Question

Assume the population has a normal distribution. Using a margin of error E = 0.012, 93% confidence level and sample proportion unknown. Calculate the minimum sample size required to estimate the population proportion (must round up).HINT: 7687, 5537, 5688, 4685

A) What is your alpha (type I error)?

B) What is the critical Z-score (I hope you divided alpha by 2)?

C) Which formula did you use?

D) This type of problem has a special rule for rounding. What is the rule for rounding?

E) What is the minimum sample size required?

Answer 1

Given data,
Assume sample proportion is 50%
margin of error is 0.012
confidence level is 93%
sample size is minimum
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.07 is = 1.812
Sample Proportion = 0.5
ME = 0.012
n = ( 1.812 / 0.012 )^2 * 0.5*0.5
= 5700.25 ~ 5701          
A.
level of significance = alpha value = 1-0.93 =0.07
B.
Z score = 1.812
C.
formula
Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
D.
rule of rounding
Examples:
3.714 = 3 near to the digit
= 3.7 near to one digit
= 3.71 near to 2 digits.
E.
minimum sample size = 5701