ANOVA


One-Way Analysis of Variance "ANOVA" is used to compare the means of two or more samples against each other. This calculation determines whether it is likely that the samples came from populations with the same mean. This is similar to a 2-Sample t-Test except that three or more samples can be examined with ANOVA.

(ANOVA) can also be used to examine multiple variables and levels at the same time, but here the focus is primarily on the One-Way (ANOVA). One-Way examines just one variable and multiple levels. For example, a team might need to determine if 3 operators are different

  • A single variable is the Operator.
  • With 3 levels or 3 Operators.
  • Measure amount of time to perform a task. 
  • Measure each operator several times.

For Example 

Measure 15 points for each operator to preform a task. Use ANOVA to make the judgment and see if all the operators' average (mean) task times are the same

Level Of Confidence

You will need to determine the the level of confidence, such as 90% or 95% for the calculaton. This depends your required level of certainity from the analysis of variation calculation.

Explaination Of Anova

This is shown graphically in the Figure below . The upper curves represent the distributions of the three operators' times (known as the populations). The exact nature of these distributions is unknown to the team, because they represent all data points for all time. However, the team can see the sample's distrubution.  Shown as the lower curves.

ANOVA examines the sample data with the aim of making an inference on the location of the population means (μ) relative to each other. It does this by breaking down the variation (using variances) in all the sample data into separate pieces, hence the name Analysis Of Variance.

ANOVA compares the size of the variation between the samples versus the variation within the samples.

Graphical Representation Of ANOVA

If the variation between the samples is large relative to the variation within the samples, then it means the samples are spread widely (between) compared with the background noise (within). This would imply that the means of the parent distributions are different

If the between variation is not large compared to the within variation, then it is likely that the means of the parent distribution are about the same. More specifically the test cannot distinguish between them. 

The result of the test is a number called the p-value, which stands for probability. A high p-value means the samples come from populations with the same mean. The reverse is also true. A low p-value tells us the populations are significantly different. In our example the p-value tells us the probability that the mean operator times are the same or different. If the p-value is low, then at least one of the mean operator times is distinguishable from the others; if the p-value is high, they all are not distinguishable.

Roadmap

The roadmap of the test analysis itself is shown graphically in Figure below

One-Way Anova Roadmap

Roadmap adapted from SBTI's Process Improvement Methodology training material.

Step 1.Identify the metric and levels to be examined (for example, three operators). Make the metric well defined and understood by the team.

Step 2.Determine the sample size. Use a sample size calculator.

Step 3.Collect the sample data set, one from each level of the variable. Follow the rules of good experimentation. If the sample size calculator determined a sample size of ten data points, then ten points need to be collected for each and every level. For example, if the variable is operator and there are three levels (three operators), then 3 x 10 = 30 data points are collected in total.

Step 4.Examine stability of all sample data sets using a Control Chart for each, typically an Individuals and Moving Range Chart (I-MR). A Control Chart identifies whether the processes are stable, having
This is important; if the processes are not stable, Then the study will give an incorrect answer.

Step 5.Examine normality of the sample data sets using a Normality Test for each.   

Step 6.Perform the ANOVA if all of the sample data sets were determined to be normal in Step 5 

Anova Calculations

It is beyond the scope of this page to show you the calculations. We recommend using Mini tab to easily conduct the calculations. Below we discuss how to interpret the results of the calculations.

Interpreting The Output

This test calculates a ratio of the signal (variation due to the variable, the "between") relative to the noise (any other variation not due to the variation, the "within"). If the signal-to-noise ratio gets large enough then this would be considered to be unlikely to have occurred purely by random chance and the variable is thus considered statistically significant.

This is achieved by looking up the signal-to-noise ratio in a reference distribution (F-Test), which returns a p-value. The p-value represents the likelihood that an effect this large could have occurred purely by random chance even if the populations were the same.

Based on the p-values, statements can be generally formed as follows:

  • Based on the data, I can say that at least one of the means is different and there is a (p-value) chance that I am wrong
  • Or based on the data, I can say that there is an important effect due to this X and there is a (p-value) chance the result is just due to chance

Example Output From An ANOVA

ANOVA results for a comparison of samples of Bob's vs Jane's vs Walt's performance (output from Minitab v14).

From The First Table In The Results:

  • The average variation due to Operator (between variation or the signal) was 40.193 units
  • The average variation due to Error (within variation or the noise ) was 0.898 units
  • The signal-to-noise ratio is therefore 40.193 ÷0.898 = 44.76.
  • The likelihood of seeing a signal-to-noise ratio this large (if the populations were perfectly aligned) is 0.000%(p-value). This is well below 0.05 (95% confidence interval). You can conclude that at least one of the trio is performing significantly differently from the others.
  • A sample of 30 data points was taken for each operator.
  • Bob's sample mean is 24.848, Jane's is 25.446, and Walt's is 27.084.
  • Bob's sample standard deviation is 0.869, Jane's is 0.988, and Walt's     is 0.981.
  • The text graph shows the 95% confidence intervals for the locations of   the population means for each of the trio.
  • The p-value of 0.000% in the upper table indicates that at least one of  the trio is performing differently from the other two. There is no overlap in 95% confidence intervals in the bottom table between Walt's performance and the other two; therefore, it is clearly Walt who has a different mean.




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