Question

Certain medical costs are approximately normally distributed. A random sample of 250 operations showed a mean of R374 and a standard deviation of R56.25. Construct a 99% confidence interval for the mean of these costs.

Answer 1

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 \((R364.82, R383.18)\).