This function performs the Levene's test for equality of variance.

`levene_test(data = NULL, x, groups, alpha = 0.05, modcv = FALSE)`

- data
a data.frame

- x
the variable in the data.frame or a vector on which to perform the Levene's test (usually strength)

- groups
a variable in the data.frame that defines the groups

- alpha
the significance level (default 0.05)

- modcv
a logical value indicating whether the modified CV approach should be used.

Returns an object of class `adk`

. This object has the following fields:

`call`

the expression used to call this function`data`

the original data supplied by the user`groups`

a vector of the groups used in the computation`alpha`

the value of alpha specified`modcv`

a logical value indicating whether the modified CV approach was used.`n`

the total number of observations`k`

the number of groups`f`

the value of the F test statistic`p`

the computed p-value`reject_equal_variance`

a boolean value indicating whether the null hypothesis that all samples have the same variance is rejected`modcv_transformed_data`

the data after the modified CV transformation

This function performs the Levene's test for equality of variance. The data is transformed as follows:

$$w_{ij} = \left| x_{ij} - m_i \right|$$

Where \(m_i\) is median of the \(ith\) group. An F-Test is then performed on the transformed data.

When `modcv=TRUE`

, the data from each group is first transformed
according to the modified coefficient of variation (CV) rules before
performing Levene's test.

“Composite Materials Handbook, Volume 1. Polymer Matrix Composites Guideline for Characterization of Structural Materials,” SAE International, CMH-17-1G, Mar. 2012.

```
library(dplyr)
carbon.fabric.2 %>%
filter(test == "FC") %>%
levene_test(strength, condition)
#>
#> Call:
#> levene_test(data = ., x = strength, groups = condition)
#>
#> n = 91 k = 5
#> F = 3.883818 p-value = 0.00600518
#> Conclusion: Samples have unequal variance ( alpha = 0.05 )
#>
##
## Call:
## levene_test(data = ., x = strength, groups = condition)
##
## n = 91 k = 5
## F = 3.883818 p-value = 0.00600518
## Conclusion: Samples have unequal variance ( alpha = 0.05 )
```