Question 1001735
Assuming I can use a normal distribution, z*SE=half width of confidence interval
125=1.96*(s/sqrt(n))
125sqrt(n)=1.96*(500)=980
sqrt(n)=980/125
n=(980/125)^2
61.47 rounded upward to 62.
I could go back to the t-table, where t df=60(0.95)=2
That would give a sample size of 64 (1000/125)^2