A Study of the Reproducibility of Measurements with HUR Leg Extension/Curl Research Line
An important property of measurements is that the results should be the same under equivalent conditions save for the unavoidable statistical fluctuations. In physiological measure- ments we do however never have “identical” conditions. Yet, if a person performs, let us say, a number of MVC1 isometric contrac- tions within a short span of time (a day or so) under similar circumstances, such that we can neglect training effects, we should expect to obtain readings of the peak force which do not vary “too much”. We will assume that the measurement values fluctuate around a mean value and then try to determine the size of the fluctuations. Our preliminary result is that the measure- ment values of peak torque (isometric and dynamic tests), and maximum angular velocity (dynamic test) will be within about 10% of the mean value in 95% of the cases. This is comparable with results cited2 for isokinetic devices which claim a variation of the measurement results (peak torque, total work, average force) of the order of 9 – 14%.
The data used in the present study is derived form a previous study3. In one of the
schemes employed in that study they tested a group of five women practisers of aerobics.
During one day of testing the women performed five isometric extension tests and five flexion tests, and five sets of dynamic extension tests with varying resistance and for both legs (left, right), using the standard HUR test menu with HUR Leg Extension/Curl.
Suppose we have obtained a series of values xi > 0 (i = 1, 2, ...., n) by repeating a measurement n times under similar conditions. One way to decribe the variation of the result is to form the parameters4
(1) x
x zi xi
ˆ
− ˆ
=
where the x with a “hat” denotes the average
1 Maximum voluntary contraction.
2 Quoted in (Männikkö and Martikainen, 2000) – see ref. below.
3 N Männikkö, V Martikainen, Isometriset ja Dynaamiset Mittaukset (Oulun diakonissalaitos, 2000). Ms.
4 Dividing with the average value make the variations for different subjects comparable.
(2)
∑
=
= n
i
xi
x n
1
ˆ 1
As an example of the data we give the results of the isometric flexion (curl) tests (left leg, maximum torque in Nm) in the following table5:
86 77 99 120 115
81 86 102 123 122
84 81 101 123 114
85 80 89 123 122
90 79 89 121 127
Each column gives the five results of each subject (N = 5). For each column we caclulate the averages (2) and then the 5 x 5 = 25 values for the z-parameter (1). This is repeated for right leg data, and extension right and left leg data. It is found that z varies between about –0.1 and 0.1. We can draw a corresponding histogram for the total data6:
Figure 1 Frequency of the variation of isometric data (totque)
5 The isometric test consists of two measurements of MVC peak torque and the better result is taken as the representative value of the test. The joint angle in flexion (curl) measurements was set to 140 degrees and in the extension measurements to 120 degrees.
6 Data consisted of 90 (= 100 – 2 x 5) data points because two incomplete tests were dropped. In the diagrams we have used all the z-points though for every test (consisting of five measurements) one z-point is determined by the four others since their sum according to the definition (1) will be zero. Using all the z- points gives us smoother curves and forces the average to be exactly zero.
A clearer picture of the statistics of the data emerges if we draw the cumulative distribution of the variation:
Figure 2 Distribution of the variation of the isometric data
This shows e.g. that in about 85 % of the measurements the variation z is less than 0.05:
i.e. the measured value does not exceed the “true mean value” by more than 5 %. The continous line in the diagram is the curve of the normal distribution (Gauss) with mean µ
= 0 and standard deviation σz = 0.045. The standard deviation was estimated from σz = s using
(3)
∑
− =
= n
i
zi
s n
1 2
1 1
As we can see from the figure the gaussian distribution seems to describe the distribution of the variation in the isometric measurement data quite well. The standard deviation σ of the measurement data can be estimated using σz = 0.045 and equ (6) in the Appendix:
µ µ σ
σ 0.05
1 1
≈
−
=
n
z
Thus, based on this data we can say that a measurement of isometric peak torque will in about 95 percent of the cases be within ±10 % (10% = 2 σ/µ)7 of the expected mean value of such measurements.
7 The 95% confidence interval of a normal distribution is given by (µ - 1.96σ, µ + 1.96σ ).
That is, if we make two measurements on a person and obtain results that differ by more than about 14% (10 √2 := 14)8, then, with a 95 percent certainty we may expect the difference to be significant and not within the “bounds of statistical variation”.
In the dynamic extension tests the peak torque and maximum angular velocity was
measured during MVC at resistances in steps between 2 and 8 bar. This was repeated five times. The torque and velocity data of the dynamic tests was processed in the same way as the torque data of the isometric tests. For peak torque data we used the measurements at 4 and 8 bar, and for velocity data we used measurements at 2 and 4 bar resistance (from both right and left leg). (Note: 2 and 4 bar data treated separately show practically the same distributions whence it makes sense to lump them together.) The distribution of the variation for the peak torque is presented in the following figure:
Figure 3 Distr. of varia. of peak torque of the dynamic tests
The normal distribution drawn in the figure has a standard deviation σz = 0.037.
For the variation of maximum angular velocities we obtain the distribution:
8 The difference of two independent random variables with normal distribution N(µ,σ ) will have a normal distribution N(0, σ √2). Formally the size of the significant difference of two measurements m1 and m2
could be expressed as 0.1 m12 +m22 .
