Table 2 One-way repeated measures ANOVA results between groups at each treadmill speed are presented
Speed (km/h) | Application | ANOVA | ||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
Moves | Accupedo | Runtastic | Yamax | True count | ||||||||||
M (SD) | Post-hoc | Zeros recorded | M (SD) | Post-hoc | Zeros recorded | M (SD) | Post-hoc | Zeros recorded | M (SD) | Post-hoc | Zeros recorded | M (SD) | ||
2 | 16.30* (33.35) | t(9) = −5.64 p < 0.001 | 7 | 27.6* (31.41) | t(9) = −4.67 p = 0.001 | 5 | 30.80* (29.30) | t(9) = −5.50 p < 0.001 | 0 | 54.40* (24.89) | t(9) = −2.916 p < 0.017 | 0 | 75.80 (9.53) | F(4,36) = 8.72 p < 0.001 |
3 | 24.64* (36.11) | t(10) = −6.51 p < 0.001 | 6 | 21.55* (23.45) | t(10) = −9.61 p < 0.001 | 4 | 70.64 (42.00) | t(10) = −2.06 p = 0.07 | 1 | 80.73 (25.88) | t(10) = −1.94 p = 0.08 | 0 | 94.73 (8.09) | F(4,40) = 14.55 p < 0.001 |
4.5 | 70.09 (59.79) | 4 | 76.55 (47.60) | 2 | 97.00 (53.93) | 0 | 104.82 (44.27) | 1 | 119.64 (11.41) | F(4,40) = 2.44 P = 0.06 | ||||
6 | 28.00* (45.55) | t(10) = −8.88 p < 0.001 | 6 | 103.18 (56.89) | t(10) = −1.90 p = 0.09 | 1 | 119.00 (55.36) | t(10) = −0.97 p = 0.36 | 0 | 138.09 (17.33) | t(10) = 0.65 p = 0.53 | 0 | 134.91 (10.38) | F(4,40) = 18.86 p < 0.001 |
8 | 80.20 * (74.59) | t(10) = −3.20 p = 0.01 | 3 | 148.36 (27.26) | t(10) = −0.06 p = 0.96 | 0 | 143.82 (53.37) | t(10) = −0.32 p = 0.75 | 0 | 165.91* (14.80) | t(10) = 2.84 p = 0.02 | 0 | 148.82 (13.92) | F(2.0,19.7) = 7.09 p = 0.005 |
9 | 80.09* (75.79) | t(10) = −3.94 p = 0.003 | 3 | 156.36 (28.21) | t(10) = −0.79 p = 0.45 | 0 | 168.27 (23.39) | t(10) = −0.79 p = 0.45 | 0 | 170.36* (12.23) | t(10) = 2.39 p = 0.04 | 0 | 162.55 (15.42) | F(1.4,14.3) = 12.56 p = 0.002 |
10 | 79.18* (71.88) | t(10) = −4.68 p = 0.001 | 1 | 167.82 (55.30) | t(10) = 0.24 p = 0.81 | 0 | 147.82 (57.70) | t(10) = −0.85 p = 0.81 | 0 | 174.18* (12.58) | t(10) = 2.31 p = 0.04 | 0 | 163.18 (19.19) | F(4,40) = 7.36 p < 0.001 |
11 | 130.90 (101.50) | 3 | 166.10 (27.01) | 0 | 144.60 (49.16) | 0 | 175.50 (14.49) | 0 | 162.70 (25.10) | F(1.5,13.5) = 1.218 p = 0.32 | ||||