As I was updating our Green Belt course material to incorporate V17 screen shots I saw examples of data that could have been tested two uniquely different methods.
Both conditions could exist at the end of an improvement project when you are demonstrating your improvement with a hypothesis test.
These examples are quite important to me as a statistician and Lean Six Sigma instructor because they show the internal consistency of the inferential statistical methods.
With every available test applied to the data, you will arrive at the correct business decision.
Continuous Data
I used a data set that our students are used to seeing. It is example 8-11 (means nothing to everyone else of course) which is a comparison of two machine outputs.
It includes two different sample sizes.
I included enough information for you to duplicate my work. Now lets compare
The standard comparison test will be a 2-sample t-test, lets try it
2-Sample t-test
It turns out that you need to make sure to use the equal variance assumption option for the t-test, since that is how ANOVA looks at comparisons.
Two-sample T for ResponseMachine N Mean StDev SE Mean Machine A 23 104.9 18.1 3.8 Machine B 9 120.9 14.8 4.9 Difference = ? (Machine A) - ? (Machine B) Estimate for difference: -16.01 95% CI for difference: (-29.92, -2.11) T-Test of difference = 0 (vs ?): T-Value = -2.35 P-Value = 0.025 DF = 30
One-Way ANOVA
Again, we are using the equal variance assumption.
One-way ANOVA: Response versus Machine
Null hypothesis All means are equal Alternative hypothesis At least one mean is different Significance level ? = 0.05
Equal variances were assumed for the analysis.
Factor Information
Factor Levels Values Machine 2 Machine A, Machine B
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value Machine 1 1659 1658.5 5.53 0.025 Error 30 8996 299.9 Total 31 10655
Model Summary
S R-sq R-sq(adj) R-sq(pred) 17.3167 15.57% 12.75% 4.87%
T-test : ANOVA Comparison
- Both provided a p-value for the difference equal to 0.025.
- The t-test t-value equaled -2.35. If you square this number you get 5.525, which equals the F-value used in the ANOVA.
This result is not extraordinary. The t-value is always the square-root of the F-value when you have only two levels and assume equal variances.
In my view, the fact that both methods provide what appears to be the exactly same result (within the rounding of the Minitab display) is very comforting. Since the p-value is a measure of the risk in rejecting the null hypothesis in error, both methods provide me the same estimate of risk. Which is good, since both methods have the same null hypothesis so I would have expected the p-value to be consistent.