# Complete Guide: How to Interpret ANOVA Results in R

A one-way ANOVA is used to determine whether or not there is a statistically significant difference between the means of three or more independent groups.

This tutorial provides a complete guide on how to interpret the results of a one-way ANOVA in R.

### Step 1: Create the Data

Suppose we want to determine if three different workout programs lead to different average weight loss in individuals.

To test this, we recruit 90 people to participate in an experiment in which we randomly assign 30 people to follow either program A, program B, or program C for one month.

The following code creates the data frame we’ll be working with:

```#make this example reproducible
set.seed(0)

#create data frame
data <- data.frame(program = rep(c('A', 'B', 'C'), each = 30),
weight_loss = c(runif(30, 0, 3),
runif(30, 0, 5),
runif(30, 1, 7)))

#view first six rows of data frame

program weight_loss
1       A   2.6900916
2       A   0.7965260
3       A   1.1163717
4       A   1.7185601
5       A   2.7246234
6       A   0.6050458
```

### Step 2: Perform the ANOVA

Next, we’ll use the aov() command to perform a one-way ANOVA:

```#fit one-way ANOVA model
model <- aov(weight_loss ~ program, data = data)
```

### Step 3: Interpret the ANOVA Results

Next, we’ll use the summary() command to view the results of the one-way ANOVA:

```#view summary of one-way ANOVA model
summary(model)

Df Sum Sq Mean Sq F value   Pr(>F)
program      2  98.93   49.46   30.83 7.55e-11 ***
Residuals   87 139.57    1.60
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
```

Here’s how to interpret every value in the output:

Df program: The degrees of freedom for the variable program. This is calculated as #groups -1. In this case, there were 3 different workout programs, so this value is: 3-1 = 2.

Df Residuals: The degrees of freedom for the residuals. This is calculated as #total observations – # groups. In this case, there were 90 observations and 3 groups, so this value is: 90 -3 = 87.

Sum Sq program: The sum of squares associated with the variable program. This value is 98.93.

Sum Sq Residuals: The sum of squares associated with the residuals or “errors.” This value is 139.57.

Mean Sq. Program: The mean sum of squares associated with program. This is calculated as Sum Sq. program / Df program. In this case, this is calculated as: 98.93 / 2 = 49.46.

Mean Sq. Residuals: The mean sum of squares associated with the residuals. This is calculated as Sum Sq. residuals / Df residuals. In this case, this is calculated as: 139.57 / 87 = 1.60.

F Value: The overall F-statistic of the ANOVA model. This is calculated as Mean Sq. program / Mean sq. Residuals. In this case, it is calculated as: 49.46 / 1.60 = 30.83.

Pr(>F): The p-value associated with the F-statistic with numerator df = 2 and denominator df = 87. In this case, the p-value is 7.552e-11, which is an extremely tiny number.

The most important value in the entire output is the p-value because this tells us whether there is a significant difference in the mean values between the three groups.

Recall that a one-way ANOVA uses the following null and alternative hypotheses:

• H0 (null hypothesis): All group means are equal.
• HA (alternative hypothesis): At least one group mean is different from the rest.

Since the p-value in our ANOVA table (.7552e-11) is less than .05, we have sufficient evidence to reject the null hypothesis.

This means we have sufficient evidence to say that the mean weight loss experienced by the individuals is not equal between the three workout programs.

### Step 4: Perform Post-Hoc Tests (If Necessary)

If the p-value in the ANOVA output is less than .05, we reject the null hypothesis. This tells us that the mean value between each group is not equal. However, it doesn’t tell us which groups differ from each other.

In order to find this out, we must perform a post hoc test. In R, we can use the TukeyHSD() function to do so:

```#perform Tukey post-hoc test
TukeyHSD(model)

\$program
B-A 0.9777414 0.1979466 1.757536 0.0100545
C-A 2.5454024 1.7656076 3.325197 0.0000000
C-B 1.5676610 0.7878662 2.347456 0.0000199
```

Here’s how to interpret the results:

• The adjusted p-value for the mean difference between group A and B is .0100545.
• The adjusted p-value for the mean difference between group A and C is .0000000.
• The adjusted p-value for the mean difference between group B and C is .0000199.

Since each of the adjusted p-values is less than .05, we can conclude that there is a significant difference in mean weight loss between each group.