# Explanation of Mood's Median Results

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• #54274

Nikki
Guest

I am very new into Six Sigma and using the tools and i would like some help interpreting the results shown below:

Mood median test for Crews Per Start
Chi-Square = 28.38 DF = 1 P = 0.000

Individual 95.0% CIs
Phase N Median Q3-Q1 —+———+———+———+—
After 38 10 1.182 0.356 (——*)
Before 15 41 1.567 0.344 (——*—-)
—+———+———+———+—
1.12 1.28 1.44 1.60

Overall median = 1.385

A 95.0% CI for median(After) – median(Before): (-0.495,-0.267)

I want to be able to explain the results to someone who knows nothing about this tool but still be able to tell them what the results are saying.

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#194541

Chris Seider
Participant

It is very similar to the ANOVA method. Look at the P-value if you accept the alpha risk you can see before result is higher than the before somewhere between 0.495 and 0.267–assuming 95% confidence.

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#194544

MBBinWI
Participant

Nikki: To add to what Chris posted, you can tell them that given the data, 95% of the time the difference between the before and after medians would be less than zero (or, there was a true reduction in the item measured with a 95% confidence).

Just curious, but why are you using medians? Do you have some small number of really out of whack values?

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#194545

Darth
Participant

To expand a bit upon my esteemed colleagues’ statements:
1. Your basic hypothesis is that there is no difference in your median value before and after. Your alternate hypothesis is that there is a difference. Assuming you would liked to have seen a difference, you would hope that the data allows you to reject your basic hypothesis and claim that whatever you did had an impact.
2. Because you are using samples, there is a chance that you will make the wrong decision. By selecting a risk of being wrong only 5% of the time (Confidence of 95%) you now have to determine what your “real” risk is of saying there was a change before and after.
3. Essentially your p value is the actual risk of being wrong in rejecting your basic hypothesis and claiming that whatever you did resulted in a statistically significant change in your median values. Since your p value is zero and thus less than your assumed risk it appears that you don’t really have any risk in claiming that your after median is statistically lower than the before risk.
It is
As my friend MBBinWI asked, why the use of the Mood’s median? It is generally used for comparing the median of more than two groups. With only two groups you would likely use a Mann-Whitney test instead. You would only use medians if the data was severely non-normal although the 2 sample t test is pretty robust to departures from normality. Parametric tests are more powerful than nonparametric tests so you have to have a good reason for using them.

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#194546

Chris Seider
Participant

@Darth He has risen…and I don’t mean on Easter.

Welcome back

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