The sample size is 150. If the sample size is greater than 30 we consider it an appropriate large sample size and can consider the population to be normally distributed. Since sample size is quite larger than 150, we assume that the sample is from the population which is normally distributed.
We are given the sample proportion p = 0.56
Sample Size = n = 150
We have to construct 95% confidence interval about the sample proportion. The z value corresponding to 95% confidence interval is 1.96. The formula to calculate the margin of error (E) is:
E= ±

Using the values in the above formula, we get:
E =±

E = ± 0.079
Therefore, the correct option is:
A. As the sample size is appropriately large, the margin of error is ±0.079.