into the formula:
\[ n = \left(\frac{1.96 \times 1000}{100}\right)^2 \]
Calculating the right-hand side step-by-step:
- Calculate \(\frac{1.96 \times 1000}{100}\):
\[ \frac{1.96 \times 1000}{100} = 19.6 \]
- Square that value:
\[ n = (19.6)^2 = 384.16 \]
Since the sample size must be a whole number, we round up:
\[ n \approx 385 \]
Thus, the economist would need a total sample size of 385 individuals to achieve a 95% confidence interval with a width of no more than R100.