How many observations for a $2 margin of error?

Question

You want to estimate a population mean to within a margin of error of $E = \$ 2 $with$ 95% $confidence. The population standard deviation is known to be$ \sigma = $16 $. Use$ z = 1.96$.

How many observations do you need?

Accepted Answer

A 95\% confidence interval for the mean reaches out about 1.96 standard errors on each side. You want that reach to be no more than \2: E = z \cdot \frac{\sigma}{\sqrt{n}}. Solve for n by rearranging. Multiply out and square both sides: n = \left(\frac{z\,\sigma}{E}ight)^2 = \left(\frac{1.96 	imes 16}{2}ight)^2 = \left(\frac{31.36}{2}ight)^2 = (15.68)^2 \approx 245.9. Since you cannot take a fraction of an observation and rounding down would miss the target, round up: n = 246. Takeaway. Required sample size is n = (z\sigma/E)^2, and you always round up: because n sits under a square ro…

How many observations for a $2 margin of error?

Your answer

More Statistics questions