The Degree of freedom is 16 and we have an alpha of .05. Using Table C.2 we are given the critical value of 2.120. The value of .096 does not exceed the critical value of 2.120. Thus, we fail to reject (aka retain) the null hypothesis.

I am having a hard time writing my conclusion so far this is what i have

An independent-sample t test was used to determine if the sample means differ from one another. We fail to reject the null that the experimental group(athletes) and the group (non-athletes) mean were equal, t (16) =.096, p greater than .05. The experiment group(athletes) mean, 4.5, was scientifically higher than the group (non-athletes) mean, 1.7.