Chi-Square Goodness of Fit Test Examples and Procedures

Overview of Chi-Square Goodness of Fit Test Examples

• The Chi-Square Goodness of Fit test is used to determine how well observed data fits an expected distribution.

• There are two primary ways to determine expected proportions:

  • Assuming all categories are equal: This treats every category as having the same probability of occurrence.

  • Based on previous research: This uses existing data or historical statistics to define what the proportions should be today.

Example 1: Assuming All Categories are Equal (The Fair Die)

• Problem Context: A gambler wants to know if a six-sided die is fair.

• Definition of a Fair Die: A die is considered "fair" if it rolls all six sides equally. This is analogous to a "fair coin" providing heads and tails with equal frequency.

• Data Collection: The gambler throws the die a total of $n = 50$ times.

• Observed Frequencies ($f_o$): These are the actual data points collected from the experiment.

  • Side 1: $8$

  • Side 2: $6$

  • Side 3: $10$

  • Side 4: $9$

  • Side 5: $6$

  • Side 6: $11$

  • Total observations ($n$): $50$.

Step 1: State the Null Hypothesis ($H_0$)

• $H_0$: The die is fair.

• In a Chi-Square test, $H_0$ states the expected proportions. For a fair six-sided die, the proportion for each side is expected to be $1/6$.

• Individual Expected Proportions: $P_1 = 1/6, P_2 = 1/6, P_3 = 1/6, P_4 = 1/6, P_5 = 1/6, P_6 = 1/6$.

• Alternative Hypothesis ($H_1$): The die is not fair (at least one side is not rolled $1/6$ of the time).

Step 2: Calculate Expected Frequencies ($f_e$) and Chi-Square Obtained ($\chi^2_{obt}$)

• Formula for Expected Frequency: $f_e = n \times \text{expected proportion}$.

• Calculation: $50 \times (1/6) = 50/6 = 8.33$.

• Because we assume all categories are equal, the $f_e$ for every side is $8.33$.

• Self-Check: The sum of the expected frequencies must equal the sample size ($n$).

  • $8.33 \times 6 = 49.98$.

  • The discrepancy (off by $0.02$) is due to rounding error from using two decimal places instead of infinity. For the purpose of the test, this is considered equal to $50$.

• Formula for Chi-Square Obtained: $\chi^2 = \sum \frac{(f_e - f_o)^2}{f_e}$

• Calculations for each category:

  • Side 1: $\frac{(8.33 - 8)^2}{8.33} = 0.01$

  • Side 2: $\frac{(8.33 - 6)^2}{8.33} = 0.65$

  • Side 3: $\frac{(8.33 - 10)^2}{8.33} = 0.34$

  • Side 4: $\frac{(8.33 - 9)^2}{8.33} = 0.05$

  • Side 5: $\frac{(8.33 - 6)^2}{8.33} = 0.65$

  • Side 6: $\frac{(8.33 - 11)^2}{8.33} = 0.85$

• Total $\chi^2_{obt}$: Adding these six values gives $2.55$.

• Calculator Warning: When squaring values, always ensure you handle negatives correctly (e.g., using parentheses) so the result is positive; a squared number cannot be negative.

Step 3: Find Degrees of Freedom ($df$) and Chi-Square Critical ($\chi^2_{crit}$)

• Degrees of Freedom formula: $df = c - 1$, where $c$ is the number of categories.

• Calculation: $6 - 1 = 5$.

• Decision Criteria: Use alpha ($\alpha$) = $0.05$.

• Looking at the distribution table for $df = 5$ and $\alpha = 0.05$, the critical value is $11.07$.

Step 4: Decision and Conclusion

• Comparison: Is \chi^2_{obt} > \chi^2_{crit}?

• 2.55 > 11.07 is False.

• Decision: Fail to reject $H_0$.

• Retain Hypothesis: We keep $H_0$, meaning there is no significant evidence to suggest the die is unfair. We did not "prove" it is fair, but we failed to find evidence to throw the assumption out.

• Conclusion: The die is rolled $1/6$ of the time for each side; the die is fair.

• APA Format: \chi^2(5, n = 50) = 2.55, p > 0.05 (Note: p > 0.05 because we failed to reject the null hypothesis).

Example 2: Proportions from Previous Research (Student Major Trends)

• Problem Context: 15 years ago, the Chaffey Psychology Department surveyed Social Science 10 students and found specific proportions of majors. An instructor today wants to see if these proportions have changed.

• Past Research Proportions ($15$ years ago):

  • Psychology: $70\%$

  • Sociology: $20\%$

  • All other majors: $10\%$

• Current Data ($n = 30$ students):

  • Psychology observed: $12$

  • Sociology observed: $10$

  • Other observed: $8$

• Objective: Use $\alpha = 0.05$ to see if student major distributions have changed.

Step 1: State the Null Hypothesis ($H_0$)

• $H_0$: The proportions today are the same as they were 15 years ago ($70\%, 20\%, 10\%$).

• This implies the status quo has not changed over the passage of time.

Step 2: Calculate Expected Frequencies ($f_e$) and Chi-Square Obtained ($\chi^2_{obt}$)

• Calculate $f_e$ based on the sample size ($n=30$):

  • Psych: $30 \times 0.70 = 21$

  • Sociology: $30 \times 0.20 = 6$

  • Other: $30 \times 0.10 = 3$

• Verification: $21 + 6 + 3 = 30$. The total expected equals the total observed.

• Initial Observation: Comparing observed ($12, 10, 8$) to expected ($21, 6, 3$), there appear to be large differences. We observe fewer Psych majors and more "Other" majors than expected.

• Chi-Square Calculation:

  • Psych: $\frac{(21 - 12)^2}{21} = 3.86$

  • Sociology: $\frac{(6 - 10)^2}{6} = 2.67$

  • Other: $\frac{(3 - 8)^2}{3} = 8.33$

• Total $\chi^2_{obt}$: $3.86 + 2.67 + 8.33 = 14.86$.

Step 3: Find Degrees of Freedom ($df$) and Chi-Square Critical ($\chi^2_{crit}$)

• Categories ($c$) = 3 (Psych, Soc, Other).

• $df = 3 - 1 = 2$.

• For $\alpha = 0.05$ and $df = 2$, consult the table: $\chi^2_{crit} = 5.99$.

Step 4: Decision and Conclusion

• Comparison: Is \chi^2_{obt} > \chi^2_{crit}?

• 14.86 > 5.99 is True.

• Decision: Reject $H_0$.

• Conclusion: The proportions of majors today are significantly different from the past research 15 years ago. Specifically, the distribution of Psych, Sociology, and Other majors in Social Science 10 has changed.

• APA Format: \chi^2(2, n = 30) = 14.86, p < 0.05 (Note: Speaker clarifies that the APA string should list sample size as $n = 30$ rather than repeating $df$).

• Since the null hypothesis was rejected, we use p < 0.05 to indicate significance.