I am trying to compute the margin of error of a mean of a group.

Suppose a have a population of 50 workers and I want to know how much I'm paying as a whole in salary for each sale they perform (salary per sale). The problem is I only have data from four workers (simply, randomly chosen) and I want to use the data from these four to get an estimate from all 50. Here is the data:

Worker 1 = 50k , 1 sale -> 50k per sale;

Worker 2 = 100k , 3 sales -> 33k per sale;

Worker 3 = 150k, 5 sales -> 30k per sale;

Worker 4 = 75k, 0 sales -> n/a;

Total= 375k, 9 sales -> 42k per sale

I also have the salary data for the other 96 workers (just not their number of sales).

I am able to get a $42 per sale estimate, but I am having trouble calculating the margin of error of that estimate because I need an salary per sale estimate for worker #4 to calculate the standard deviation and standard error (via the MOE formula for mean).

Does anyone know how I can calculate the MOE for the salary per sale? FYI: in my real situation my n is bigger and N bigger than 50 but I wanted to simplify things for illustration.

Thanks in advance!

  • $\begingroup$ In this situation it is plausible to suppose that (a) you could look up the salaries of all workers in the population and (b) you have access to data about all nine individual sales. Would this be the case? $\endgroup$ – whuber Dec 11 '14 at 3:13
  • 1
    $\begingroup$ Yes, I do have the salaries of all 100 workers. I just don't have the sales of all 100 workers (only four of them). And, yes, I do have the data about the nine individual sales from the four workers. $\endgroup$ – Rob Dec 11 '14 at 20:06
  • $\begingroup$ Then please, Rob, edit your question to include that information: you are likely to get better solutions that way. $\endgroup$ – whuber Dec 11 '14 at 20:25

The good news: the mean that you want to calculate is actually a ratio, and ratio estimation is a well studied topic in survey sampling. The bad news is that with only four observations, no probabilistically valid margin of error is possible.

To compute a margin of error (bound on error), one starts with a standard error (SE). There are at least three methods for computing a standard error for a ratio estimate. None of the solutions will be good with only four observations. Not only is the standard error likely to be poor, but the estimate itself will be biased (Cochran, 1977, p. 160).

A margin of error adds a multiplier to the standard error, so that the interval $\pm$ K SE) hopefully has good coverage probabilities, e.g. fails to cover the true value only 5% of the time. This is effectively a "confidence interval". However, the the actual distribution of a sample ratio with small $n$ islikely to be asymmetric and non-normal. So, even when $n$ is sufficiently large to get a good standard error, a larger sample size is needed to get a good interval:

The first step towards getting a defensible margin of error is therefore: Take a larger sample.

In what follows below, I consider just simple random sampling, but other sampling designs might be better, given that you know salaries for every worker (Cochran, p. 172).

Some details:

Denote $x_i$ = salary for sampled worker $i$ and $y_i$ = number of sales. Then $\overline{x}$ and $\overline{y}$ are the sample means for the $n$ = 4 observations. If $X$ and $Y$ are the totals in the population (you know $X$), you want to estimate the average salary per sale or:

$$ R = \frac{X}{Y} $$

the ratio of sample means The natural estimate of $R$ is ratio of sample totals, which is the same as the ratio of sample means

$$ \widehat{R} = \frac{x}{y} = \frac{\overline{x}}{\overline{y}} $$

In your case $\widehat{R}= 375/9= 41.67$, as you found. With a small sample size, this is likely to be seriously biased.

Solutions for standard error.

From your question, you appear to think that computation of a standard error involves the individual worker ratios $x_i/y_i$. This is not so for the methods below.

Solution 1. Linearization

The approximate variance of $\widehat{R}= \frac{x}{y} = \frac{\overline{x}}{\overline{y}}$ is:

\begin{equation} \verb+var+ \widehat{R} = \left(1- \frac{n}{N}\right)\frac{\sum_{i=1}^n (y_i - \widehat{R}x_i)^2}{n (n-1)\overline{y}^2} \end{equation}

If the population mean $\overline{Y}$ is known, it should be substituted for $\overline{y}$.

This method requires "large" n (Cochran, 1977, pp; 32; 160-163).

