An article in Bioelectromagnetics (“Electromagnetic Effects on Forearm Disuse Os
ID: 3024461 • Letter: A
Question
An article in Bioelectromagnetics (“Electromagnetic Effects on Forearm Disuse Osteopenia: A Randomized, Double-Blind, Sham-Controlled Study,” Vol. 32, 2011, pp. 273 – 282) describes a randomized, double-blind, sham-controlled, feasibility and dosing study to determine if a common pulsing electromagnetic field (PEMF) treatment could moderate the substantial osteopenia that occurs after forearm disuse. Subjects were randomized into four groups after a distal radius fracture, or carpal surgery requiring immobilization in a cast. Active of identical sham PEMF transducers were worn on a distal forearm for 1, 2, or 4h/day for 8 weeks starting after cast removal (“baseline”) when bone density continues to decline. Bone mineral density (BMD) and bone geometry were measured in the distal forearm by dual energy X-ray absorptiometry (DXA) and peripheral quantitative computed tomography (pQCT). The data below are the percent losses in BMD measurements on the radius after 16weeks for patients wearing the active or sham PEMF transducers for 1, 2, or 4h/day (data were constructed to match the means and standard deviations read from a graph in the paper).
Sham
PEMF 1h/day
PEMF 2h/day
PEMF 4h/day
4.51
5.32
4.73
7.03
7.95
6.00
5.81
4.65
4.97
5.12
5.69
6.65
3.00
7.08
3.86
5.49
7.97
5.48
4.06
6.98
2.23
6.52
6.56
4.85
3.95
4.09
8.34
7.26
5.64
6.28
3.01
5.92
9.35
7.77
6.71
5.58
6.52
5.68
6.51
7.91
4.96
8.47
1.70
4.90
6.10
4.58
5.89
4.54
7.19
4.11
6.55
8.18
4.03
5.72
5.34
5.42
2.72
5.91
5.88
6.03
9.19
6.89
7.50
7.04
5.17
6.99
3.28
5.17
5.70
4.98
5.38
7.60
5.85
9.94
7.30
7.90
6.45
6.38
5.46
7.91
What are the degrees of freedom for the Error?
Sham
PEMF 1h/day
PEMF 2h/day
PEMF 4h/day
4.51
5.32
4.73
7.03
7.95
6.00
5.81
4.65
4.97
5.12
5.69
6.65
3.00
7.08
3.86
5.49
7.97
5.48
4.06
6.98
2.23
6.52
6.56
4.85
3.95
4.09
8.34
7.26
5.64
6.28
3.01
5.92
9.35
7.77
6.71
5.58
6.52
5.68
6.51
7.91
4.96
8.47
1.70
4.90
6.10
4.58
5.89
4.54
7.19
4.11
6.55
8.18
4.03
5.72
5.34
5.42
2.72
5.91
5.88
6.03
9.19
6.89
7.50
7.04
5.17
6.99
3.28
5.17
5.70
4.98
5.38
7.60
5.85
9.94
7.30
7.90
6.45
6.38
5.46
7.91
Explanation / Answer
There are total 4 groups so k=4
Each group has 20 observations so total number of observations is N= 4*20 =80
Therefore dgree of freedom of error is : df = N-k =80-4=76
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.