Figure 4 Distr. of varia. of max. angular velocity of the dyn. data
In this case the standard deviation of the superimposed normal distribution curve is σz = 0.041. In figure 3 and 4 we see that the agreement with the normal distribution is not perfect. This could suggest that the measurements does not exactly follow a normal distribution around a mean value but instead shows a higher concentration around the zero variation point z = 0. One might speculate – though this data does not warrant any far reaching conclusions in this respect (the bumps are pretty much within the statistical fluctuations) - e.g. that in the dynamic tests the motoric program activated also stabilizes the results. Submaximal efforts which perhaps allow for greater control of the motion could also stabilize the results. To test this more data would be needed.
The esimation of of the standard deviation σz using (3) is also fraught with some uncertainty. Indeed, if xi were independent normal variables with the distribution N(0, σz), then the sum
∑
== N
i i z
x
1 2 2
2 1
χ σ
would obey the Chi-square distribution. If the degrees of freedom N is greater than 30 the Chi-square distribution is well approximated by a normal distribution N
(
N, 2N)
. Inour case we obtain xi from zi by dropping every fifth variable since according to (1) the sum of the five measurements of every test is zero; thus, only 4 in 5 are independent variables. With these considerations we get N = (4/5) 100 = 80 and
[
80 1.96 2 80 80 1.96 2 80]
0.95Prob − ⋅ ≤χ2 ≤ + ⋅ =
Thus the 95% confidence interval for the estimate of the standard deviation
σ χ x N
s N z
N
i i x
2
1
1 2
=
=
∑
=
will be
− +
80 96 2 . 1 1 80,
96 2 . 1
1 z
z σ
σ
which is (0.85 σz, 1.15 σz); i.e. the standard variation is within ±15% of the estimate, where we for σz can use the estimate sx or (3) which will be of the same size (around 0.04). The point of this is only to show that no big variations in the estimate of the standard deviation are to be expected.
In conclusion: Based on the data from the aerobic test group we may expect that one can make measurements of maximum torque (isometric test), peak torque and maximum angular velocity (dynamic test) with HUR Leg extension/curl such that the variation is less than about ±10 % in 95 percent of the cases. This is e.g. on the same level as has been reported for measurements with isokinetic machines.
Tests with other groups of subjects will most likely show the same pattern if the measurements are done properly9. No big change in the range of variation is to be expected, but only new data will tell more about this. For instance, subjects with poor motor control may be expected to show larger variations in the results.
Mathematical Appendix
Suppose the independent random variables xi (i = 1, 2, ...., n) have the normal distribution N(µ, σ) with µ > 2σ > 0, then the probability density function (pdf) for
(4) x
x z x
ˆ
1− ˆ
=
is given quite accurately by
9 The test subjects should e.g. have enough time to familiarize themselves with the equipment before the measurements.
(5)
− +
−
− +
−
= n
z n
z
e n z n z n
f
2 2 2 2
2 3
1 1 2
1 2
1 1 1 1
) 2 (
σ µ
σ π µ
The following picture (Figure 5) shows a simulated test series based on generating a series of 5 (n = 5) normally distributed numbers x (µ = 100, σ = 5) and calculating the z- numbers (1). This is repeated 100 times giving 500 datapoints whose distribution is plotted in the figure. The continous line in the picture is obtained using n = 5, µ = 100, and σ = 5 (this corresponds to the case σz = 0.045) by numerically computing the distribution of (5). Apparently it fits the simulated distribution very well.
Figure 5 Distribution of simulated data. Continous line is the distribution computed from (5).
For small |z| (that is, for z2 << n) (5) approaches a normal distribution with the standard deviation σz given by
(6) z n
1−1
= µ σ σ
It follows that (5) is well approximated by a normal distribution if we have10
10 The condition says that the mean value must be considerable larger than the standard deviation of the average value. This is also an obvious requirement for (4) to be a meaningful description of variation.
(7) 1 2 2
2 n>>
σ µ
In the present case (with σz about 0.05 and n = 5) we get for the left hand side in (7) a number around 1000 so this condition is well satisfied. This supports the use of (4) as a measure of variation and the use of the normal distribution in characterizing the
distribution of the variation. A curious property of the distribution (5) is that it lacks second moment; i.e, the standard deviation is not defined for it (the integral diverges).
This is of no practical consequence here because very big z-values (in the “tail” of (5)) are physically impossible/irrelevant. Mathematically one may also adopt an alternative definition of variation given by
(7)
∑
== −
n
i i i i
n x x z x
1
1 2
ˆ
which may be expected to be mathematically “better behaved” in certain respects. The variable (7) has in general a complicated formula for the probablility density function.
E.g. in the case n = 2 the variable (the square of that in (7))
(8)
( )
2 2
2
2 1
y x
y z x
+
= −
takes values in the range 0 to 1. If x and y are independent random variables with the same N(µ,σ)-distribution, the probability density function for z in (8) becomes
(9)
( ) ∫ ( ) ∑
≥( )
−
− +
−
−
−
= −
= −
0 1 2
4 1
2 1
! 2
! 1
1 4 1
) 1 (
2 2
2
n
n u n
z u
n z n z z e e
u du z z z e
f σ
µ π
π
σ µ σ
σ µ µ
This distribution is of interest only for small values of µ/σ, contrary to (5).
