To construct a 99% confidence interval for the mean of the medical costs, we will use the formula for the confidence interval based on a sample mean. The formula is:
\[ CI = \bar{x} \pm z \left(\frac{s}{\sqrt{n}}\right) \]
Where:
- \(\bar{x}\) is the sample mean
- \(z\) is the z-score corresponding to the desired confidence level
- \(s\) is the sample standard deviation
- \(n\) is the sample size
Given values:
- Sample mean (\(\bar{x}\)) = R374
- Sample standard deviation (s) = R56.25
- Sample size (n) = 250
Step 1: Find the z-score for 99% confidence level
For a 99% confidence level, the z-score can be found using a z-table or standard normal distribution table. The z-score corresponding to 99% confidence is approximately 2.576.
Step 2: Calculate the standard error (SE)
\[ SE = \frac{s}{\sqrt{n}} = \frac{56.25}{\sqrt{250}} \approx \frac{56.25}{15.8114} \approx 3.56 \]
Step 3: Calculate the margin of error (ME)
\[ ME = z \cdot SE = 2.576 \cdot 3.56 \approx 9.16656 \approx 9.17 \]
Step 4: Construct the confidence interval
\[ CI = \bar{x} \pm ME = 374 \pm 9.17 \]
This gives us:
- Lower limit: \(374 - 9.17 \approx 364.83\)
- Upper limit: \(374 + 9.17 \approx 383.17\)
Step 5: Write the confidence interval
Thus, the 99% confidence interval for the mean of these costs is approximately:
\[ [364.83 ; 383.17] \]
Matching with the provided options:
The closest option that matches our calculated confidence interval is:
\[ [364.8356 ; 383.1644] \]
This is the correct choice.