A survey of 78 students found that 34% were in favor of raising tuition to pave new parking lots. The standard deviation of the sample proportion is 5.2%. How large a sample (to the nearest person) would be required to reduce this standard deviation to 3.6%?
Answer:
Consider,
SE = sqrt(pq/n)
substitute values
0.036 = sqrt(0.34*(1-0.34)/n)
0.036^2 = 0.34*0.66/n
n = 0.34*0.66/0.036^2
= 173.148
sample size n = 173
A survey of 78 students found that 34% were in favor of raising tuition to pave...