To calculate the upper limit of a 99% confidence interval for the average income, we can use the formula for the confidence interval for the mean when the population standard deviation is known:
\[ \text{Confidence Interval} = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \]
Where:
- \(\bar{x}\) is the sample mean
- \(z\) is the z-score corresponding to the desired confidence level
- \(\sigma\) is the population standard deviation
- \(n\) is the sample size
Given:
- Sample mean, \(\bar{x} = R15000\)
- Population standard deviation, \(\sigma = R1000\)
- Sample size, \(n = 50\)
- For a 99% confidence level, the z-score (from the z-table) is approximately 2.576.
Now, we can calculate the margin of error:
\[ \text{Margin of Error} = z \left( \frac{\sigma}{\sqrt{n}} \right) \]
Substituting in the values:
\[ \text{Margin of Error} = 2.576 \left( \frac{1000}{\sqrt{50}} \right) \]
Calculating \(\sqrt{50}\):
\[ \sqrt{50} \approx 7.071 \]
Now substituting this into the margin of error calculation:
\[ \text{Margin of Error} = 2.576 \left( \frac{1000}{7.071} \right) \approx 2.576 \times 141.421 = 364.16 \]
Now we can find the upper limit of the confidence interval:
\[ \text{Upper Limit} = \bar{x} + \text{Margin of Error} = 15000 + 364.16 \approx 15364.16 \]
Thus, the upper limit of the 99% confidence interval for the average income is approximately:
\[ \text{Upper Limit} \approx R15364.16 \]