SOLUTION: Find the sample size. Margin of error = $125, confidence level = 95%, standard deviation = $500

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Question 1001735: Find the sample size.
Margin of error = $125, confidence level = 95%, standard deviation = $500
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
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