It's believed that as many as $21 \%$ of adults over 50 never graduated from high school. We wish to see if this percentage is the same among the 25 to 30 age group. What sample size would allow us to increase our confidence level to $95 \%$ while reducing the margin of error to only $2 \%$ ?
$n=\square$ (Round up to the nearest integer.)
Answer
Since we need to round up to the nearest integer, the final sample size is \(\boxed{1594}\)
Step 1 ：We are given that 21% of adults over 50 never graduated from high school. We want to see if this percentage is the same among the 25 to 30 age group. We want to increase our confidence level to 95% and reduce the margin of error to only 2%.
Step 2 ：The formula for sample size in this case is: \(n = \frac{{Z^2 * p * (1-p)}}{{E^2}}\)
Step 3 ：Where: \(n\) is the sample size, \(Z\) is the z-score corresponding to the desired confidence level (for a 95% confidence level, \(Z = 1.96\)), \(p\) is the estimated population proportion (in this case, 21% or 0.21), \(E\) is the desired margin of error (in this case, 2% or 0.02).
Step 4 ：Substituting the given values into the formula, we get: \(n = \frac{{(1.96)^2 * 0.21 * (1-0.21)}}{{(0.02)^2}}\)
Step 5 ：Calculating the above expression, we get \(n = 1594\)
Step 6 ：Since we need to round up to the nearest integer, the final sample size is \(\boxed{1594}\)
