# Table 3 Calculation of between-group effect sizes, based on information provided in eight randomized sham-controlled studies included in a previous review on pressure pain threshold changes after spinal manipulation

First author
Year
1) What is the number of participants of the experimental group (NE) and the control group (NC) and are they equal?
- NE = NC
NE ≠ NC
2) Are the standard deviations of the experimental group (SDE) and the control group (SDC) equal?
- SDE = SDC
- SDE ≠ SDC
3) Which type of effect size coefficient should be used?
- Cohen’s d coefficient (if NE = NC and SDE = SDC or SDE ≠ SDC)
- Hedges’ g coefficient (if NE ≠ NC and SDE ≠ SDC)
4) Give the equations that would be used:
- Effect size (d/g)
- Standard deviation of PPT values used to calculate the effect size (SD* or SD pooled)
- Standard deviation of the effect size (SD(d))
- Confidence interval of the effect size (95% CI)
What are the reported mean PPT values for the experimental group (ME) and for the control group (MC) with their standard deviation (+/− SD), at each follow-up time (units)? At each follow-up time, what are the:
- Effect size (d/g),
- Its standard deviation (SD(d))
- Its confidence interval
(95% CI)
- p value between groups
Effect sizes of clinical significant findings at each follow-up:
0.2 to 0.49 (small)
- 0.5 to 0.79 (medium)
- 0.8 to 1.00 (large)
Ruiz Saez
2007
1) NE = NC = 36
2) SDE ≠ SDC
3) Cohen’s d coefficient
4)
-$$d=\frac{M_E-{M}_C}{SD^{\ast }}$$
- $$\mathrm{SD}\ast =\sqrt{\frac{{SD_E}^2+{SD_C}^2}{2}}$$
- $$\mathrm{SD}(d)=\sqrt{\frac{N_E+{N}_C}{N_E\times {N}_C}+\frac{d^2}{2\left({N}_E+{N}_C\right)}}$$
- 95% CI: [d − 1.96 × SD (d); d + 1.96 × SD(d)]
-T0:
ME = + 1.35 +/−  0.5 Kg/cm2
MC = + 1.27 +/−  0.4 Kg/cm2
-T + 5:
ME = + 1.38 +/−  0.5 Kg/cm2
MC = +  1.15 +/−  0.4 Kg/cm2
-T + 10:
ME = + 1.39 +/−  0.5 Kg/cm2
MC = + 1.1 +/−  0.5 Kg/cm2
-T0:
-d = 0.17
-SD(d) = 0.24
- 95% CI: [− 0.29; + 0.63]
- p value: NS
-T + 5:
-d = 0.51
-SD(d) = 0.24
- 95% CI: [+ 0.04; + 0.98]
- p < 0.01
-T + 10:
-d = 0.58
-SD(d) = 0.24
- 95% CI: [+ 0.11; +  1.05]
-p < 0.01
T0: small
T + 5: medium
T + 10: medium
Srbely
2013
1) NE = NC = 18
2) SDE ≠ SDC
3) Cohen’s d coefficient
4)
-$$d=\frac{M_E-{M}_C}{SD^{\ast }}$$
-$$\mathrm{SD}\ast =\sqrt{\frac{{SD_E}^2+{SD_C}^2}{2}}$$
-$$\mathrm{SD}(d)=\sqrt{\frac{N_E+{N}_C}{N_E\times {N}_C}+\frac{d^2}{2\left({N}_E+{N}_C\right)}}$$
- 95% CI: [d − 1.96 × SD (d); d + 1.96 × SD(d)]
-T + 1:
ME = + 34.4 +/−  9.6 N
MC = +  30.7 +/−  7.5 N
