Question

The article "Application of Surgical Navigation to Total Hip Arthroplasty" (T. Ecker and S. Murphy, Journal of Engineering in Medicine, 2007: 699–712) reports that in a sample of 123 hip surgeries of a certain type, the average surgery time was 136.9 minutes with a standard deviation of 22.6 minutes. Find a 95% confidence interval for the mean surgery time for this procedure. Find a 99.5% confidence interval for the mean surgery time for this procedure. A surgeon claims that the mean surgery time is between 133.9 and 139.9 minutes. With what level of confidence can this statement be made? Approximately how many surgeries must be sampled so that a 95% confidence interval will specify

Answer 1

a) The 95% confidence interval for the mean is (132.9, 140.9).

b) The 99.5% confidence interval for the mean is (131.1, 142.7).

c) Confidence level = 85.6%

Explanation:

a) We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=136.9.

The sample size is N=123.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

`s_M=(s)/(√(N))=(22.6)/(√(123))=(22.6)/(11.091)=2.038`

The t-value for a 95% confidence interval and 122 degrees of freedom is t=1.98.

The margin of error (MOE) can be calculated as:

`MOE=t\cdot s_M=1.98 \cdot 2.038=4.034`

Then, the lower and upper bounds of the confidence interval are:

`LL=M-t \cdot s_M = 136.9-4.034=132.9\\\\UL=M+t \cdot s_M = 136.9+4.034=140.9`

The 95% confidence interval for the mean is (132.9, 140.9).

b) We have to calculate a 99.5% confidence interval for the mean.

The t-value for a 100% confidence interval is t=2.859.

The margin of error (MOE) can be calculated as:

`MOE=t\cdot s_M=2.859 \cdot 2.038=5.826`

Then, the lower and upper bounds of the confidence interval are:

`LL=M-t \cdot s_M = 136.9-5.826=131.1\\\\UL=M+t \cdot s_M = 136.9+5.826=142.7`

The 99.5% confidence interval for the mean is (131.1, 142.7).

c) We have a confidence interval that is (133.9, 139.9).

This corresponds to a margin of error of:

`MOE=(UL-LL)/(2)=(139.9-133.9)/(2)=(6)/(2)=3`

The critical value for this margin of error is:

`t=MOE/s_M=3/2.038=1.472`

This critical value of t=1.472 for 122 degrees of freedom corresponds to a confidence level of 85.6%. See picture attached.