Question

An economist is interested in studying the incomes of consumers in a particular country. The population standard deviation is known to be R1000. A random sample of 50 individuals resulted in a mean income of R15000. What total sample size would the economist need to use for a 95% confidence interval if the width of the interval should not be more than R100?

Answer 1

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:

1. Calculate \(\frac{1.96 \times 1000}{100}\):

\[
\frac{1.96 \times 1000}{100} = 19.6
\]

2. 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.