-T + 5:
ME = + 37.5 +/−  11.9 N
MC = + 28.7 +/−  6.0 N
-T + 10:
ME = + 37.9 +/−  14.4 N
MC = + 28.9 +/−  6.3 N
-T + 15:
ME = + 34.3 +/−  11.5 N
MC = + 28.6 +/− 7.0 N
-T + 1:
-d = 0.42
-SD(d) = 0.34
- 95% CI: [− 0.24; + 1.08]
-p < 0.01
-T + 5:
-d = 0.93
-SD(d) = 0.35
− 95% CI: [+ 0.24; + 1.62]
-p < 0.01
-T + 10:
-d = 0.80
-SD(d) = 0.35
- 95% CI: [+ 0.12; + 1.48]
-p < 0.01
-T + 15:
-d = 0.59
-SD(d) = 0.34
- 95% CI: [−  0.08; + 1.26]
-p < 0.01
T + 1: small
T + 5: large
T + 10: large
T + 15: medium
Fernandez de la Penas
2008
1) NE = NC = 10
2) SDE ≠ SDC
3) Cohen’s d coefficient
4)
- $$d=\frac{M_E-{M}_C}{SD^{\ast }}$$
- $$\mathrm{SD}\ast =\sqrt{\frac{{SD_E}^2+{SD_C}^2}{2}}$$
- $$\mathrm{SD}(d)=\sqrt{\frac{N_E+{N}_C}{N_E\times {N}_C}+\frac{d^2}{2\left({N}_E+{N}_C\right)}}$$
- 95% CI: [d − 1.96 × SD (d); d + 1.96 × SD(d)]
-T + 5:
(dominant side/dominant side)
ME = + 387.6 +/−  70.9 kPa/s
MC = + 312.3 +/−  47.7 kPa/s
-T + 5:
-d = 1.24
-SD(d) = 0.49
- 95% CI: [+ 0.28; + 2.20]
-p < 0.05
T + 5: large
Fernandez de la Penas
2007
1)NE = NC = 15
2) SDE ≠ SDC
3) Cohen’s d coefficient
4)
- d = $$\frac{M_E-{M}_C}{SD^{\ast }}$$
- SD*= $$\sqrt{\frac{{SD_E}^2+{SD_C}^2}{2}}$$
- SD(d)=$$\sqrt{\frac{N_E+{N}_C}{N_E\times {N}_C}+\frac{d^2}{2\left({N}_E+{N}_C\right)}}$$
- 95% CI: [d − 1.96 × SD (d); d + 1.96 × SD(d)]
-T + 5:
ME = + 2.9+/− 0.6 Kg/cm2
MC = + 2.3+/− 0.5 Kg/cm2
-T + 5:
-d = 1.08
-SD(d) = 0.48
− 95% CI: [+ 0.14; + 2.02]
-p < 0.01
T + 5: large
Hamilton
2007
1) NE ≠ NC
- NE = 35
- NC = 25
2) SDE ≠ SDC
3) Hedge’ g coefficient
4)
$${\mathrm{SD}}_{\mathrm{pooled}}=\sqrt{\frac{\left({\mathrm{N}}_{\mathrm{E}}\hbox{-} 1\right){\mathrm{SD}}_{{\mathrm{E}}^2}+\left({\mathrm{N}}_{\mathrm{C}}\hbox{-} 1\right){\mathrm{SD}}_{{\mathrm{C}}^2}}{{\mathrm{N}}_{\mathrm{E}}+{\mathrm{N}}_{\mathrm{C}}\hbox{-} 2}}$$
- $$\boldsymbol{g}=\frac{{\boldsymbol{M}}_{\boldsymbol{E}}-{\boldsymbol{M}}_{\boldsymbol{C}}}{{\boldsymbol{SD}}_{\boldsymbol{Pooled}}}$$
-$$\mathrm{SD}\left(\mathrm{g}\right)=\sqrt{\frac{N_E+{N}_C}{N_E\times {N}_C}+\frac{d^2}{2\left({N}_E+{N}_C\right)}}$$
- 95% CI: [d − 1.96 × SD (d); d + 1.96 × SD(d)]
-T + 5:
ME = +  398.06 +/−  133.51 kPa/s
MC = + 368.44 +/−  208.16 kPa/s
-T + 30:
ME = + 374.58 +/−  127.50 kPa/s
MC = + 368.68 +/−  192.62 kPa/s
-T + 5:
-g = 0.17
-SD(g) = 0.26
− 95% CI: [− 0.34; +  0.68]
-p < 0.01
-T + 30:
-g = 0.03
-SD(g) = 0.26
− 95% CI: [− 0.48; + 0.54]
-p value: NS
T + 5: small
T + 30: small
Yu
2012
1) NE = NC = 30
2) SDE ≠ SDC
