---
# System prepended metadata

title: Chi-Square Explained

---

Reference: https://chat.openai.com/share/8b953344-98a2-4431-a765-66a01affd28f

The chi-square test is a statistical test that determines if there is a significant association between two categorical variables in a sample. It is used in various fields, including biology, marketing, social sciences, and more, for tasks such as testing independence, homogeneity, and goodness of fit.

### Types of Chi-Square Tests

1. **Chi-Square Test of Independence**: It tests whether two categorical variables are independent of each other or not. It is often used with contingency tables.

2. **Chi-Square Test of Homogeneity**: It tests whether different samples have the same distribution of a categorical variable.

3. **Chi-Square Goodness of Fit Test**: It tests whether the observed sample distribution fits a given theoretical distribution.

### How It Works

The chi-square statistic is calculated as

$$
X^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}
$$

where: 

$$(X^2): Chi-square statistic$$
$$(O_{ij}): Observed frequency$$
$$(E_{ij}): Expected frequency if variables are independent$$ 


### Example

Let's take an example where we want to test the independence between two categorical variables: Gender (Male, Female) and Preference (Tea, Coffee).

Here is the observed data in a contingency table:

|         | Tea | Coffee | Total |
|---------|-----|--------|-------|
| Male    | 30  | 10     | 40    |
| Female  | 5   | 45     | 50    |
| Total   | 35  | 55     | 90    |

We will calculate the expected frequencies under the assumption that Gender and Preference are independent, and then use the chi-square formula to calculate the chi-square statistic.

### Calculation

The expected frequency for each cell is calculated as


$$ E_{ij} = \frac{{\text{{row total}} \times \text{{column total}}}}{\text{{grand total}}}$$

For example, for Male-Tea, the expected frequency is $$( \frac{{40 \times 35}}{90} = 15.56 )$$.

We'll compute the chi-square statistic using the formula above.

Let's calculate it.

### Results

With a p-value of $$(1.30 \times 10^{-9})$$, which is less than the common alpha level of 0.05, we reject the null hypothesis that gender and preference are independent. This means there is a significant association between gender and preference for tea or coffee in the sample data.

### References

1. Agresti, A. (2007). An Introduction to Categorical Data Analysis (2nd ed.). Wiley-Interscience. 
   - This book provides a comprehensive introduction to methods for categorical data analysis, including the chi-square test.

2. McDonald, J.H. (2014). Handbook of Biological Statistics (3rd ed.). Sparky House Publishing.
   - This handbook is a practical guide to statistical methods for analyzing biological data, with a section dedicated to the chi-square test.

### Note

Remember, the chi-square test has some assumptions, such as the expectation that no cell in the table has an expected count less than 5. It is always essential to check these assumptions before interpreting the results.

I hope this helps! Let me know if you have further questions or need additional clarification on any points.

The p-value is a measure that helps us understand the evidence against a specified statistical model, such as a null hypothesis. In the context of the chi-square test, the p-value is calculated based on the chi-square statistic and the degrees of freedom of the test. The p-value indicates the probability of obtaining test results at least as extreme as the ones observed during the test, assuming that the null hypothesis is true.

### How the P-Value is Calculated

1. **Calculate the Chi-Square Statistic**: This is calculated using the formula

$$
X^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}
$$

where:
$$(X^2\): Chi-square statistic$$
$$(O_{ij}): Observed frequency$$
$$(E_{ij}): Expected frequency if variables are independent$$

2. **Determine the Degrees of Freedom**: The degrees of freedom depend on the number of categories in the categorical variables being tested. For a test of independence or homogeneity,

$$
\text{df} = (\text{number of rows} - 1) \times (\text{number of columns} - 1)
$$

3. **Compute the P-Value**: Using the chi-square statistic and the degrees of freedom, the p-value is calculated. It's the area under the chi-square distribution curve to the right of the calculated chi-square value.

### Relationship

- **Chi-Square Statistic and Expected Frequencies**: The chi-square statistic is a measure of the difference between observed and expected frequencies. A larger chi-square value indicates a larger difference between observed and expected frequencies, suggesting that the variables are not independent.

- **Chi-Square Statistic and P-Value**: The chi-square statistic is used to calculate the p-value. A larger chi-square value will result in a smaller p-value. The p-value tells us the probability of observing a chi-square statistic as extreme as, or more extreme than, the statistic calculated from the sample data, under the assumption that the null hypothesis is true.

