The Chi-square Test is a non-parametric statistical test used to examine whether distributions of categorical variables differ from one another. It’s widely used to test the association between two categorical variables in a contingency table. The test compares the observed frequencies in each category of a contingency table with the expected frequencies, which are calculated based on the assumption that there is no association between the variables.
Chi-square Test is applicable in a wide range of disciplines, including sociology, marketing, and education, to test hypotheses about associations or differences in categorical data distributions.
The Chi-square Test of Independence is a non-parametric statistical test used to determine if there is a significant association between two categorical variables. This test assesses whether observed frequencies in a contingency table differ significantly from expected frequencies, which are calculated under the assumption of independence between the variables.
The test statistic for the Chi-square Test of Independence is calculated as follows:
\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \] where \(O_i\) represents the observed frequency, and \(E_i\) represents the expected frequency in each category.
The test statistic follows a chi-squared distribution with \((r-1)\) times \((c-1)\) degrees of freedom, where \(r\) is the number of rows and \(c\) is the number of columns in the contingency table.
A researcher wants to determine if there is an association between gender (male, female) and preference for a new product (like, dislike). The data collected is as follows:
| Gender | Like | Dislike |
|---|---|---|
| Male | 20 | 10 |
| Female | 30 | 40 |
Observed Frequencies (O): The observed frequencies are given in the table:
| Gender | Like | Dislike | Row Totals |
|---|---|---|---|
| Male | 20 | 10 | 30 |
| Female | 30 | 40 | 70 |
| Column Totals | 50 | 50 | 100 |
Expected Frequencies (E): The expected frequencies are calculated based on the assumption of independence. The expected frequency for each cell is calculated using the formula: \[ E_{ij} = \frac{(R_i \times C_j)}{N} \] where \(R_i\) is the row total, \(C_j\) is the column total, and \(N\) is the grand total.
For Male and Like: \[ E_{11} = \frac{(30 \times 50)}{100} = 15 \]
For Male and Dislike: \[ E_{12} = \frac{(30 \times 50)}{100} = 15 \]
For Female and Like: \[ E_{21} = \frac{(70 \times 50)}{100} = 35 \]
For Female and Dislike: \[ E_{22} = \frac{(70 \times 50)}{100} = 35 \]
The expected frequencies are:
| Gender | Like (E) | Dislike (E) |
|---|---|---|
| Male | 15 | 15 |
| Female | 35 | 35 |
Chi-square Test Statistic: The test statistic is calculated using the formula: \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]
For Male and Like: \[ \chi^2_{11} = \frac{(20 - 15)^2}{15} = \frac{25}{15} = 1.67 \]
For Male and Dislike: \[ \chi^2_{12} = \frac{(10 - 15)^2}{15} = \frac{25}{15} = 1.67 \]
For Female and Like: \[ \chi^2_{21} = \frac{(30 - 35)^2}{35} = \frac{25}{35} = 0.71 \]
For Female and Dislike: \[ \chi^2_{22} = \frac{(40 - 35)^2}{35} = \frac{25}{35} = 0.71 \]
The total chi-square statistic is: \[ \chi^2 = 1.67 + 1.67 + 0.71 + 0.71 = 4.76 \]
Degrees of Freedom (df): The degrees of freedom for the test is calculated as: \[ df = (r - 1) \times (c - 1) \] where \(r\) is the number of rows and \(c\) is the number of columns. In this case: \[ df = (2 - 1) \times (2 - 1) = 1 \]
Critical Value and P-value: The critical value for \(\chi^2\) at \(\alpha = 0.05\) and 1 degree of freedom can be found in chi-square distribution tables. The critical value is 3.841.
Compare the calculated \(\chi^2\) value with the critical value:
In this case, \(\chi^2 = 4.76\) which is greater than 3.841, so we reject the null hypothesis.
Alternatively, you can calculate the p-value using a chi-square distribution calculator or software. For \(\chi^2 = 4.76\) with 1 degree of freedom, the p-value is approximately 0.029.
Since the p-value (0.029) is less than the significance level (\(\alpha = 0.05\)), we reject the null hypothesis. There is sufficient evidence to conclude that there is a significant association between gender and preference for the new product.
