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 is the upper limit in a 99% confidence interval for the average income?

1 answer

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:

\[ \text{Upper Limit} \approx R15364.16 \]