- **P-Value and Decision Making**: The p-value is compared to a predetermined significance level (often 0.05). If the p-value is less than this significance level, we reject the null hypothesis in favor of the alternative hypothesis. In the context of the chi-square test, this would mean concluding that there is a significant association between the two categorical variables.

### Example

In the previous example, we calculated a chi-square statistic of approximately 36.82 with 1 degree of freedom. We then used these values to calculate the p-value. Because the p-value was much smaller than the typical alpha level of 0.05, we rejected the null hypothesis, concluding that there is a significant association between the two variables (gender and preference for tea or coffee).

### Reference

- Everitt, B.S., & Skrondal, A. (2010). The Cambridge Dictionary of Statistics (4th ed.). Cambridge University Press.
   - This statistical dictionary provides clear and concise definitions of key statistical terms, including p-value and chi-square statistic, and can be a useful reference for understanding these concepts.

I hope this explanation clarifies the concept of the p-value, its calculation, and its relationship with the chi-square statistic and expected frequencies. If you have further questions, feel free to ask!


### Hypothesis Testing

You've touched on an important aspect of hypothesis testing. In practice, it's more common to compare the p-value to the significance level (\(\alpha\)) rather than comparing the chi-square statistic to the critical value, although both methods are mathematically equivalent and will lead to the same conclusion.

### Comparing p-value to \(\alpha\)

This approach is straightforward and widely used in practice. You calculate the p-value from the chi-square test and compare it to your chosen significance level (\(\alpha\)). If the p-value is less than \(\alpha\), you reject the null hypothesis.

### Comparing chi-square statistic to the critical value

This is an alternative method. You compare the calculated chi-square statistic to the critical value from the chi-square distribution given the degrees of freedom and chosen significance level. If the calculated chi-square statistic is greater than the critical value, you reject the null hypothesis.
Code Reference:https://chat.openai.com/share/918b9a6d-7092-4924-8535-b2c91627bea8
```
# Define the significance level (e.g., 0.05)
alpha <- 0.05

# Degrees of freedom (df) from your chi-square test
df <- 2  # Replace with the appropriate df from your test

# Calculate the critical value from the chi-square distribution table
critical_value <- qchisq(1 - alpha, df)

# Your calculated chi-square statistic from the test
calculated_chi_square <- 1.5858  # Replace with your calculated value

# Compare the calculated chi-square statistic with the critical value
if (calculated_chi_square > critical_value) {
  cat("There is a significant difference. Reject the null hypothesis.\n")
} else {
  cat("There is no significant difference. Fail to reject the null hypothesis.\n")
}

```

### Why p-value comparison is more common

1. **Interpretability**: The p-value is easier to interpret. It represents the probability of observing a test statistic as extreme, or more extreme, than the one calculated from your sample data, assuming the null hypothesis is true. A smaller p-value indicates stronger evidence against the null hypothesis.

2. **Consistency**: Using the p-value for hypothesis testing is consistent across various types of statistical tests, not just the chi-square test. It provides a uniform way to interpret the results of these tests.

3. **Adjustment for Multiple Testing**: When performing multiple tests, it's often necessary to adjust p-values to control the Type I error rate. This process is more intuitive when working directly with p-values.

### Revision to the R Code

Here is how you might revise the code to use the p-value instead of the critical value for decision making:

```r
# Perform chi-squared test
chi_squared_result <- chisq.test(contingency_table)

# Print the results
print(chi_squared_result)

# Extract p-value from chi_squared_result
p_value <- chi_squared_result$p.value
alpha <- 0.05  # You can replace this with your desired significance level

# Compare the p-value with alpha
if (p_value < alpha) {
  cat("There is a significant difference. Reject the null hypothesis.\n")
} else {
  cat("There is no significant difference. Fail to reject the null hypothesis.\n")
}
```

### Limitations

The chi-square test is a powerful tool for analyzing categorical data, but it has several limitations and assumptions that need to be considered. Here are some of the key limitations:

### 1. **Categorical Data Only**
The chi-square test can only be used for categorical or nominal data. It is not appropriate for continuous or ordinal data unless they have been categorized.

