A random sample of 144 patients at a local clinic who had cataracts removed during the past year showed an average age of 60 years, with a standard deviation of eight years. A construction of a 95% confidence interval for the population mean is required. What is the marginal error?

Answer 1

To calculate the margin of error (ME) for a confidence interval, we can use the formula:

\[ ME = z^* \left( \frac{\sigma}{\sqrt{n}} \right) \]

Where:

• \( z^* \) is the z-score corresponding to the desired confidence level (for 95% confidence, \( z^* \approx 1.96 \)).
• \( \sigma \) is the sample standard deviation.
• \( n \) is the sample size.

Given:

• \( \sigma = 8 \) years
• \( n = 144 \)

Now, we can calculate the standard error (SE):

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{8}{\sqrt{144}} = \frac{8}{12} = \frac{2}{3} \approx 2.67 \]

Next, we calculate the margin of error:

\[ ME = z^* \times SE = 1.96 \times 2.67 \approx 5.23 \]

Thus, the margin of error for the 95% confidence interval is approximately 5.23 years.