r cruciate ligament was 32 patients. The outcome measure is the difference in qu
ID: 3276418 • Letter: R
Question
r cruciate ligament was 32 patients. The outcome measure is the difference in quadriceps between the injured and uninjured knees for each individual. The observed sample mean of the difference scores was 0.60 Nm peak force (measured in Nm/kg) /kg with a sample standard deviation 0.40 Nm/kg. During your lit review you found an article where another lab conducted a similar study using a sample of 16 patients. The other lab found a mean difference score of 0.48 Nm/kg with a sample standard deviation of o.32. 1) (4 points) Which lab (yours or the other lab) measured the mean difference score with a smaller precision? Hint: you want to look at the estimated standard errors. 2) (4 points) Assuming the observed mean difference and sample standard deviation from your study are the true population values. How likely is it for a single (n 1) ACL injury patient to have an observed difference score of 0.48 or greater? (Use the standard normal distribution for this problem).Explanation / Answer
1)for your la b std error =std deviation/(n)1/2 =0.40/(32)1/2 =0.0707
for other lab std error =std deviation/(n)1/2 = 0.32/(16)1/2 =0.08
your lab measures the mean difference score with a smaller precision due to lower std error of mean differnence.
2) as we know that z score =(X-mean)/ std deviation
therefore P(X>0.48)=1-P(X<0.48)=1-P(Z<(0.48-0.60)/0.40) =1-P(Z<-0.3)=1-0.3821=0.6179
3)for 16 scores std error of mean =std deviation/(n)1/2 =0.40/(16)1/2 =0.10
therefore P(X>0.48)=1-P(X<0.48)=1-P(Z<(0.48-0.60)/0.10) =1-P(Z<-1.2)=1-0.1151 =0.8849
4)as we can see they are different. The reason is that for average of 16 values should be more centric towards mean due to lower variablility then for a single value,
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