I need to calculate sample size $n$, so that $SE$ of the mean would be 4 points. There are $55$ units in total in the $I$ group. $N_2 = 80$ in the $II$ group and $N_3=65$ in the $III$ group. The sample strata sizes are approximately proportional to their proportion in population, i.e., $\frac{n_1}{N_1} \approx \frac{n_2}{N_2} \approx \frac{n_3}{N_3} \approx \frac{1}{4}$. Here is the data:
$I $ group - 80 ,68, 72, 85, 90, 62, 61, 92, 85, 87, 91, 81, 79, 83,
$II$ group - 85, 82, 48, 75, 53, 73, 65, 78, 49, 69, 72, 81, 53, 59, 68, 52, 71, 61, 59, 42,
$III$ group - 42, 32 ,36 ,31 ,65 ,29 ,43, 19, 53, 14, 61, 31, 42, 30, 39, 32.
Calculated mean for the stratified sample: $\bar{y}_{str} = 59. 989$. The st. dev. for the first group sd($I$) = $10.254$, sd($II$) = $12.578$, and sd($III$) = $14.643$. I need to calculate the sample size so that the standard error of the mean would equal $4$ points. I need to use proportional allocation. The confidence level is 95%.
How to do this?