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I have a list of patients and their footfall specific velocities. Each patient has come in for two visits: the original and the retest visit. At each visit patients were asked to walk at six different self selected speeds ranging from very slow to very fast.

I would like to see two things. 1.) if each individual patient walks at similar velocities across all the trials between the two visits and 2.) if each trial alone is similar between the two visits.

This is the beginning of a data frame (stack.data) for one patient:

 Velocity    Visit Trial
1   1.02660 original     1
2   0.60908 original     2
3   0.75979 original     4
4   1.08750 original     5
5   1.35840 original     6
6   1.00520 original     1
7   0.64521 original     2
8   0.79415 original     4
9   1.23540 original     5
10  1.42090 original     6
11  0.80487 original     1
12  0.66805 original     2
...

I have thought to do this two ways. The first is is a T test between all the velocities, and then another T test between each trials original retest visits. I don't believe I can do a paired T test because there are a different number of footfalls in each visits trials. This t test looks like this:

t.test(stack.data$Velocity[stack.data$Visit == "original"], stack.data$Velocity[stack.data$Visit == "retest"])

I am concerned that doing this for each trial between original and retest will give artificially significant p values though, so I have also done a repeated measures ANOVA like in this post.

aov_velocity = aov(Velocity ~ Visit + Error(Trial / Visit), data = stack.data)

Calling summary of aov_velocity gives:

Error: Trial
          Df Sum Sq Mean Sq F value Pr(>F)
Visit      1 0.0838  0.0838   0.087  0.787
Residuals  3 2.8887  0.9629               

Error: Trial:Visit
          Df  Sum Sq Mean Sq F value   Pr(>F)    
Visit      1 0.23728 0.23728   77.96 0.000908 ***
Residuals  4 0.01217 0.00304                     

I believe Error: Trial:Visit is giving the p value for the Velocity between visits by trial, and I can say they are significantly different. I was reading this though and it said to use the Error portion when "Trial" is a random effect. Since patients are asked to walk at a certain speed, but that speed is open to their interpretation is that a random or fixed effect? If it is a fixed effect is this formula better suited:

aov(Velocity ~ (Visit*Trial) + Error(Velocity/(Visit*Trial)), data=stack.data)

Also if either of these is a valid way to do repeated measure ANOVA what would be the best post hoc test to perform where you are able to tell that a patient is walking at a different speed overall due to the difference between trial x original and trial x retest?

Thank you for your help, please let me know if more information would be helpful, or if this is posted in the wrong place.

kjetil b halvorsen
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Brennan
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  • "Similar" in what sense? Is this a question on equivalence tests? – Michael M Apr 13 '20 at 19:17
  • Yes, I believe so. I would like to know the best way to say if a patient is walking over the same range of speeds between the two visits when they are instructed to walk at a certain speed. – Brennan Apr 13 '20 at 19:38

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