# Comparing normal with non-normal

I have two sets of data for setup time of a machine, one is setup time when fixture is running alone, another is when another fixture is running alongside.

For fixture running alone the data I gathered is not normal although there are only 12 data points.

For parallel running fixture the setup is normally distributed with p0.089 and sample size 10

I wanted to test the null hypothesis that there is no difference in setup time for these two scenarios, would comparing the median of these two data the right way to do it or not.

I understand the sample size is too low, but this experiment is for a product where you do not get many data points. they wanted just an estimate to start working, so all I wanted to do is atleast follow the right way of doing it.

Thanks

• If you only have 10 data points and a test of normality yields a p-value of 0.089, I would not put much stock in the idea that those data are normally distributed. You have 2 non-normal samples (which is typical of durations). Sep 28, 2014 at 2:37
• You should think very carefully whether median setup time is of more interest than mean setup time (or indeed whether some other measure would be more suitable). Sep 28, 2014 at 6:46
• Have you plotted anything? All these tests etc. are silly without a basic visual understanding of what you have in front of you. Sep 28, 2014 at 16:36
• May 12, 2021 at 7:58

With so few data points, I wouldn't get caught up in complicated formulas, test statistics and P-values. Instead, I'd keep it simple by just asking a few questions:

• What is the average setup time for the two alternatives? Does running standalone result in a faster/slower time than running concurrently with another machine?

• What is the maximum setup time? The average may not be very useful if, in the worst case, one alternative takes twice as long as the other.

• Similarly, what is the standard deviation or interquartile range of setup times? The idea is to see how much variability or volatility there is in setup times. If one alternative is slightly faster on average, but could sometimes be much slower, that's important to know.

The objective is to arrive at a better qualitative understanding of how setup time can vary, and from that, decide how to proceed. With such a small dataset, any quantitative test you could perform is unlikely to tell you anything you couldn't figure out for yourself.

• 1. running standalone does result in faster time than running concurrently with another machine because there is only one operator in both cases. Sep 28, 2014 at 23:28
• 1. running standalone does result in faster time than running concurrently with another machine because there is only one operator in both cases. 2. 138-235 seconds is range for one with median 157 and 200-239 seconds is for another with median 204 I also understood similar thing first, as these are operators time and they tend to very by big numbers but apart from 2 cases in both scenarios most of the points in one setup is way smaller than other. I just needed something statistical to show to my advisor, in companies they want just good estimate, but you know schools and professors Sep 28, 2014 at 23:43

If you have only 10 data points and your goal is to know whether the two sets of samples come from the same distribution, I would not even check for normality.

As I see it, your options are:

• Nonparametric test for equal medians (e.g., Wilcoxon sum rank test, see @Glen_b comments though)
• Nonparameteric test for equal distributions (e.g., Kolmogorov-Smirnov)

Each of the tests above assumes something different on the target distribution, so you should see what you is logical to assume in your case.

Edited - I suggested the wrong tests before

• There are some other, smaller problems I haven't mentioned, but they're not worth a downvote. I may take those issues up if the answer is amended. Sep 28, 2014 at 6:48
• @Glen_b, you are absolutely correct. I was too quick to draw. Thanks! Sep 28, 2014 at 7:58
• +1. However, the Wilcoxon-Mann-Whitney (rank sum test/U-test) is not actually a test for equality of medians unless you make the additional assumption that the distributions are identical apart from possible location shift (an assumption not required for the test to be valid). If you do make that assumption, it's a test for equality of almost any sensible location parameter (modes, upper quartiles, 10th percentiles, trimmed means, and even means or midrange if they exist). A lot of basic books seem to have this wrong (and it's not such a big deal in many cases). Sep 28, 2014 at 8:46
• See the last paragraph of this section. What it actually tests is $P(X>Y)=\frac{1}{2}$ (against the alternative of inequality). Sep 28, 2014 at 9:10