Answer to Question #109489 in Statistics and Probability for İbrahim

Question

4. The concern of a study by Beynnon et al. (A-4) were nine subjects with chronic anterior
cruciate ligament (ACL) tears. One of the variables of interest was the laxity of the anteroposterior,
where higher values indicate more knee instability. The researchers found that among subjects
with ACL-deficient knees, the mean laxity value was 17.4 mm with a standard deviation of
4.3 mm.
(a) What is the estimated standard error of the mean?
(b) Construct the 99 percent confidence interval for the mean of the population from which the nine
subjects may be presumed to be a random sample.
(c) What is the precision of the estimate?

Answer 1

(a) $SE_{\bar x}=\frac{s}{\sqrt{n}}=\frac{4.3}{\sqrt{9}}=1.4333.$

(b) $99\%CI=(\bar x-t_{0.005,n-1}SE_{\bar x},\bar x+t_{0.005,n-1}SE_{\bar x})=$

$=(17.4-3.3554*1.4333,17.4+3.3554*1.4333)=(12.5907,22.2092).$

(c) Precision of the estimate$=t_{0.005,n-1}SE_{\bar x}=3.3554*1.4333=4.8092.$