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 80% confidence interval for the population mean is required. What is the lower limit of the interval?

1 answer

To construct an 80% confidence interval for the population mean based on the given sample data, we can use the formula for the confidence interval for the population mean when the population standard deviation is known. The formula is:

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

Where:

  • \(\bar{x}\) is the sample mean.
  • \(z^*\) is the z-score that corresponds to the desired confidence level.
  • \(\sigma\) is the sample standard deviation.
  • \(n\) is the sample size.

From the data provided:

  • Sample mean (\(\bar{x}\)) = 60
  • Sample standard deviation (\(\sigma\)) = 8
  • Sample size (\(n\)) = 144

Step 1: Find the z-score for 80% confidence level.

For an 80% confidence interval, we need to find the z-score that corresponds to the middle 80% of the standard normal distribution. This leaves 20% for the tails, or 10% in each tail.

Using a standard normal distribution table or calculator:

  • The z-score for 0.10 in the upper tail (or the z-score for 90% cumulative probability) is approximately \(z^* = 1.28\).

Step 2: Calculate the standard error (SE).

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

Step 3: Calculate the margin of error (ME).

\[ ME = z^* \times SE = 1.28 \times 2.67 \approx 3.42 \]

Step 4: Construct the confidence interval.

Now we can calculate the lower and upper limits of the confidence interval:

  • Lower limit: \[ \bar{x} - ME = 60 - 3.42 \approx 56.58 \]

  • Upper limit: \[ \bar{x} + ME = 60 + 3.42 \approx 63.42 \]

Final Result: The lower limit of the 80% confidence interval for the population mean is approximately \(56.58\).