Eye-Tracking Training Improves Inhibitory Control in Children with Attention-Deficit/Hyperactivity Disorder

Abstract

Disinhibition is a common sign among children with attention-deficit/hyperactivity disorder (ADHD). The present study examined the effect of computerized eye-tracking training to improve inhibitory control in ADHD children. Thirty-two ADHD children (mean age = 8.4 years) were recruited. Half of the participants underwent 240 min of eye-tracking training over two weeks (i.e., experimental group), while the other half did not receive any training (i.e., control group). After training, the experimental group exhibited significant improvements in neuropsychological tests of inhibition, such as faster reaction time in the incongruent condition of the Flanker test, more unique designs in the Category Fluency and Five-Point Tests, and a faster completion time in Trail 2 of the Children's Color Trail Test. The control group did not show significant changes in any of these tests. Our findings support the use of eye-tracking training to improve the inhibitory control of ADHD children.

Keywords: ADHD; cognitive training; eye-tracking; inhibition; mental flexibility.

Figure 1

(a) Inhibitory control index. (b) Mean reaction time in the experimental (n = 16) and control groups (n = 16). “Pre” and “Post” refer to the test performance of participants before and after real training/no training period, respectively. Error bars represent 1 standard error ± the mean. Asterisks indicate the level of significance of t-tests (two-tailed). * p < 0.05, ** p < 0.01.

Figure 2

Performance in the (a) Category Fluency Test and Five-Point Test, in addition to the (b) Children’s Color Trails Test (CCTT) in the experimental (n = 16) and control (n = 16) groups. “Pre” and “Post” refer to the test performance of participants before and after real training/no training period, respectively. Error bars represent 1 standard error ± the mean. Asterisks indicate the level of significance of paired t-tests (two-tailed). ** p < 0.01, * p < 0.05, ⁺ p = 0.05.