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

Question 1001735: Find the sample size.
Margin of error = $125, confidence level = 95%, standard deviation = $500
Answer by Boreal(15235) (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
RELATED QUESTIONS
FINDING SAMPLE SIZE. Q1. margin of error:$125, confidence level:95%,standard... (answered by stanbon)

Find the margin of error (E) of a confidence estimate interval, when the confidence level (answered by Boreal)

I am stuck on how to solve this. Use the given margin of error, confidence level, and... (answered by Theo)

Find the margin of error E that corresponds to the given statistics and confidence level: (answered by Boreal)

Use the given data to find the minimum sample size required to estimate the population... (answered by Boreal)

Determine the necessary sample size for a confidence interval around a sample mean: Set... (answered by Boreal)

Keep Confidence Level 90%. For Sample size 4, Sample Mean 26 and sample standard... (answered by stanbon)

A population mean is to be estimated from the sample described. Round your answer to one... (answered by Boreal)

Assuming a standard deviation of 6 hours, what is the required sample size if the error... (answered by stanbon)