Rule of Thumb:.
Think of standardized residuals like z-scores: values greater than +2 or less than -2 are the βstatistically interestingβ parts of your table that are driving your p-value.
Section 6.6 Chi-Square Test
Now, we have done some investigation on our data through contingency tables and visualizations. It is time to run what is called a chi-square test. This allows us to see if two categorical variables are related to each other.
chisq.test(contingency_table)
Pearson's Chi-squared test data: contingency_table X-squared = 7.7759, df = 2, p-value = 0.02049
Whether we performed it on our contingency table, we get:
-
X^2- This is the test statistic. This is telling us how much different our observed is vs our expected.
-
df - Degrees of freedom = (rows - 1)*(columns - 1). For 3 treatments and 2 outcomes, df = (3-1)*(2-1) = 2.
-
p-value- This tells us if it is statistically significant or not.
We see that our X^2 value is 7.78, our df is 2, and our p-value is 0.0204871. Since our p-value is less than .05, we can rightly say that there is a significant relationship between treatment plans and outcomes.
Now that we know it is significant, we need to dive a little deeper. We are understanding that cranberry is looking like the leader, but we can become more solidified on that stance.
chi_square_test <- chisq.test(contingency_table)
# All the parts of the chi square can be called.
chi_square_test$statistic
chi_square_test$parameter
chi_square_test$p.value
chi_square_test$method
chi_square_test$data.name
# What our data already is
chi_square_test$observed
# What our data would look like if there was no relationship
chi_square_test$expected
# Difference between observed and expected
# Positive is more than expected
# Negative is less than expected
# Bigger number = bigger difference
# Represent how many standard deviations
chi_square_test$residuals
# Standard Residuals
# More like an actual z-score
chi_square_test$stdres
X-squared
7.77592
df
2
[1] 0.0204871
[1] "Pearson's Chi-squared test"
[1] "contingency_table"
Infection
Treatment No Yes
Control 32 18
Cranberry 42 8
Lactobacillus 30 20
Infection
Treatment No Yes
Control 34.66667 15.33333
Cranberry 34.66667 15.33333
Lactobacillus 34.66667 15.33333
Infection
Treatment No Yes
Control -0.4529108 0.6810052
Cranberry 1.2455047 -1.8727644
Lactobacillus -0.7925939 1.1917591
Infection
Treatment No Yes
Control -1.001671 1.001671
Cranberry 2.754595 -2.754595
Lactobacillus -1.752924 1.752924
We can also visualize our observed versus our expected counts, as visualizations are key to understanding our data.
# 1. Extract observed and expected counts from our test object
obs_exp_data <- data.frame(
Treatment = rep(rownames(chi_square_test$observed), 2),
Infection = rep(colnames(chi_square_test$observed), each = 3),
Observed = as.vector(chi_square_test$observed),
Expected = as.vector(chi_square_test$expected)
)
# 2. Pivot the data to a 'long' format for ggplot
plot_data <- obs_exp_data |>
pivot_longer(cols = c(Observed, Expected),
names_to = "Type",
values_to = "Counts")
# 3. Create the plot
ggplot(plot_data, aes(x = Treatment, y = Counts, fill = Type)) +
geom_bar(stat = "identity", position = "dodge", alpha = 0.8) +
facet_wrap(~Infection) +
scale_fill_manual(values = c("Expected" = "grey70", "Observed" = "#CC0000")) +
labs(
title = "Observed vs. Expected Infection Counts",
subtitle = "The grey bars show what we would expect by random chance alone",
x = "Treatment Group",
y = "Number of Participants"
) +
theme_minimal()
Our deep dive uncovered some very important things about our experiment:
-
Expected: all of the values were different from what we the actual observations were. If there was no relationship between the two variables, it was expected that each treatment would have about 35 people not infected, and 15 people infected
-
Residuals/stdres: this is where we really start to tie everything together. With the residuals, we are looking for if it is positive/negative, and then strength in relation to the other. Cranberry if the only one of the three where there are less people infected than anticipated, almost 2 standard deviations less than expected. Both other treatments have more people infected than expected.
-
Note: Standardized residuals behave like z-scores β values beyond Β±2 suggest cells contributing most strongly to the overall ΟΒ² statistic.
-
We tie this information together to get a very strong case for cranberry. There is just one piece of the puzzle left to bring this home.
