Chi-Squared Test for Goodness-of-Fit Study Notes

Chi-Squared Test for Goodness-of-Fit

  • Overview of Chi-Squared Test for Goodness-of-Fit

    • Chi-squared test is a type of hypothesis test used to determine if there is a significant difference between expected frequencies and observed frequencies in categorical data.

    • The test is typically grounded in probability experiments where different outcomes are expected based on specific hypotheses.

  • Hypotheses in Chi-Squared Testing:

    • Null Hypothesis (H0): States that there is no significant difference between the expected and observed probabilities.

    • For a fair die, the null hypothesis would state that the probability of rolling each side (1 to 6) is equal (i.e., $P = \frac{1}{6}$ for each side).

    • Alternative Hypothesis (H1): Proposes that there is a significant difference between the expected and observed probabilities.

    • For example, it might suggest that the die is not fair and the rolls vary from the rough expectation of $P = \frac{1}{6}$.

  • Understanding Expected Frequencies

    • Expected frequencies refer to the number of occurrences that would be expected if the null hypothesis were true.

    • If rolling a die 30 times, the expected frequency for rolling a 1 would be calculated as:

    • E1=Total Rolls×P(1)=30×16=5E_1 = \text{Total Rolls} \times P(1) = 30 \times \frac{1}{6} = 5

  • Observed Frequencies

    • Observed frequencies refer to the actual counts of occurrences from the experiment (the results of the die rolls).

  • Calculation of Chi-Squared Test Statistic ($\chi^2$)

    • The formula for calculating the chi-squared value is:
      χ2=∑(O<em>i−E</em>i)2Ei\chi^2 = \sum \frac{(O<em>i - E</em>i)^2}{E_i}
      where:

    • $O_i$ = observed frequency of outcome i

    • $E_i$ = expected frequency of outcome i

    • The summation is done over all possible outcomes.

    • Example of computation based on rolling a die:

    • Browse through the observed and expected values!

  • Distribution of Chi-Squared Test Statistic

    • The chi-squared statistic is distributed according to the chi-squared distribution, which is skewed and defined by the degrees of freedom (df).

    • For a die with 6 faces, degrees of freedom would be calculated as:

    • df=k−1df = k - 1

      • Where k = number of outcomes (for a fair die, k=6, thus $df = 6 - 1 = 5$).

  • Rejecting or Failing to Reject the Null Hypothesis

    • The rejection criterion involves comparing the calculated chi-squared statistic with a critical value from a chi-squared distribution table based on chosen significance level and degrees of freedom.

    • For instance, with a significance level of $\alpha = 0.025$ and $df = 5$, the critical value ($\chi_{\alpha}^2$) can be looked up in a table, yielding 12.833.

    • If the calculated chi-squared statistic exceeds this critical value, we reject the null hypothesis (

    • If $\chi^2 \geq \chi{\alpha}^2$ then reject $H0$).

  • Example Experiment

    • Performing an experiment by rolling a die 36 times; the observed occurrences were:

    • 1: 5

    • 2: 5

    • 3: 6

    • 4: 8

    • 5: 5

    • 6: 7

    • Calculate observed differences from expected:

      • E.g. for die rolls: $E = 6$ for each side.

    • Compute $\chi^2$ using formula:

    • χ2=(5−6)26+(5−6)26+(6−6)26+(8−6)26+(5−6)26+(7−6)26\chi^2 = \frac{(5-6)^2}{6} + \frac{(5-6)^2}{6} + \frac{(6-6)^2}{6} + \frac{(8-6)^2}{6} + \frac{(5-6)^2}{6} + \frac{(7-6)^2}{6}

    • Sum gives $\chi^2 = 1.33$.

  • Conclusion

    • Since calculated chi-squared (1.33) is less than the critical value (12.833), we fail to reject the null hypothesis.

    • Result suggests that the die is likely fair, as expected frequencies are consistent with observed frequencies based on the rolls.