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I've recently come across an old Excel sheet that is used to help assess whether someone has collected enough samples when performing a time study (essentially trying to figure out the distribution of time required to complete a task). In the data column (labeled "time"), the person performing the time study enters the length of time it took to complete a task, with each row representing one trial. The sheet contains the following formula, labeled "Current % Error":

\begin{equation}\frac{t \sigma}{\mu \sqrt{n-1}}\end{equation}

where $t$ is a t-score which varies based on sample size, $\mu$ is sample average, and $\sigma$ is sample standard deviation.

My question is, what algorithm is this? I know that we can measure confidence intervals using the similar equation $\mu \pm \frac{t \sigma}{\sqrt{n-1}}$, but I'm not sure what's being calculated when we divide by the mean. What does this correspond to, if anything?

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  • $\begingroup$ The spreadsheet has two apparent (small) errors: the $\sqrt{n-1}$ should be $\sqrt{n}$ and the critical $t$ value should be $1.656.$ $\endgroup$
    – whuber ♦
    Commented Dec 12, 2013 at 21:35

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The clue as to what is intended (even though there are some errors) is right there in the label.

$\frac{\sigma}{\sqrt{n}}$ is the standard error of the mean.

$\frac{t \sigma}{\sqrt{n}}$ is then the half-interval width.

The ratio of that to $\mu$ is the interval as a proportion of the mean.

If you wanted to specify the mean within a certain percentage error (at some confidence level), you specify that percentage and you can figure out a sample size from it.

However, the phrasing of things in the sheet suggests that the percentage error and the confidence level are being conflated; they're two different things and needn't have the same value.

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  • $\begingroup$ You wrote: "$\frac{t \sigma}{\mu \sqrt{n}}$ is then the half-interval width". Did you mean "$\frac{t \sigma}{\sqrt{n}}$"? As you state in the following sentence, the ratio of that to $\mu$ is the interval as a proportion of the mean. I just want to make sure I didn't miss something. $\endgroup$
    – eykanal
    Commented Dec 13, 2013 at 20:17
  • $\begingroup$ @eykanal you're right, that's caused by editing it from saying something else and not doing it properly; now fixed. $\endgroup$
    – Glen_b
    Commented Dec 13, 2013 at 23:38

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