Impact of Six Weeks of Aerobic Training on School-Age Children's Body Mass Index
Abstract
The purpose of this study was to find out the effect of six weeks Aerobic training on body mass index of school going children.
For this purpose forty male Aerobic trainers of kamargaon were selected as subjects, the age of the subjects was ranged between 14 to 16 years age.
All the subjects were equally distributed in equal number into two homogeneous groups and named as experimental (A) and control (b) group. The experimental group had undergone the training programme for a period of six weeks.
The data pertaining to the study collected on the selected subject’s just height and weight for body mass index were recorded before and at the end of six weeks experimental treatment.
To find out the training effect‘t’-test was employed to determine the significant mean difference between the pre-test and post test scores of both the groups.
The data were further treated with‘t’-test to find out the significant difference in between the post test scores of experimental and control group .
The level of significance was set at 0.05 level of confidence the difference between the pre-test and post –test means of control group did not show any significant difference as the calculate‘t’ value of 0.05 was lesser than the tabulated t-value of 2.093 at 0.05 level of confidence.
The difference between the pre-test and post test means of experimental group was not significant at 0.05 level as the calculated t-value of 0.26 was lesser than the tabulated t0.05(19) = 2.093.
The finding of table 3 revealed that there was insignificant difference in between the post tests of experimental group and control group. Because the obtained‘t’- ratio value of 0.98 was lesser than the tabulated‘t’ value of t0.05(18)=2.093.
The difference between the pre-test mans of control group and post test means of experimental group also did not show statistically significant ally , because obtained ‘t’- ratio value of 1.341 was lesser than the tabulated ‘t’ value of t0.05(18)= 2.093 .