Question

In an article published in journal of medicine, it is reported that in a sample of 123 hip surgeries of a certain type, the average time was 139 minutes with standard deviation 22.6 minutes. Find a 95% confidence interval for the mean surgery time.

Answer 1

Answer:

135 to 143 minutes.

Step-by-step explanation:

Standard deviation (s) = 22.6 minutes

Sample size (n) = 123

Sample average (X) = 139 minutes

The lower and upper bounds of a confidence interval are given by:

Lower bound:

$L= X-z*\frac{s}{\sqrt{n}}$

Upper bound:

$U= X+z*\frac{s}{\sqrt{n}}$

For a confidence interval of 95%, the z-score value is 1.96.

Therefore, the 95% confidence interval for the mean surgery time is:

Lower bound:

$L= 139-1.96*\frac{22.6}{\sqrt{123}}\\L=135$

Upper bound:

$U= 139+1.96*\frac{22.6}{\sqrt{123}}\\U=143$

The 95% confidence interval is 135 to 143 minutes.