To find out more about the X^2 value, it is important to answer the question:.
What percentage of the X^2 value is each contribution responsible for?
Section 6.8 Contribution
If we want to hone in on this more, we can take the contributions and turn them into percentages.
# Calculate contribution to chi-square statistic
contributions <- ((chi_square_test$observed-chi_square_test$expected)^2)/chi_square_test$expected
contributions
Infection Treatment No Yes Control 0.2051282 0.4637681 Cranberry 1.5512821 3.5072464 Lactobacillus 0.6282051 1.4202899
The formula for calculating the chi-square statistic is:.
\(X^2= ((observed-expected)^2)/expected\)
percent_contributions <- contributions / chi_square_test$statistic * 100
percent_contributions
Infection Treatment No Yes Control 2.637993 5.964158 Cranberry 19.949821 45.103943 Lactobacillus 8.078853 18.265233
Through this, we discovered that about 65% of the X^2 value is due to the cranberry treatment.
You already know that visualizations really help paint the picture, and chi-square contributions are not exempt. For this, the
pheatmap package ([D.1.18]) comes in handy with the pheatmap() function..
library(pheatmap)
# Create heatmap for percentage contributions
pheatmap(percent_contributions,
display_numbers = TRUE,
cluster_rows = FALSE,
cluster_cols = FALSE,
main = "% Contribution to Chi-Square Statistic")
That glaring, deep red box? That is the cranberry yes cell!
Tying all of this together, we have discovered:
-
Cranberry has the most people that were not infected (least amount infected).
-
The chi-square test shows a significant relationship between treatment and outcome.
-
When looking at residuals, cranberry was the only treatment with less people infected than expected.
-
When looking at chi-square contributions, cranberry had the largest contribution, with about 65% of the test statistic being accounted for by cranberry.
Now, how impactful is this overall? What is the effect of treatment overall on outcome?
