Question

Based on historical data in Oxnard college, we believe that 45% of freshmen do not visit their counselors regularly. For this year, you would like to obtain a new sample to estimate the proportiton of freshmen who do not visit their counselors regularly. You would like to be 98% confident that your estimate is within 3% of the true population proportion. How large of a sample size is required? Do not round mid-calculation.

Answer 1

Solution:
Given in the question
P(freshmen do not visit their counselors regularly) = 0.45
Confidence level = 0.98
level of significance($\alpha$) = 1 - confidence level = 1 - 0.98 = 0.02
$\alpha$ /2 = 0.02/2 = 0.01
From Z table we found Zalpha/2 = 2.32635
Margin of error (E) = 0.03
Sample size required can be calculated as
Sample size(n) = (Z$\alpha$/2 /E)^2 * (p*(1-p)) = (2.32635/0.03)^2 * (0.45*(1-0.45)) = (77.575)^2 * 0.2475 = 1488.274 or 1489
So sample size required = 1489