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Comments (5)

IndrajeetPatil avatar IndrajeetPatil commented on June 9, 2024

It's correct. What's being reported are results from Welch's ANOVA:

image

For more, see the details section here:

https://indrajeetpatil.github.io/ggstatsplot/reference/ggbetweenstats.html

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M-Colley avatar M-Colley commented on June 9, 2024

Thanks! I then believe the error is due to for GroupE that all values are the same. Do you, by any chance, know what I can/should do in this case?

Kind regards

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IndrajeetPatil avatar IndrajeetPatil commented on June 9, 2024

I am not sure what you mean by error here. Do you mean the fact that the values are all NA?

If yes, then I would require to see the data that is producing these results. You can send it to me in an email.

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IndrajeetPatil avatar IndrajeetPatil commented on June 9, 2024

The problem is that there is no variation in trust scores for GroupE and so Welch's ANOVA fails.

df <- structure(
  list(
    group = structure(
      c(
        6L, 6L, 6L, 6L, 6L, 6L, 6L,
        6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 4L, 3L, 5L, 2L, 3L,
        2L, 3L, 3L, 4L, 3L, 4L, 5L, 3L, 1L, 2L, 1L, 5L, 1L, 4L, 2L, 3L,
        1L, 4L, 5L, 3L, 3L, 4L, 3L, 3L, 2L, 2L, 2L, 1L, 1L, 5L, 1L, 4L,
        4L, 2L, 1L, 1L, 2L, 2L, 1L, 4L, 2L, 4L, 2L, 2L, 3L, 4L, 5L, 2L,
        2L, 1L, 5L, 5L, 1L, 3L, 4L, 3L, 3L, 5L, 4L, 5L, 1L, 4L, 1L, 4L,
        4L, 1L, 3L, 2L, 4L, 5L, 5L, 4L, 4L, 4L, 2L, 1L, 5L, 1L, 3L, 5L,
        2L, 4L, 1L, 3L
      ),
      levels = c(
        "GroupA", "GroupB", "GroupC", "GroupD",
        "GroupE", "GroupF"
      ),
      class = "factor"
    ),
    t = c(
      5, 4, 4, 4, 5,
      3.5, 3, 4, 3, 4, 5, 4.5, 4.5, 3.5, 4.5, 5, 4, 4, 2, 5, 5, 1.5,
      4.5, 4, 4, 3, 5, 5, 2, 5, 5, 4, 4, 4, 5, 5, 2, 5, 4, 4, 4, 5,
      5, 2.5, 4.5, 2, 2, 5, 5, 4, 2, 3, 5, 5, 5, 5, 5, 2.5, 4.5, 4,
      3, 4, 3, 5, 4, 4, 5, 4, 2.5, 5, 5, 4, 5, 5, 5, 5, 5, 2, 1, 2.5,
      5, 2.5, 5, 5, 4, 5, 3, 4, 5, 3, 3, 4.5, 5, 5, 2, 5, 3.5, 5, 5,
      5, 3, 4, 5, 4, 5, 4, 2
    )
  ),
  row.names = c(NA, -107L),
  class = c("tbl_df", "tbl", "data.frame")
)

oneway.test(t ~ group, data = df)
#> 
#>  One-way analysis of means (not assuming equal variances)
#> 
#> data:  t and group
#> F = NaN, num df = 5, denom df = NaN, p-value = NA

Created on 2023-09-09 with reprex v2.0.2

So you can instead run Fischer's ANOVA by assuming equal variance:

library(ggstatsplot)
ggbetweenstats(df, group, t, var.equal = TRUE)

Created on 2023-09-09 with reprex v2.0.2

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M-Colley avatar M-Colley commented on June 9, 2024

Thank you very very much!!!

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