P-values : Hypothesis test 
My try:

Is it correct?
The reason I am asking this because , when I tried to calculate the value of t by std . dev and mean , I got a different value from 1.9527887

If "mean" is taken to be the mean of $15$ differences between the $15$ measured cut-on voltages and the hypothetical average of $0.6$, then my software agrees with the output in the posted question above.
However, in that case I would say the problem is not expressed clearly.
One should have
$$
T = \frac{\overline X - 0.6}{s/\sqrt{15}}.
$$
If $\overline X-0.6 = 0.0453333$ then this is
$$
T = \frac{0.0453333}{0.08991/\sqrt{15}} = 1.952787
$$
which agrees with what is reported.  Then one finds that $\Pr(|T| > 1.962787) = 0.07113305$ where $T$ has a t-distribution with $14=15-1$ degrees of freedom, and that is greater than $0.01$, so at the $0.01$ level one does not reject the null hypothesis.
My software reports $\Pr(T>1.952782) = 0.03556652$, and that is what must be multiplied by $2$ to get $0.07113305$.  You shouldn't be multipllying $0.07113305$ by $2$.
But I would would have thought that the output meant $\overline X = 0.0453333$ rather than $\overline X - 0.6 = 0.0453333$.  That is what is confusingly stated.
