Chi-Square Test Calculator

Data Visualization

Chi-Square Test Calculator

What is a Chi-Square Test?

The chi-square test is a statistical method used to determine if there is a significant association between two categorical variables. It's commonly used in hypothesis testing to evaluate how likely it is that an observed distribution is due to chance.

Formula and Its Meaning

The chi-square statistic is calculated using the following formula:

\[\chi^2 = \sum_{i=1}^{n} \frac{(O_i - E_i)^2}{E_i}\]

Where:

  • \(\chi^2\) is the chi-square statistic
  • \(O_i\) is the observed frequency for category i
  • \(E_i\) is the expected frequency for category i
  • \(n\) is the number of categories

This formula measures the discrepancy between the observed data and what we would expect if there were no association between the variables.

Calculation Steps

  1. Determine the observed frequencies for each category.
  2. Calculate the expected frequencies for each category.
  3. For each category, calculate \(\frac{(O_i - E_i)^2}{E_i}\).
  4. Sum these values to get the chi-square statistic.
  5. Determine the degrees of freedom (df = number of categories - 1).
  6. Use a chi-square distribution table or calculator to find the p-value.

Example and Visual Representation

Let's consider a simple example where we're testing if a die is fair. We roll the die 60 times and get the following results:

Face Value Observed (O) Expected (E)
1510
2810
3910
41210
51510
61110

Calculating the chi-square statistic:

\[\chi^2 = \frac{(5-10)^2}{10} + \frac{(8-10)^2}{10} + \frac{(9-10)^2}{10} + \frac{(12-10)^2}{10} + \frac{(15-10)^2}{10} + \frac{(11-10)^2}{10} = 6.5\]

With 5 degrees of freedom (6 categories - 1), we can look up the p-value in a chi-square distribution table or use a calculator. In this case, the p-value is approximately 0.26.

1 2 3 4 5 6 Expected

This bar chart visualizes the observed frequencies (blue bars) compared to the expected frequency (red dashed line) for each face value of the die.

Since the p-value (0.26) is greater than the common significance level of 0.05, we fail to reject the null hypothesis. This means we don't have strong evidence to conclude that the die is unfair based on these results.