# One Way ANOVA (Analysis of Variance)

## Non-Statisticians Can Perform ANOVA Single-Factor Analysis in Excel Using the QI Macros

### One-Way ANOVA Example

Imagine you manufacture paper bags and you want to improve the tensile strength of the bag. You suspect that changing the concentration of hardwood in the bag will change the tensile strength. You measure the tensile strength in pounds per square inch (PSI). So, you decide to test this at 5%, 10%, 15% and 20% hardwood concentration levels. These "levels" are also called "treatments."

Since we are only evaluating a single factor (hardwood concentration) this is called one-way ANOVA. Imagine that we've conducted these trials at each of the four levels of hardwood concentration.

### To perform a One-Way Analysis of Variance in the QI Macros follow these steps:

1. Just click and drag over the data to select it.

2. Now click on the QI Macros, Statistical Tools and ANOVA: Single factor.

3. The QI Macros will prompt you for the significance level you desire.  While the default is 0.05 (95% confident), in this example we want to be even more certain, so we use 0.01 (99% confident).

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The QI Macros will perform the calculations and analyze the results for you.

### What's Cool About QI Macros One-Way ANOVA?

When you run ANOVA, you don't have to think. Unlike other statistical software, the QI Macros is the only SPC software that compares the p-value (0.000) to the signficance (0.01) and tells you to "Reject the Null Hypothesis because p<0.01" and that the "Means are Different".

The QI Macros also loads test data for each kind of Statistical test on your PC at c:\qimacros\testdata\ANOVA.xls.

### Interpreting ANOVA One-Way Test Results Yourself

The QI Macros automatically compares the p value to a, but you might want to know how to do this manually.

To conduct the one-way ANOVA test, you need to randomize the trials (assumption #1) and the data must be normal (assumption #2). The "null" hypothesis assumes that there is no difference between the hardwood concentrations.

The null hypothesis is that the means are equal:

H0: Mean1 = Mean2 = Mean3 = Mean4

The alternate hypothesis is that at least one of the means is different:

Ha: At least one of the means is different
 If Then test statistic > critical value  (i.e. F> Fcrit) Reject the null hypothesis test statistic < critical value  (i.e. F< Fcrit) Cannot Reject the null hypothesis p value < a Reject the null hypothesis p value > a Cannot Reject the null hypothesis

The P-value of 0.000 is less than the significance level (0.01), so we can reject the null hypothesis and safely assume that hardwood concentration affects tensile strength.

F (19.60521) is greater than F crit (4.938193), so again, we can reject the null hypothesis.

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Now we can look at the average tensile strength and variances:

The average tensile strength increases, but we cannot say for certain which means differ. The variance at the 15% level looks substantially lower than the other levels. We might need to do additional analysis.

If we reran the one way ANOVA test with just 10% and 15%, we'd discover there is no statistically significant difference between the two means.

Based on this analysis, if we were aiming for a tensile strength of 15 PSI or greater, the 10% level might be more cost effective.

Another way to look at this data might be to use a box and whisker diagram which shows the distribution of each level:

The 10% and 15% levels are close, but 10% shows more variability with some of the bags below the 15 PSI target.

This example came from Montgomery's Intro to SPC book.