sample size used?

Question

The poll conducted by media and other agencies use a 95% confidence.

Where n is the sample size. Suppose =0.36 is the proportion of those in the sample who approve of the way the President is handling the economy. If the margin of error is ±3%, the sample size used in the study was?

 

please help me and thank you for helping me get to the answer.

a-735

b-840

c-984

d-649

e-526

 

Answer 1

If the true proportion is 36% then for sample size n, we have the standard error is:

 

SE(p) = √[(0.36)(1-0.36)/n] = √(0.2304/n) = 0.48 / √n

 

Using the normal distribution approximation (CLT), the margin of error is

 

m_95% = 1.96SE = 0.9408 / √n

 

We want this to be 0.03 so

 

0.9408 / √n = 0.03

 

√n = 0.9408 / 0.03 = 31.36

 

n = 983.4496

 

We must round up, since rounding down gives slightly too much error.

 

So 984 people were sampled.