Answer:
The p-value of the test is 0.0853 > 0.05, which means that there is not enough evidence to reject the manufacturer's claim based on this observation.
Step-by-step explanation:
A manufacturer of nails claims that only 4% of its nails are defective.
At the null hypothesis, we test if the proportion is of 4%, that is:

At the alternative hypothesis, we test if the proportion is more than 4%, that is:

The test statistic is:

In which X is the sample mean,
is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
4% is tested at the null hypothesis
This means that 
A random sample of 20 nails is selected, and it is found that two of them, 10%, are defective.
This means that 
Value of the test statistic:



P-value of the test and decision:
Considering an standard significance level of 0.05.
The p-value of the test is the probability of finding a sample proportion above 0.1, which is 1 subtracted by the p-value of z = 1.37.
Looking at the z-table, z = 1.37 has a p-value of 0.9147
1 - 0.9147 = 0.0853
The p-value of the test is 0.0853 > 0.05, which means that there is not enough evidence to reject the manufacturer's claim based on this observation.