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

135 to 143 minutes.

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*(s)/(√(n))`

Upper bound:

`U= X+z*(s)/(√(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*(22.6)/(√(123))\\L=135`

Upper bound:

`U= 139+1.96*(22.6)/(√(123))\\U=143`

The 95% confidence interval is 135 to 143 minutes.