# Data for the Chi-square Test of Independence
data <- matrix(c(20, 10, 30, 40), nrow = 2, byrow = TRUE)
# Perform the test
result <- chisq.test(data)
# Output the result
print(result)
Pearson's Chi-squared test with Yates' continuity correction
data: data
X-squared = 3.8571, df = 1, p-value = 0.04953import numpy as np
from scipy.stats import chi2_contingency
# Data for the Chi-square Test of Independence
data = np.array([[20, 10], [30, 40]])
# Perform the test
chi2, p, dof, expected = chi2_contingency(data)
# Output the result
print(f"Chi2 Statistic: {chi2}")Chi2 Statistic: 3.8571428571428577P-value: 0.04953461343562668Degrees of Freedom: 1Expected Frequencies:[[15. 15.]
[35. 35.]]The Chi-square Goodness-of-Fit Test is used to determine whether a sample distribution matches an expected distribution. This test compares the observed frequencies of categories to the frequencies expected under a specified theoretical distribution.
The test statistic for the Chi-square Goodness-of-Fit Test is calculated as follows: \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \] where \(O_i\) represents the observed frequency, and \(E_i\) represents the expected frequency for each category.
The test statistic follows a chi-squared distribution with \(k-1\) degrees of freedom, where \(k\) is the number of categories.
A company wants to know if the observed sales distribution across four regions (North, South, East, West) matches their expected distribution. The expected distribution is equal across all regions. The observed sales are as follows:
| Region | Observed Sales |
|---|---|
| North | 50 |
| South | 60 |
| East | 40 |
| West | 50 |
The company will use the Chi-square Goodness-of-Fit Test to analyze the data.
Observed Frequencies (O): The observed frequencies are given in the table:
| Region | Observed Sales (O) |
|---|---|
| North | 50 |
| South | 60 |
| East | 40 |
| West | 50 |
Expected Frequencies (E): The expected frequencies are calculated based on the assumption that the sales are equally distributed across all regions. Since the total number of observations is 200 (50 + 60 + 40 + 50 = 200) and there are four regions, the expected frequency for each region is:
\[ E = \frac{\text{Total Sales}}{\text{Number of Regions}} = \frac{200}{4} = 50 \]
So, the expected frequencies are:
| Region | Expected Sales (E) |
|---|---|
| North | 50 |
| South | 50 |
| East | 50 |
| West | 50 |
Chi-square Test Statistic: The test statistic is calculated using the formula:
\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]
For North: \[ \chi^2_{North} = \frac{(50 - 50)^2}{50} = \frac{0}{50} = 0 \]
For South: \[ \chi^2_{South} = \frac{(60 - 50)^2}{50} = \frac{100}{50} = 2 \]
For East: \[ \chi^2_{East} = \frac{(40 - 50)^2}{50} = \frac{100}{50} = 2 \]
For West: \[ \chi^2_{West} = \frac{(50 - 50)^2}{50} = \frac{0}{50} = 0 \]
The total chi-square statistic is:
\[ \chi^2 = 0 + 2 + 2 + 0 = 4 \]
Degrees of Freedom (df): The degrees of freedom for the test is calculated as:
\[ df = k - 1 \]
where \(k\) is the number of categories (regions). In this case:
\[ df = 4 - 1 = 3 \]
Critical Value and P-value: The critical value for \(\chi^2\) at \(\alpha = 0.05\) and 3 degrees of freedom can be found in chi-square distribution tables. The critical value is 7.815.
Compare the calculated \(\chi^2\) value with the critical value:
In this case, \(\chi^2 = 4\) which is less than 7.815, so we do not reject the null hypothesis.
Alternatively, you can calculate the p-value using a chi-square distribution calculator or software. For \(\chi^2 = 4\) with 3 degrees of freedom, the p-value is approximately 0.261.
Since the p-value (0.261) is greater than the significance level (\(\alpha = 0.05\)), we do not reject the null hypothesis. There is insufficient evidence to conclude that the observed sales distribution significantly differs from the expected distribution.
# Observed sales
observed <- c(50, 60, 40, 50)
# Expected sales (equal distribution)
expected <- rep(sum(observed) / length(observed), length(observed))
# Perform the test
result <- chisq.test(observed, p = expected / sum(expected))
# Output the result
print(result)
Chi-squared test for given probabilities
data: observed
X-squared = 4, df = 3, p-value = 0.2615import numpy as np
from scipy.stats import chisquare
# Observed sales
observed = np.array([50, 60, 40, 50])
# Expected sales (equal distribution)
expected = np.full(len(observed), np.mean(observed))
# Perform the test
chi2, p = chisquare(f_obs=observed, f_exp=expected)
# Output the result
print(f"Chi2 Statistic: {chi2}")Chi2 Statistic: 4.0P-value: 0.26146412994911034Observed Frequencies:[50 60 40 50]Expected Frequencies:[50. 50. 50. 50.]