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SouravSeptember 10, 2024

how are the expected counts calculated when a chi-square goodness-of-fit test is conducted?

how are the expected counts calculated when a chi-square goodness-of-fit test is conducted?

 

Sourav
SouravSeptember 10, 2024

Answer

In a Chi-Square goodness-of-fit test, expected counts are calculated based on the null hypothesis that the observed data fits a specified distribution. Here’s a detailed, step-by-step process for calculating the expected counts:

  1. Determine the Hypothesized Distribution:
    • Identify the theoretical distribution that you expect the data to follow under the null hypothesis. This could be a uniform distribution, a normal distribution, or any other theoretical distribution relevant to your data.
  2. Calculate the Total Number of Observations:
    • Sum the observed frequencies across all categories to find the total number of observations. Let this total be denoted by NN.
  3. Determine the Proportions or Probabilities:
    • For each category, determine the proportion or probability that the null hypothesis predicts for that category. These proportions or probabilities are typically derived from the theoretical distribution. Denote these proportions as pip_i, where ii represents each category.
  4. Calculate the Expected Counts:
    • Multiply the total number of observations NN by the proportion pip_i for each category. This gives the expected count for each category. Mathematically, this can be expressed as: Ei=N×piE_i = N \times p_i

    where EiE_i is the expected count for the ii-th category.

Example Calculation:

Assume you are conducting a Chi-Square goodness-of-fit test to determine if a six-sided die is fair. You roll the die 60 times and want to test if each face appears equally often.

  1. Total Observations:

    N=60N = 60

  2. Expected Proportions:
    • For a fair die, each face should appear with equal probability, so pi=16p_i = \frac{1}{6} for each face.
  3. Calculate Expected Counts:
    • For each face, the expected count EiE_i is: Ei=N×pi=60×16=10E_i = N \times p_i = 60 \times \frac{1}{6} = 10

    Therefore, each face of the die is expected to appear 10 times.

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