4. Knee-Jerk Reaction. The administrator of a physical therapy facility has foun
ID: 3134862 • Letter: 4
Question
4. Knee-Jerk Reaction. The administrator of a physical therapy facility has found that postop- erative performance scores on a knee flexibility test have tended to follow a Normal distribution with a standard deviation = 4. For a simple random sample of ten patients who have recently had knee surgery, the scores are as follows: 101, 92, 94, 88, 52, 93, 76, 84, 72, and 98. (You may use JMP here.)
(a) Calculate a 90% confidence interval for the unknown population mean, .
(b) Free response submission. In the context of this problem, what does represent?
(c) Free response submission. Give the interpretation of the confidence interval that you calculated in part (a).
(d) Free response submission. Five years ago, the mean knee flexibility test score was 80. Based on your confidence interval, do you think there is enough statistical evidence to suggest the mean knee flexibility test score has changed? Explain your answer.
(e) Free response submission. Suppose that knee flexibility test scores do not follow a Normal distribution. Are the results in the part (a) still trustworthy? Explain your answer.
Explanation / Answer
a)
Note that
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.05
X = sample mean = 85
z(alpha/2) = critical z for the confidence interval = 1.644853627
s = sample standard deviation = 4
n = sample size = 10
Thus,
Lower bound = 82.91940645
Upper bound = 87.08059355
Thus, the confidence interval is
( 82.91940645 , 87.08059355 ) [ANSWER]
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b)
u represents the population mean postoperative performance scores for patients who recently had knee injury.
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c)
We are 90% confident that the population mean postoperative performance scores for patients who recently had knee injury is between 82.91940645 and 87.08059355.
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d)
As 80 is not inside the interval, yes, there is evidence to suggest the mean knee flexibility test score has changed.
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e)
No, because we only have a very small sample size, n = 10, so we cannot really approximate the sample mean to be normally distributed by central limit theorem.
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