Home › Forums › General Forums › New to Lean Six Sigma › Mood Median Results Interpretation
This topic contains 4 replies, has 4 voices, and was last updated by Robert Butler 4 years, 8 months ago.
Hi,
I’d like some assistance with interpreting the results below as in my limited knowledge of stats the groups appear to be significantly different according to the graphical output but with a P value of zero there is no significance. Can anybody please explain.
Mood Median Test: % Total versus Sprayed – Non Sprayed
Mood median test for % Total
Chi-Square = 14.78 DF = 1 P = 0.000
Sprayed – Individual 95.0% CIs
Non Sprayed N Median Q3-Q1 +———+———+———+——
Non Sprayed 192 199 0.0800 0.1000 (—–*
Sprayed 65 26 0.0400 0.0700 (——*—–)
+———+———+———+——
0.030 0.045 0.060 0.075
Overall median = 0.0700
A 95.0% CI for median(Non Sprayed) – median(Sprayed): (0.0200,0.0500)
Why do you think a P of 0.0000 means there’s no significance?
Put simply what this means that there is a 0.0000% chance of your data distribution appearing by chance. Therefore your two data sets are different.
@chrisrichardson1973 – It seem that you’re using Minitab, so I’ll give you the definition for the Mood’s Median Test from their help file (something you really should become familiar with so you wont have to post these questions):
Stat > Nonparametrics > Mood’s Median Test
Mood’s median test can be used to test the equality of medians from two or more populations and, like the Kruskal-Wallis Test, provides an nonparametric alternative to the one-way analysis of variance. Mood’s median test is sometimes called a median test or sign scores test. Mood’s median test tests:
H0: the population medians are all equal versus H1: the medians are not all equal
An assumption of Mood’s median test is that the data from each population are independent random samples and the population distributions have the same shape. Mood’s median test is robust against outliers and errors in data and is particularly appropriate in the preliminary stages of analysis. Mood’s median test is more robust than is the Kruskal-Wallis test against outliers, but is less powerful for data from many distributions, including the normal.
===================
You always need to start with an understanding of the null hypothesis, and from that you can interpret your statistical info. You seemed to think that a P value of 0.0 indicates no significance, when the test is whether two or more sets of data have statistically the same median (or the alt hyp, do not have the same median). Since the P value is low, the null must go (therefore they are statistically different). Whether that statistical difference is practically different is up to you to decide. The more data you have, the more likely to find a statistical difference.
Chris-
If you’re using Minitab 16, I recommend using the Assistant (Assistant > Hypothesis Tests > 2-Sample t) to do this test. The 2-sample t-test is very robust to non-normality despite it being frequently cited as an assumption (while it is an assumption in the mathematical sense, testing has shown it is not very important), but is a much more powerful test. Of course in your case power doesn’t seem to be relevant since you are showing significance.
In any event the Assistant will give you a good interpretation of what the results are telling you. It will present things graphically as well and provide “plain English” guidance.
Good luck,
Joel
Joel, just so you have a little more ammunition the next time you have to address the the shopworn claims about normality and t-tests I’d recommend getting a copy of The Design and Analysis of Industrial Experiments 2nd Edition by Davies. He has an excellent discussion of the issue on pages 52-54. Indeed, the title of the chapter subsection is “Effect of Departures from Normality on the t-test”. As you noted – it’s not much.
© Copyright iSixSigma 2000-2017. User Agreement. Any reproduction or other use of content without the express written consent of iSixSigma is prohibited. More »