Answer:
Option C: fail to reject the null hypothesis; accept the shipment.
Explanation:
The question is incomplete. It should say:
"A port inspector working for a trading company needs to decide whether to accept a large shipment of products of a certain foreign country. He will decline the shipment if there is compelling evidence that the proportion of the defective products is more than 5 percent (0.05). Suppose that a random sample of 1000 products from this shipment has yielded p=0.06. At the 5 % level of significance, what decision should the port inspector make?
A) reject the null hypothesis; reject the shipment
B) reject the null hypothesis; accept the shipment
C) fail to reject the null hypothesis; accept the shipment
D) fail to reject the null hypothesis; reject the shipment"
We have to perform an hypothesis test on proportion, with a sginificance level of 0.05.
The null and alternative hypothesis are:

The standard deviation is

Now we can calculate the z value:

For a one-side test with z=1.357, the P-value is 0.08739.
The P-value is greater than the significance level, so the null hypothesis can not be rejected.
The inspector has no compelling evidence that the proportion of defective products is more than 5%, so he has to accept the shipment.