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.

a. Find a 95% confidence interval for the mean surgery time for this procedure.

b. Find a 99.5% confidence interval for the mean surgery time for this procedure.

c. 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?

d. Approximately how many surgeries must be sampled so that a 95% confidence interval will specify the mean to within ±3 minutes?

e. Approximately how many surgeries must be sampled so that a 99% confidence interval will specify the mean to within ±3 minutes?

Solution 6E

Step1 of 6:

Let us consider a random variable X it represents the surgery time. Here random variable X follows normal distribution with mean standard deviation and n = 123.

That is

The probability density function of normal distribution is given by

, .

Where,

x = random variable

= mean of X

= variance of X

= standard deviation os X

= mathematical constant and its value is 3.14

Here our goal is:

a).We need to find 95% confidence interval for the mean surgery time for this procedure.

b).We need to find 99.5% confidence interval for the mean surgery time for this procedure.

c).A surgeon claims that the mean surgery time is between 133.9 and 139.9 minutes.We need to check what level of confidence can this statement be made?

d).We need to find how many surgeries must be sampled so that a 95% confidence interval will specify the mean to within ±3 minutes?

e).We need to find how many surgeries must be sampled so that a 99% confidence interval will specify the mean to within ±3 minutes?

Step2 of 6:

a).

Here we have to find 95% CI, let us take .

Now,

= 0.025

Z-scores(are to the right) is given by

is obtained from standard normal table(area under normal curve). In standard normal table we have to see where 0.9750 value falls, it falls in row 1.9 under column 0.06.

Hence,

Now 95% confidence interval for the mean surgery time for this procedure is given by

(132.9059, 140.8940)

Hence, 95% confidence interval for the mean surgery time for this procedure is

(132.9059, 140.8940).

Step3 of 6:

b).

Here we have to find 99.5% CI, let us take .

Now,

= 0.0025

Z-scores is given by

is obtained from standard normal table(area under normal curve). In standard normal table we have to see where 0.9975 value falls, it falls in row 2.8 under column 0.01.

Hence,

Now 95% confidence interval for the mean surgery time for this procedure is given by

(131.1740, 141.8999)

Hence, 99.5% confidence interval for the mean surgery time for this procedure is

(131.1740, 141.8999).

Step4 of 6:

c).

We have and n = 123.

In a given information we have upper bound 139.9

Now,

139.9-136.9 = (2.0377)

3 = (2.0377)

=

= 1.4722

Hence, = 1.4722.

Z-scores is given by , and this value is obtained from standard normal table(area under normal curve). In standard normal table we have to see in row 1.4 under column 0.07.

= 0.9292

Area to the right of Z is 1 - 0.9292

= 0.0708.

Now,

= 0.1416

Therefore level is (1-) = 1 - 0.1416

= 0.8584.

Hence level is 0.8584.

Step5 of 6:

d).

We know that (from part (a))

= 3

= 3

= 218.0150.

Hence, n = 218.0150.

Step6 of 6:

e).

We know that , s = 22.6.

= 3

= 3

= 374.8354.

Hence, n = 374.8354.