When Testing for a Difference Between the Means of a Treated

When testing for a difference between the means of a treated population and an untreated population,the computer display below is obtained.Explain what the P-value of 0.0768 means in this context. $$\begin{array} { | l | l | l | l | } \hline & \text { t-Test: Two Sample for Means } & & \\\hline 1 & & \text { Variable 1 } & \text { Variable } 2 \\\hline 2 & \text { Mean } & 65.10738 & 66.18251 \\\hline 3 & \text { Known V ariance } & 8.102938 & 10.27387 \\\hline 4 & \text { Observations } & 50 & 50 \\\hline 5 & \text { Hypothesized Mean Difference } & 0 & \\\hline 6 & \text { t } & - 1.773417 & \\\hline 7 & \text { P(T<=t) one-tail } & 0.0384 & \\\hline 8 & \text { TCritical one-tail } & 1.644853 & \\\hline 9 & \text { P(T<=t) two-tail } & 0.0768 & \\\hline 10 & \text { tCritical two-tail } & 1.959961 & \\\hline\end{array}$$