Question

Given the population standard deviation is known as R500. A random sample of 50 individuals resulted in a mean of R10000. What is the width of the 90% confidence interval?

Answer 1

To calculate the width of the 90% confidence interval for the population mean, we'll use the formula for the confidence interval when the population standard deviation is known. The formula is:

\[
CI = \bar{x} \pm z \left(\frac{\sigma}{\sqrt{n}}\right)
\]

Where:
- \(\bar{x}\) = sample mean
- \(z\) = z-score corresponding to the desired confidence level
- \(\sigma\) = population standard deviation
- \(n\) = sample size

1. **Determine the parameters:**
- Sample mean, \(\bar{x} = R10000\)
- Population standard deviation, \(\sigma = R500\)
- Sample size, \(n = 50\)

2. **Find the z-score for a 90% confidence level:**
- For a 90% confidence interval, the z-score (which corresponds to the critical value) can be found using standard normal distribution tables. The z-score for 90% confidence level is approximately \(1.645\).

3. **Calculate the standard error (SE):**
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{500}{\sqrt{50}} \approx \frac{500}{7.0711} \approx 70.71
\]

4. **Calculate the margin of error (ME):**
\[
ME = z \cdot SE = 1.645 \cdot 70.71 \approx 116.67
\]

5. **Calculate the width of the confidence interval:**
The width of the confidence interval is \(2 \times ME\):
\[
\text{Width} = 2 \times 116.67 \approx 233.34
\]

Thus, the width of the 90% confidence interval is approximately **R233.34**.