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.

1 answer

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:

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