Question

For a simple random sample of forty patients who have recently had knee surgery, the administrator of a physical therapy facility finds a 90% CI of (5.76, 7.25) for the average postoperative performance scores on a knee flexibility test.

1. (b) Let’s say I now want to be 99% confident. What value will be in the center of the 99% confidence interval?

2. (c) We are not told whether knee flexibility scores follow a normal distribution. Is the CI we computed still valid (i.e. trustworthy)? Explain why or why not.

Answer 1

Solution:

Given:

Sample size = n = Number of patients who have recently had knee surgery = 40

90% confidence interval = ( 5.76 , 7.25 )

Part b) Let’s say I now want to be 99% confident. What value will be in the center of the 99% confidence interval?

Center for 90% confidence interval is same as center for 99% confidence interval, since center is the sample mean and sample mean does not change even if we change the confidence level.

Thus center = Sample mean for 90% confidence interval:

Add both the limits and divide it by 2 to get Sample mean.

Thus we get:

Thus center of the 99% confidence interval is =

Part c) We are not told whether knee flexibility scores follow a normal distribution. Is the CI we computed still valid (i.e. trustworthy)? Explain why or why not.

Yes, the CI we computed still valid, since sample size = n = 40 > 30, is large sample and hence we can apply central limit theorem which assumes for large sample , sampling distribution of sample means is approximately Normal.