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 90% confidence interval for the mean of these costs.

Answer 1

To construct a 90% confidence interval for the mean of the medical costs, we will use the formula for the confidence interval for the mean when the population standard deviation is unknown:

\[
\text{Confidence Interval} = \bar{x} \pm z^* \left(\frac{s}{\sqrt{n}}\right)
\]

where:
- \(\bar{x}\) is the sample mean,
- \(z^*\) is the z-score corresponding to the desired confidence level,
- \(s\) is the sample standard deviation,
- \(n\) is the sample size.

Given data:
- Sample mean (\(\bar{x}\)) = R374
- Sample standard deviation (\(s\)) = R56.25
- Sample size (\(n\)) = 250

### 1. Find the z-score for a 90% confidence level.

For a 90% confidence interval, we split the remaining 10% (0.10) into two tails, giving us 0.05 in each tail. Using a z-table, we find that the critical z-value for 0.05 in the upper tail is approximately:

\[
z^* \approx 1.645
\]

### 2. Calculate the standard error (SE) of the mean.

The standard error is calculated using the formula:

\[
SE = \frac{s}{\sqrt{n}} = \frac{56.25}{\sqrt{250}}
\]

Calculating \(\sqrt{250}\):

\[
\sqrt{250} \approx 15.81
\]

So the standard error becomes:

\[
SE = \frac{56.25}{15.81} \approx 3.56
\]

### 3. Calculate the margin of error (ME).

Now we can calculate the margin of error:

\[
ME = z^* \times SE = 1.645 \times 3.56 \approx 5.85
\]

### 4. Construct the confidence interval.

Now, we can create the confidence interval using the sample mean and the margin of error:

\[
\text{Confidence Interval} = \bar{x} \pm ME = R374 \pm R5.85
\]

This gives us:

\[
\text{Lower limit} = 374 - 5.85 \approx R368.15
\]
\[
\text{Upper limit} = 374 + 5.85 \approx R379.85
\]

### Final Result:

The 90% confidence interval for the mean of the medical costs is approximately:

\[
\text{CI} = (R368.15, R379.85)
\]