Following cardiac surgery, patients are encouraged to exercise regularly (assume that regular exercise is defined as exercising on 3 or more days per week). A physician suspects that patients exercise regularly immediately following cardiac surgery but tend to reduce, even stop exercising completely, over time. An investigation is planned to estimate the mean number of weeks that patients exercise regularly following cardiac surgery. Assume that the standard deviation (s.d) in the number of weeks cardiac patients exercise regularly following surgery is 6.3 weeks.
(a) If a sample of 40 cardiac patients is followed, and the number of weeks in which each patient exercised regularly is recorded, what is the probability that the sample mean will be no more than 1 week higher than the true mean?
Under central limit therom,
standard error = s.d./sqrt(n) = 6.3/sqrt(40) = 0.99
X = true mean + 1
Z = (X - true mean) / 0.99 = (true mean + 1 = true mean) / 0.99 = 1/0.99
(b) Find the probability that the sample mean is at least two weeks less than the true mean for a sample of 40 cardiac patients.
X = true mean - 2
Z = (X - true mean) / 0.99 = (true mean -2 - true mean) / 0.99 = -2/0.99
(c) If the sample is increased to 100 cardiac patients, what is the probability that the sample mean will be no more than 1 week higher than the true mean?
same as question
Can anyone help me check my answers to see whether is correct or not? Please correct and explain if wrong…
Best Answer
Hint: If each patient's exercise record (the number of weeks he or she exercises regularly following surgery) follows a normal distribution $N(\mu,\sigma^2)$, the sample mean of 40 patients will follow $N(\mu, \sigma^2/40)$ (Why?). Here, $\mu$ is what you call the true mean, and $\sigma$ (not $\sigma^2$) is the standard deviation of $N(\mu,\sigma^2)$ (which is 6.3 weeks). Hope this helps.