Solution 2: Jackknife

The jackknife is one of several methods intended to reduce small sample bias (Cochran, p. 174; 178).

Solution 3: Bootstrap

The bootstrap (Efron, 1986) is another approach to getting a non-parametric standard error (Lohr, 2007, p 384). It works by re-sampling from the sample of four with replacement. There are a total of $4^4= 64$ possible samples, but in over 20% of them, at most two of the original four workers will appear. **Added:**This may well be the best of the three methods (and there are others; see Cochran's book), but, to repeat, none will work well with only four observations.


Cochran, William G. 1977. Sampling techniques. New York: Wiley. Wolter, Kirk M. 2007. Introduction to variance estimation. New York: Springer.

Efron, B., and R. Tibshirani. 1986. Bootstrap methods for standard errors, confidence intervals, and other measures of statistical accuracy. Statistical science 1, no. 1: 54-75.

  • 1
    $\begingroup$ This is a very good and detailed answer. I proposed an edit to the sampling variance estimator, but @gung pointed out that I should leave a comment mentioning this and let the answerer do that (Sorry, this is definitely more courteous and it is probably in some SE or CV guideline that I missed). Nonetheless, my edit was approved, but for some reason there is a term missing on the denominator ($n \overline{x}^2 $) which I had included in my edit. I will not edit this further unless recommended otherwise. Sorry about the confusion! $\endgroup$ – Raphael Nishimura Dec 15 '14 at 3:25
  • $\begingroup$ Thanks very much, Raphael. Fixed now. For some reason my last edit didn't take either. I had an additional sentence at the end, which I've also inserted. $\endgroup$ – Steve Samuels Dec 16 '14 at 3:32
  • 1
    $\begingroup$ Actually, because you defined the ratio estimator as $\widehat{R} = \frac{x}{y}$ (instead of the standard $\widehat{R} = \frac{y}{x}$), I think its Taylor Series sampling variance estimator should be $var( \widehat{R}) = \left(1- \frac{n}{N}\right) \frac{1}{n \overline{y}^2} \frac{\sum_{i=1}^n (x_i - \widehat{R}y_i)^2}{n-1}$. $\endgroup$ – Raphael Nishimura Dec 16 '14 at 4:09

Second Answer: harmonic mean of the individual ratios

Above I solved the problem for the ratio of total salary to total sales. (Y:X) for the population. As you pointed out, you can't compute an arithmetic mean for individual ratios $ R_i = y_i/x_i$ because of the zeros in the denominator of some.

However, you can compute the harmonic mean of the individual ratios and get a confidence interval for that. The harmonic mean for the population is defined as:

$$ \overline{R}_h =\frac{1}{\frac{1}{N} \sum_{i=1}^N x_i/y_i} $$

The harmonic mean requires only the $x_i/y_i$ ratios, which are computable when $x_i =0$.

  1. Find individual ratios for to sales/salary, the inverse of the original ratios: Call these $ z_i = x_i/y_i$.

  2. Compute $\bar{z}$, the sample mean of the $z_i$

  3. Compute the standard error $se_{\bar{z}}$ for $\overline{z}$ by the usual formula.

  4. Compute upper and lower confidence intervals for the true Z mean: $\bar{z} \pm 3.2 \times se_{\bar{z}}$. Call these $z_L$ and $z_U$. (The 3.2 is the t-multiplier for 95% confidence and 3 degrees of freedom; you might well choose a lower level of confidence).

  5. Convert $\bar{z}$, $z_L$ , and $z_U$ to the harmonic mean for the $r_i$'s and corresponding interval endpoints:

Harmonic mean:

$$ \overline{r}_h = 1/\bar{z} $$

Lower endpoint of CI for harmonic mean

$$ r_{hL} = 1/z_U $$

Upper endpoint of CI for harmonic mean

$$ r_{hU} = 1/z_L $$

It is up to you to defend the harmonic mean; you might do better presenting the Z scale results. The Z mean is likely to be closer to approximate normality than the ratio estimate in the first answer, because it depends only on the distribution of a sum. However with $n=4$, the confidence interval will still be on very shaky ground. I repeat my original advice: draw a larger sample.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.