Suppose we're testing the hypothesis that p=0.6 against the hypothesis that p>0.6.
If we have a sample of 100 and we have 70 successes, so p hat, the sample proportion, is 0.7, and we want to know the p-value of this result, we are effectively working out P(X>=70) if Ho is true.
Under Ho, X is distributed bin (n=100, p=0.6) and approx normal (np=60, npq= 24). Using the continuity correction (we want to include 70 so we use 69.5):
P(X>=70)=P(Z>(69.5-60)/sq rt (24))=P(Z>1.94)=0.026
Alternatively, under Ho, p hat is distributed approx normal (p=0.6, sq rt pq/n = sq rt (0.6x0.4/100)), so, using the continuity correction:
P(phat >=0.7)=P(Z>(0.695-0.6)/sq rt (0.6x0.4/100))= P(Z>1.94)=0.026
Notice that 70 is above the mean of 60 (and 0.7 is above the mean of 0.6) so we correct downwards towards the mean. If the sample result had been 30, we would have wanted P(X<=30) which would mean finding P(X<30.5) because we want to include 30, so we correct upwards towards the mean.
You do this even if you aren't asked for the p-value. For example, the test statistic in the example above would be 1.94, which you would then compare with the critical value (5%) of 1.6449.
