To construct a 99% confidence interval for the mean of the medical costs, we can use the formula for the confidence interval for the mean, which is given by:
\[ \text{Confidence Interval} = \bar{x} \pm z \left( \frac{s}{\sqrt{n}} \right) \]
where:
- \(\bar{x}\) is the sample mean,
- \(z\) is the z-score associated with the desired confidence level,
- \(s\) is the standard deviation of the sample, and
- \(n\) is the sample size.
Given:
- Sample mean (\(\bar{x}\)) = R374
- Standard deviation (\(s\)) = R56.25
- Sample size (\(n\)) = 250
- Confidence level = 99%
Step 1: Find the z-score for a 99% confidence level
For a 99% confidence interval, the z-score can be found from a standard normal distribution table (or using an inverse normal distribution function). The z-score that corresponds to the two-tailed 99% confidence level is approximately 2.576.
Step 2: Calculate the standard error
The standard error (SE) is calculated as follows:
\[ \text{SE} = \frac{s}{\sqrt{n}} = \frac{56.25}{\sqrt{250}} \approx \frac{56.25}{15.81} \approx 3.56 \]
Step 3: Calculate the confidence interval
Now, we can plug in the values into the confidence interval formula:
\[ \text{Confidence Interval} = 374 \pm 2.576 \times 3.56 \]
Calculating the margin of error:
\[ 2.576 \times 3.56 \approx 9.18 \]
Therefore, the confidence interval is:
\[ 374 \pm 9.18 \]
This produces:
\[ \text{Lower limit} = 374 - 9.18 \approx 364.82 \] \[ \text{Upper limit} = 374 + 9.18 \approx 383.18 \]
Conclusion
The 99% confidence interval for the mean of the costs of the operations is approximately \((R364.82, R383.18)\).