Online marketers often need to choose a "winning" variation among, say, 2 possible variations.


  • PageVariation1 had 1000 sessions with 20 conversions (2.0% response rate) and
  • PageVariation2 had 800 sessions with 8 conversions (1.0% response rate).

I'm using the "unpooled" Two-Proportion z-Test for estimating the P-value. In the example above it is 0.9615 - so I might conclude that there's a 96% that PageVariation1 has a higher response rate. Let's say it's good enough to declare it as "winner".

Problem is that some sessions and conversions weren't tracked to the PageVariation level - i.e. I know there were, say, additional 200 sessions and 3 conversions, but those can't be allocated to any of the above. Intuitively it means that there is higher uncertainty due to tracking/measurement issues. Consequently the P-value should be lower.


  • Which test should be used (and how) in order to calculate the new P-value?
  • 2
    $\begingroup$ Your p-value is not 0.9615: it is 0.07698. This is because your z-statistic is -1.7685 and a standard normal variate has a 0.07698 chance of exceeding 1.7685 in absolute value. Note that this p-value is high enough that in many applications it would not be considered significant evidence of a difference anyway. The additional uncertainty in your situation should raise, not lower, this p-value. $\endgroup$
    – whuber
    Commented Apr 28, 2011 at 16:06

2 Answers 2


It can help to survey the scene: that is, to understand the consequences for your decision of the possible states of affairs.

The missing information can be described by two integers: $m$, the number of additional sessions for "PageVariation2", and $k$, the number of additional conversions for PageVariation2. For each possible combination of $m$ and $k$ ($k = 0, 1, 2, 3$ and $0 \le m \le 200-k$) we can compute what the p-value would be.

Plot of p-values vs. m by k

In this plot the graph of the p-value versus $m$ for $k=3$ is shown in blue (the highest one), for $k=2$ in red, for $k=1$ in orange, and for $k=0$ in green. For reference, a (two-sided) p-value of 0.05 is shown with a black line.

It is evident that the p-values can range tremendously from indicating strong significance (for $m=197$ and $k=0$) to indicating no significance at all (for small $m$ and $k = 1, 2, 3$). This tells us that your assumptions about the missingness completely determine the significance or lack of significance of the results.

If you think it highly likely that $m$ is large--so that the majority of missing results are for PageVariation2--and that $k$ is small--so that the majority of missing conversions are for PageVariation1--then you can conclude the differences in conversion rates are not likely due to chance and proceed accordingly. If you do not have sufficient information to justify these assumptions, then the data do not provide strong evidence for a significant difference. That is, what you have observed could be the result of random variation rather than due to some intrinsic difference between the two variations.


A few quick points:

  • You could draw on the general literature on missing data.
  • If the 200 sessions with missing data are known to be completely random, then you could choose to ignore these observations and just analyse the data where you have complete data.
  • In general it is important to think through what mechanisms might explain the missing data.
  • You may find it useful to read up on strategies for dealing with missing data.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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