To construct a 95% confidence interval for the mean of the medical costs, we can use the formula for the confidence interval for a population mean when the population standard deviation is unknown (which is our case, since we have a sample standard deviation). The formula is given by:
\[ \bar{x} \pm z \times \left( \frac{s}{\sqrt{n}} \right) \]
Where:
- \(\bar{x}\) = sample mean
- \(z\) = z-value corresponding to the desired confidence level
- \(s\) = sample standard deviation
- \(n\) = sample size
For a 95% confidence interval, the z-value is approximately 1.96 (this can be found in standard z-tables or using statistical software).
Given:
- Sample mean (\(\bar{x}\)) = R374
- Sample standard deviation (\(s\)) = R56.25
- Sample size (\(n\)) = 250
Now, we need to calculate the standard error (SE):
\[ SE = \frac{s}{\sqrt{n}} = \frac{56.25}{\sqrt{250}} \]
Calculating \(\sqrt{250}\):
\[ \sqrt{250} \approx 15.81 \]
Now we calculate the standard error:
\[ SE = \frac{56.25}{15.81} \approx 3.56 \]
Next, we can calculate the margin of error (ME):
\[ ME = z \times SE = 1.96 \times 3.56 \approx 6.98 \]
Finally, we can construct the confidence interval:
\[ \text{Confidence Interval} = \bar{x} \pm ME = 374 \pm 6.98 \]
Calculating the lower and upper bounds:
- Lower bound: \(374 - 6.98 \approx 367.02\)
- Upper bound: \(374 + 6.98 \approx 380.98\)
Thus, the 95% confidence interval for the mean of these costs is approximately:
\[ (367.02, 380.98) \]
This means we are 95% confident that the true mean of the medical costs lies between R367.02 and R380.98.