### 2. **Expected Frequency Assumption**
Every cell in the contingency table should have an expected frequency of 5 or more for the chi-square approximation to be valid. This is especially a concern in small samples or when there are many categories.

### 3. **Independence of Observations**
The observations should be independent of each other. This means that the sampling method should be simple random sampling. This assumption is violated if the data are dependent.

### 4. **Fixed Marginals**
In the test of homogeneity, it is assumed that the marginal totals are fixed, which is often not the case in real-world data.

### 5. **Mutually Exclusive Categories**
Each participant/observation should belong to one and only one category for both variables. Overlapping or nested categories can violate the assumptions of the test.

### 6. **Large Sample Size**
Chi-square tests can be sensitive to sample size. A very large sample size can detect a statistically significant result even for a small effect size or deviation from the expected frequencies. On the other hand, a small sample size might not provide a reliable result.

### 7. **No Information on Strength or Direction of Association**
The test indicates whether an association exists between the categorical variables but does not provide information on the strength or direction of the association.

### 8. **Assumption of Randomness**
The data should be collected randomly to ensure that the sample is representative of the population. Any bias in data collection can lead to inaccurate results.

### 9. **Two-Dimensional Tables**
The traditional chi-square tests are applicable for two-dimensional tables. For multi-dimensional tables, the interpretation becomes complex and might require other sophisticated methods or modifications.

### 10. **Cannot Test for Causal Relationships**
Like many statistical tests, the chi-square test can indicate an association between two variables but cannot prove causation.

### References

1. **Agresti, A. (2007). An Introduction to Categorical Data Analysis (2nd ed.). Wiley-Interscience.**
    - This comprehensive book offers detailed insights into methods for categorical data analysis, including the chi-square test, and discusses its assumptions and limitations.

2. **McDonald, J.H. (2014). Handbook of Biological Statistics (3rd ed.). Sparky House Publishing.**
    - This handbook provides practical insights into statistical methods for biological data, including a section on the chi-square test, and discusses when and how to use it effectively.

3. **Siegel, S., & Castellan, N. J. (1988). Nonparametric Statistics for The Behavioral Sciences (2nd ed.). McGraw-Hill.**
    - This classic text offers a deep dive into nonparametric statistics, including the chi-square test, and is a great resource for understanding the limitations and applications of these methods in behavioral sciences.

4. **Everitt, B. S., & Skrondal, A. (2010). The Cambridge Dictionary of Statistics (4th ed.). Cambridge University Press.**
    - A dictionary of statistical terms, including the chi-square test, offering concise definitions and explanations which can be a helpful resource for understanding the limitations associated with the test.

5. **Field, A. (2013). Discovering Statistics Using IBM SPSS Statistics (4th ed.). Sage Publications.**
    - Andy Field's book is a comprehensive guide to understanding statistics using SPSS, including a detailed section on chi-square tests, their assumptions, and limitations.

Each of these resources can provide a deeper understanding of the chi-square test, its applications, and the constraints and considerations that should be taken into account when interpreting the results of the test.


### Conclusion

While the chi-square test is widely used and easy to perform, understanding its limitations and assumptions is crucial for interpreting the results accurately. Always consider these factors and, if necessary, look for alternative methods or supplementary analyses to validate your findings.

### Assumptions
Reference:https://chat.openai.com/share/ff59f22a-a3cc-4a7c-8cb3-23a0aab8e4c6
```
# Sample data for ice cream flavor preference and gender
ice_cream_data <- data.frame(
  Gender = c("Male", "Female", "Male", "Female", "Male", "Female", "Male", "Female"),
  Flavor = c("Vanilla", "Vanilla", "Chocolate", "Chocolate", "Strawberry", "Strawberry", "Other", "Other")
)

# Create a contingency table
contingency_table <- table(ice_cream_data)

# Check the contingency table
print(contingency_table)

# Calculate expected values for the chi-square test
expected_values <- chisq.test(contingency_table)$expected

# Check if assumptions are met
assumption1 <- sum(expected_values >= 5) / length(expected_values) >= 0.8
assumption2 <- all(expected_values >= 1)
sample_size_needed <- sum(dim(contingency_table)) * 5

cat("Assumption 1 (80% cells with expected >= 5): ", assumption1, "\n")
cat("Assumption 2 (No expected < 1): ", assumption2, "\n")
cat("Sample size needed to meet assumptions: ", sample_size_needed, "\n")

