Sample size for a target interval width

Question

A 95% confidence interval for a mean has a full width of 8 at a sample size of n = 64. You want the width to be 5, keeping the spread and confidence level fixed. What sample size is required?

Accepted Answer

Setup. Width, like the margin, scales as 1/\sqrt{n}. So \frac{w{	ext{new}}}{w{	ext{old}}} = \sqrt{\frac{n{	ext{old}}}{n{	ext{new}}}} \quad\Rightarrow\quad n{	ext{new}} = n{	ext{old}}\left(\frac{w{	ext{old}}}{w{	ext{new}}}ight)^{2}. Compute. n{	ext{new}} = 64 	imes \left(\frac{8}{5}ight)^{2} = 64 	imes 2.56 = 163.84. Round up: n = 164. Interpret. Because the target width does not divide evenly, round up so the width is at most 5. Takeaway. To reach a target width, scale the sample by the square of the width ratio and round up. Same square-root law, whether you speak in half-widt…

Sample size for a target interval width

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