3) Cohen’s d coefficient
4)
- d = $$\frac{M_E-{M}_C}{SD^{\ast }}$$
- SD*= $$\sqrt{\frac{{SD_E}^2+{SD_C}^2}{2}}$$
- SD(d)=$$\sqrt{\frac{N_E+{N}_C}{N_E\times {N}_C}+\frac{d^2}{2\left({N}_E+{N}_C\right)}}$$
- 95% CI: [d − 1.96 × SD (d); d + 1.96 × SD(d)]
-L5-S1 PD side
T0:
ME = + 5.64+/− 1.13 Kg/cm2
MC = + 4.85+/− 1.12 Kg/cm2
-L5-S1 OPD side
T0:
ME = + 5.56+/− 1.17 Kg/cm2
MC = + 4.91+/− 1.13 Kg/cm2
-L5 dermatome PD side
T0:
ME = + 4.77+/− 0.96 Kg/cm2
MC = + 4.14+/− 1.13 Kg/cm2
-L5 dermatome OPD side
T0:
ME = + 4.63+/− 0.95 Kg/cm2
MC = + 4.09+/− 0.82 Kg/cm2
-L5-S1 PD side
T0
-d = 0.70
-SD(d) = 0.27
− 95% CI: [+ 0.18; +  1.22]
-p < 0.05
-L5-S1 OPD side
T0
-d = 0.56
-SD(d) = 0.26
− 95% CI: [+ 0.04; +  1.08]
-p < 0.05
- L5 dermatome PD side
T0
-d = 0.60
-SD(d) = 0.26
− 95% CI: [+ 0.08; +  1.12]
-p < 0.05
- L5dermatome OPD side
T0
-d = 0.60
-SD(d) = 0.26
− 95% CI: [+ 0.08; +  1.12]
-p < 0.05
-L5-S1 PD side
T0: medium
-L5-S1 OPD side
T0: medium
- L5 dermatome PD side
T0: medium
- L5dermatome OPD side
T0: medium
Thomson
2009
1) NE ≠ NC
- NE = 19
- NC = 13
2) SDE ≠ SDC
3) Hedge’g coefficient
4)
- g = $$\frac{M_E-{M}_C}{SD_{Pooled}}$$
$${\mathrm{SD}}_{\mathrm{pooled}}=\sqrt{\frac{\left({\mathrm{N}}_{\mathrm{E}}\hbox{-} 1\right){\mathrm{SD}}_{{\mathrm{E}}^2}+\left({\mathrm{N}}_{\mathrm{C}}\hbox{-} 1\right){\mathrm{SD}}_{{\mathrm{C}}^2}}{{\mathrm{N}}_{\mathrm{E}}+{\mathrm{N}}_{\mathrm{C}}\hbox{-} 2}}$$
-$$\mathrm{SD}\left(\mathrm{g}\right)=\sqrt{\frac{N_E+{N}_C}{N_E\times {N}_C}+\frac{d^2}{2\left({N}_E+{N}_C\right)}}$$
- 95% CI: [d − 1.96 × SD (d); d + 1.96 × SD(d)]
(Approximate data)
-T0:
ME = + 2.2 +/−  1.1 Kg/cm2
MC = + 2.1 +/−  0.8 Kg/cm2
-T0:
-g = 0.10
-SD(g) = 0.36
− 95% CI: [− 0.61; + 0.81]
-p value: NS
T0: small
Fryer
2004
1) NE = NC = 32
2) SDE ≠ SDC
3) Cohen’s d coefficient
4)
-$$d=\frac{M_E-{M}_C}{SD^{\ast }}$$
-$$\mathrm{SD}\ast =\sqrt{\frac{{SD_E}^2+{SD_C}^2}{2}}$$
-$$\mathrm{SD}\left(\mathrm{g}\right)=\sqrt{\frac{N_E+{N}_C}{N_E\times {N}_C}+\frac{d^2}{2\left({N}_E+{N}_C\right)}}$$
- 95% CI: [d − 1.96 × SD (d); d + 1.96 × SD(d)]
-T0:
ME = + 216.51 +/−  90.50 kPa
MC = + 244.64 +/−  91.59 kPa
-T0:
-d = 0.30
-SD(d) = 0.25
− 95% CI: [− 0.19; + 0.79]
-p value: NS
T0: medium
1. NS: not significant; T0: Values at baseline; T + 1: Values after one minute; T + 5: Values after five minutes; T + 10: Values after ten minutes; T + 15: Values after fifteen minutes; T + 30: Values after thirty minutes; PD: Pelvic Deficiency; OPD: Opposite Pelvic Deficiency