```

### Stacked Bar Chart
Reference: https://chat.openai.com/share/9c85ebc6-78a4-49bd-98b7-b8a655e728f7
```
# Load required libraries
library(ggplot2)

# Sample Data
data <- data.frame(
  Gender = factor(rep(c("Male", "Female", "Non-Binary"), times = c(100, 100, 100))),
  Readmis = factor(sample(c("Yes", "No"), 300, replace = TRUE))
)

# Print the data
print(head(data))

# Creating a contingency table
contingency_table <- with(data, table(Gender, Readmis))

# Print the contingency table
print(contingency_table)

# Performing Chi-Square Test of Independence
test_result <- chisq.test(contingency_table)

# Print the test result
print(test_result)

# Plotting a Stacked Bar Chart using ggplot2
ggplot(data, aes(x=Gender, fill=Readmis)) +
  geom_bar(position="stack") +
  labs(title="Stacked Bar Chart of Gender vs Readmission",
       x="Gender",
       y="Frequency") +
  scale_fill_manual(values=c("red", "blue")) +
  theme_minimal()

```
### Scatter Plot
Reference:https://chat.openai.com/share/0944ff40-646b-47dd-9fd2-1a6d83aebf28
```
# Load required libraries
library(ggplot2)

# Generate sample data
set.seed(123)
age <- rnorm(100, mean = 35, sd = 10) # normally distributed age data
income <- exp(rnorm(100, mean = log(50000), sd = 0.7)) # log-normally distributed income data

# Calculate Spearman's rank correlation coefficient
cor.test(age, income, method = "spearman")

# Create a scatter plot
ggplot(data = data.frame(age, income), aes(x = age, y = income)) +
  geom_point() +
  geom_smooth(method = "loess", color = "blue") + # Add a smoothing line
  labs(x = "Age", y = "Income", title = "Scatter Plot of Age vs Income") +
  theme_minimal()

```
### Histogram, Box Plot, Bar Chart, and Pie Chart
Reference: https://chat.openai.com/share/b93dd976-84d5-4562-bd5c-f1e8cc6736a4
```
# Load required libraries
library(ggplot2)
library(dplyr)

# Create a sample data frame
data <- data.frame(
  age = c(21, 35, 40, 50, 65, 29, 31, 55, NA, 60),
  income = c(50000, NA, 75000, 60000, 80000, 56000, 48000, 71000, 68000, 62000),
  gender = c("Male", "Female", "Male", NA, "Female", "Male", "Female", "Female", "Male", "Female"),
  region = c("North", "South", "East", "West", "East", "South", "North", "East", "West", NA)
)

# Clean and prepare data: remove NA values for demonstration simplicity
cleaned_data <- na.omit(data)

# Univariate analysis and visualization

# Continuous variable 1: Age
cat("Age Summary Statistics:\n")
print(summary(cleaned_data$age))

# Plotting histogram for Age
ggplot(cleaned_data, aes(x=age)) +
  geom_histogram(binwidth=10, fill="skyblue", color="black") +
  ggtitle("Age Distribution") +
  xlab("Age") +
  ylab("Frequency")

# Continuous variable 2: Income
cat("\nIncome Summary Statistics:\n")
print(summary(cleaned_data$income))

# Plotting boxplot for Income
ggplot(cleaned_data, aes(y=income)) +
  geom_boxplot(fill="pink") +
  ggtitle("Income Distribution") +
  ylab("Income")

# Categorical variable 1: Gender
cat("\nGender Frequency:\n")
print(table(cleaned_data$gender))

# Plotting bar chart for Gender
ggplot(cleaned_data, aes(x=gender, fill=gender)) +
  geom_bar() +
  ggtitle("Gender Distribution") +
  xlab("Gender") +
  ylab("Frequency")

# Categorical variable 2: Region
cat("\nRegion Frequency:\n")
print(table(cleaned_data$region))

# Plotting pie chart for Region
region_freq <- as.data.frame(table(cleaned_data$region))
ggplot(region_freq, aes(x="", y=Freq, fill=Var1)) +
  geom_bar(width = 1, stat = "identity") +
  coord_polar("y") +
  ggtitle("Region Distribution")

```