In this blog, it’s continued on discussing variances test for multiple samples ( > 2 samples ). It’s called as Test for Equal Variances in Minitab.

If the multiple sample data distributions do not follow Normal distribution, Levene’s test should be adopted. Vice versa, Bartlett’s test should be adopted.

Below is a case study that illustrates the use of Test for Equal Variances.

*The process engineer would like to know further if different raw material ( type A, B, C & D ) give significant impact on the standard deviation of plastic breaking strength. He run the experiment by collecting the samples data for ALL material type reference below table:*

A | 7.2 | 7.3 | 6.8 | 7.0 | 9.1 | 8.2 | 8.5 | 6.6 | 6.9 | 8.4 |

B | 9.0 | 7.9 | 8.7 | 8.9 | 8.9 | 8.1 | 8.2 | 7.6 | 8.3 | 7.5 |

C | 7.8 | 7.9 | 8.0 | 7.4 | 8.5 | 7.3 | 8.8 | 8.6 | 8.5 | 6.8 |

D | 7.0 | 7.1 | 8.4 | 6.4 | 6.9 | 6.6 | 7.4 | 6.7 | 7.1 | 7.4 |

So, the **practical question** here was “Are the variations of breaking strength of plastic using different raw material type the same?”

It was then translated to **statistical question** by means of Hypothesis Statement.

Ho : Null Hypothesis => σ_{A }= σ_{B} = σ_{C }= σ_{D}

Ha : Alternate Hypothesis => σ_{A }≠ σ_{B} ≠ σ_{C }≠ σ_{D}

Next, conduct **Statistical Analysis**. In Minitab, click **Stat>ANOVA>Test for Equal Variance **and fill up relevant information.

Then, click **OK** to see the analysis result from Minitab.

The Pvalue of Bartlett’s test and Levene’s test are 0.522 and 0.442 respectively. Regardless of what data distribution of the sample data, both Pvalue shows more than 0.05. Fail to reject Ho (assuming Alpha risk of 5% i.e. 0.05 )

**Statistically**, it was to say the breaking strength standard deviation of plastic were the same for all raw material type ( A, B, C & D ).

**Practically**, it was concluded NO significant difference in variation between breaking strength of plastic parts using different type of raw material ( A, B, C & D ).