It's believed that as many as $19 \%$ 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.
a) How many of this younger age group must we survey in order to estimate the proportion of non-grads to within $6 \%$ with $90 \%$ confidence?
$n=\square$ (Round up to the nearest integer.)

Step 1 ：We are given that as many as $19 \%$ 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.

Step 2 ：We want to estimate the proportion of non-grads to within $6 \%$ with $90 \%$ confidence. This means we want a margin of error of $6 \%$ and a Z-score of $1.645$ (which corresponds to a $90 \%$ confidence level).

Step 3 ：We can use the formula for sample size in a proportion estimation: \(n = \frac{{Z^2 \cdot p \cdot (1-p)}}{{E^2}}\), where \(Z\) is the Z-score, \(p\) is the estimated proportion, and \(E\) is the desired margin of error.

Step 4 ：Substituting the given values into the formula, we get \(n = \frac{{(1.645)^2 \cdot 0.19 \cdot (1-0.19)}}{{(0.06)^2}}\).

Step 5 ：Calculating the above expression, we find that \(n \approx 115.5\).

Step 6 ：Since we can't have a fraction of a person, we round up to the nearest whole number to get \(n = 116\).

Step 7 ：So, the required sample size to estimate the proportion of non-grads to within $6 \%$ with $90 \%$ confidence is \(\boxed{116}\).