For z score, you are taking the sample value subtracting population mean and dividing it by std deviation. Is that correct so far?

Now, the "sample value" is defined by an equation. In my scenario, I have a bunch of stores that are reviewed by their customers. The customer amount varies. One store has 2 customers who give review while other store has 19 people who give reviews. For the 1st store in my example, both customers give 5/5. while the 2nd store customers, avg score given by all 19 customers is 4.

Sample value is calculated as score/customer count. So for the 1st store, it will be 5/2 while the other store will be 4/19. If I use that in z score eq. it will mean that first store is doing much better than the last store. I would like to know how to make the score "proportional""normalized""fair" etc...

Oh my many gods, I can't believe I ended like that. Thanks for pointing that out.

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    $\begingroup$ "If I use that in z score eq. it-" - is there more? $\endgroup$ – davidlowryduda Aug 9 '12 at 17:02
  • $\begingroup$ @user1569220 : If you click on "edit" below your question, you can edit it further. You stopped in the middle of a sentence. $\endgroup$ – Michael Hardy Aug 9 '12 at 18:38

The $z$ score in your context is the sample mean minus the population mean divided by the standard deviation of the estimate which in your case would be the standard error of the mean which is the population standard deviation divided by the square root of the sample size $n$. The motivation for calling this a standard normal random variable is that the sample mean is either normally distributed or approximately normally distributed by the central limit theorem.

Technically the population standard deviation must be known. In the case where a sample estimate is used and the data are normally distributed the statistic formed would actually have a student $t$ distribution with $n-1$ degrees of freedom. However if $n$ is large the $t$ will be close to a standard normal. For small $n$ it will be symmetric about $0$ but have heavier tails than the standard normal.

Another issue in the computation is whether or not you are considering a straight average from the observed data or a weighted average. This all depends on what you consider the population to be. The definition of the population can affect the population mean and the population standard deviation.

  • $\begingroup$ Sample value = Score/Count; Mean = Avg of Sample Value of the entire population; Std Dev = Stddev of Sample Value of the entire population...if I understand this correctly, the average is based on observed data. $\endgroup$ – user1569220 Aug 9 '12 at 18:59
  • $\begingroup$ @user1569220 I just meant that since the individual stores have differing number of customer responses do you want to weight the customers equally over are the stores or do you want to weigh the stores evenly inspite of the imbalance in number of customers. $\endgroup$ – Michael R. Chernick Aug 9 '12 at 19:06
  • $\begingroup$ I would like to weigh the customers equally. I think it would be more fair rather than just making each store equal. $\endgroup$ – user1569220 Aug 9 '12 at 19:09
  • $\begingroup$ Okay That is generally what is done. $\endgroup$ – Michael R. Chernick Aug 9 '12 at 19:12
  • $\begingroup$ So again sorry, what do you recommend I should do? $\endgroup$ – user1569220 Aug 9 '12 at 19:42

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