# 10.4 - Minitab: One-Way ANOVA

10.4 - Minitab: One-Way ANOVA

In one research study, 20 young pigs are assigned at random among 4 experimental groups. Each group is fed a different diet. (This design is a completely randomized design.) The data are the pigs' weights in kg after being raised on these diets for 10 months. We wish to ask whether mean pig weights are the same for all 4 diets.

• $$H_0: \mu_1 = \mu_2 = \mu_3 = \mu_4$$
• $$H_a:$$ Not all $$\mu$$ are equal
Data from study of pigs weights
Feed_1 Feed_2 Feed_3 Feed_4
60.8 68.3 102.6 87.9
57.1 67.7 102.2 84.7
65.0 74.0 100.5 83.2
58.7 66.3 97.5 85.8
61.8 69.9 98.9 90.3

Contained in the Minitab file: ANOVA_ex.mpx

Note that in this file the data were entered so that each group is in its own column. In other words the responses are in a separate column for each factor level. In later examples, you will see that Minitab will also conduct a one-way ANOVA if the responses are all in one column with the factor codes in another column.

## Minitab® – One-Way ANOVA

To perform an Analysis of Variance (ANOVA) test in Minitab:

1. Open the Minitab file: ANOVA_ex.mpx
2. From the menu bar, select Stat > ANOVA > One-Way.
3. Click the drop-down menu and select Responses are in a separate column for each factor level.
4. Enter the variables Feed_1Feed_2Feed_3, and Feed_4 to insert them into the Responses box.
5. Choose the Comparisons button and check the box next to Tukey. Under Results also select Tests.
6. OK and OK

The result should be the following output:

##### Method
Null hypothesis All means are equal At least one mean is different $$\alpha=0.05$$

Equal variances were assumed for the analysis

##### Factor Information
Factor Levels Values
Factor 4 Feed_1, Feed_2, Feed_3, Feed_4
##### Analysis of Variance
Factor 3 4703.2 1567.73 206.72 0.000
Error 16 121.3 7.58
Total 19 4824.5
##### Model Summary
2.75386 97.48% 97.01% 96.07%
##### Means
Factor N Mean StDev 95% CI
Feed_1 5 60.68 3.03 (58.07, 63.29)
Feed_2 5 69.24 2.96 (66.63, 71.85)
Feed_3 5 100.340 2.164 (97.729, 102.951)
Feed_4 5 86.38 2.78 (83.77, 88.99)

Pooled StDev = 2.75386

##### Grouping Information Using the Tukey Method and 95% Confidence
Factor N Mean Grouping
Feed_3 5 100.34 A
Feed_4 5 86.38   B
Feed_2 5 69.24     C
Feed_1 5 60.68       D

Means that do not share a letter are significantly different.

Difference of
Levels
Difference
of Means
SE of
Difference