# 10.4 - Minitab Express: One-Way ANOVA

10.4 - Minitab Express: 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 Express file:

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 Express will also conduct a one-way ANOVA if the responses are all in one column with the factor codes in another column.

## MinitabExpress – One-Way ANOVA

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

1. Open the ANOVA_ex.MTW data set.
2. From the menu bar, select Statistics ANOVA One-Way ANOVA.
3. Click the drop-down menu and select "Responses are in a separate column for each factor level".
4. Double-click on the variables Feed_1Feed_2Feed_3, and Feed_4 to insert them into the "Responses" box.
5. Click the comparisons tab and check the box next to "Tukey (family error rate)".
6. Click OK.

The result should be the following output:

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 4703.188 1567.72933 206.72 <0.0001
Error 16 121.340 7.58375
Total 19 4824.528
Model Summary
2.75386093 97.48% 97.01% 96.07%
Means
Factor N Mean StDev 95% CI
Feed_1 5 60.680 3.028 (58.069, 63.291)
Feed_2 5 69.240 2.958 (66.629, 71.851)
Feed_3 5 100.3400 2.1640 (97.7292, 102.9508)
Feed_4 5 86.380 2.782 (83.769, 88.991)

Pooled StDev = 2.75386093

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.

Tukey Simultaneous Tests for Differences of Means
Difference of Levels Difference of Means SE of Difference 95% CI T-Value Adjusted P-Value
Feed_2-Feed_1 8.560 1.742 (3.572, 13.548) 4.91 0.0008
Feed_3-Feed_1 39.660 1.742 (34.672, 44.648) 22.77 <0.0001
Feed_4-Feed_1 25.700 1.742 (20.712, 30.688) 14.76 <0.0001
Feed_3-Feed_2 31.100 1.742 (26.112, 36.088) 17.86 <0.0001
Feed_4-Feed_2 17.140 1.742 (12.152, 22.128) 9.84 <0.0001
Feed_4-Feed_3 -13.960 1.742 (-18.948,-8.972) -9.02 <0.0001

Individual confidence level = 98.87%

Video Walkthrough

Select your operating system below to see a step-by-step guide for this example.

 [1] Link ↥ Has Tooltip/Popover Toggleable Visibility