To compare the means of two or more populations on a continuous CT characteristic. Since we don't know the population means, an analysis of data samples is required. ANOVA (Analysis of Variance) is usually used to determine if there is a statistically significant change in the mean of a CT characteristic under two or more conditions introduced by one factor (see concept Factor and Levels).
Reference: Juran's Quality Control Handbook
A. Null (H0) and alternative (Ha) hypotheses where the means (m1, m2, …, mg) of the g levels of the factor are compared. There is only one alternative hypotheses: which is at least the means of two levels are significantly different. (Note: these hypotheses correspond to a fixed effects model, for more information see reference above.)
B. Model where yij is the (ij)th observation of the CT characteristic (i = 1, 2, …, g and j = 1, 2, …, n, for g levels of size n), m is the overall mean, ti is the ith level effect and eij is an error component.
C. Minitab session window output.
D. Descriptive Statistics – Sample sizes, means, standard deviations (StDev), and the Pooled Standard Deviation, which is a combination of the standard deviations of the g groups.
E. Confidence Intervals around the mean for each individual level.
F. ANOVA Table (see part 2).
The assumptions for using this tool is that the data comes from independent random samples taken from normally distributed populations, with the same variance. When using ANOVA, we are using a model where its adequacy has to be verified using residual analysis (see tool Residual Plots).
Define problem and state the objective of the study.
State Null and Alternative Hypothesis.
Establish sample size (see tool Sample Size – Continuous Data – One Way ANOVA).
Select random samples.
Measure the CT characteristic.
Analyze data with Minitab (part 1 of 2):
In order to fully use the Minitab functions associated with ANOVA it is recommended that the data be stacked into one column and a second column to contain the group codes. Using the function under Manip > Stack/Unstack > Stack Columns.
See Part 2 for continuation of the application cookbook.
To summarize the results of an analysis of variance calculation in a table.
A. Source – Indicates the different variation sources decomposed in the the table. "Factor" represents the variation introduced between the factor levels. The "Error" is the variation within each of the factor levels. The "Total" is the total variation in the CT characteristic.
B. DF – The number of degrees of freedom related to each sum of square (SS). They are the denominators of the estimate of variance.
C. SS – The sums of squares measure the variability associated with each source. They are the variance estimate's numerators. "Factor" SS is due to the change in factor levels. The larger the difference between the means of a factor levels, the larger the factor sum of squares will be. The "Error" SS is due to the variation within each factor level. The "Total" SS is the sum of the Factor and Error sum of squares (see tool Sum of Squares).
D. MS – Mean Square is the estimate of the variance for the factor and error sources. Computed by MS = SS/DF.
E. F – The ratio of the mean square for the "Factor" and the mean square for the "Error".
F. P-Value – This value has to be compared with the alpha level (a) and the following decision rule is used: if P < a,, reject H0 and accept Ha, with (1-P)100% confidence; if P a,, don't reject H0.
The assumptions for using this tool is that the data comes from independent random samples taken from normally distributed populations, with the same variance. When using it we are using a model where its adequacy has to be verified using residual analysis (see tool Residual Plots).
1. Analyze the data with Minitab (part 2 of 2):
Verify the assumption of equality of variance for all the levels with the function under Stat > ANOVA > Homogeneity of Variance (to interpret this analysis, see tool Homogeneity of Variance Tests).
2. Make a Statistical decision from the session window output of Minitab. Either accept or reject H0. When H0 is rejected we can conclude that there is a significant difference between the means of the levels.
3. Translate statistical conclusion into practical decision about the CT characteristic.
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