Question

The article Application of Surgical Navigation to Total Hip Arthroplasty (T. Ecker and S. Murphy,Journal of Engineering in Medicine, 2007:699712) reports that in a sample of 121 hip surgeries of acertain type, the average surgery time was 136.9 minutes with a standard deviation of 22.6 minutes.(a) Find a 99% confidence interval for the mean surgery time for this procedure. (5 points )(b) Ecker and Murphy want to use their data to plan a new more extensive study. Approximately how many surgeries must be sampled so that a 95% confidence interval will specify the mean towithin±3 minutes? (5 points)(c) Approximately how many surgeries must be sampled so that a 99% confidence interval will specifythe mean to within±3 minutes? (5 points)

Answer 1

a) The confidence interval is calculated as:

μ = M ± t(s_M)

where:

M = sample mean
t = t statistic determined by confidence level
s_M = standard error = √(s²/n)

here T-distribution is used since population standard deviation is unknown.

M = 136.9
t = 2.62 (computed using T-table shown below at 99% confidence level and Degree of freedom=n-1=121-1=120)
s_M = √(22.6²/121) = 2.05

μ = M ± t(s_M)
μ = 136.9 ± 2.62*2.05
μ = 136.9 ± 5.378

99% CI [131.522, 142.278].

b) The Minimum sample is calculated as:

Where Z score at 95% confidence level is computed using the Z table shown below as Z=1.96

Where E= margin of error=3

thus at a 95% confidence interval, 218 minimum surgeries are required.

c) again at 99% confidence level the Z score is 2.58 thus minimum sample for margin of error=3

thus at a 99% confidence interval, 378 minimum surgeries are required.

The T-table, The Z table: