In: Statistics and Probability
It is believed that 1/4 of adults over 50 years old graduated from high school. We want to see if the percentage is the same for 30 year olds. How many should we survey in order to estimate the proportion of non-grads within 0.06 with 90 % confidence? Suppose we have no idea of the % of non-grads, what would be the "n" value?
Solution,
Given that,
= 0.25
1 - = 1 - 0.25 = 0.75
margin of error = E = 0.06
At 90% confidence level
= 1 - 90%
= 1 - 0.90 =0.10
/2
= 0.05
Z/2
= Z0.05 = 1.645
sample size = n = (Z / 2 / E )2 * * (1 - )
= (1.645 / 0.06 )2 * 0.25 * 0.75
= 140.93
sample size = n = 141