24 Goodness-of-Fit Test
In this section, we address how to test whether a categorical distribution fits a specified probability distribution. We have \(n\) observations categorized into \(K\) distinct categories:
| Category | Observed Count |
|---|---|
| 1 | \(O_1\) |
| 2 | \(O_2\) |
| 3 | \(O_3\) |
| \(\vdots\) | \(\vdots\) |
| \(K\) | \(O_K\) |
Each \(O_i\) represents the number of observations in category \(i\), for \(i = 1, 2, \dots, K\).
We let \(P_i\) denote the theoretical probability of an observation falling into category \(i\), where the sum of all category probabilities must satisfy:
\[ \sum_{i=1}^{K} P_i = 1 \]
We want to test whether the actual observed distribution matches a known or hypothesized distribution with fixed probabilities \(P^*_1, P^*_2, \dots, P^*_K\).
Null Hypothesis (\(H_0\)): The data follow the expected probabilities
\(H_0: P_1 = P^*_1,\ P_2 = P^*_2,\ \dots,\ P_K = P^*_K\)
Alternative Hypothesis (\(H_1\)): The data do not follow the expected probabilities.
\(H_0:\) at least one of the \(P_i\) differs from \(P^*_i\).
If the null hypothesis is true, we expect the number of observations in each category to be: \[ E_i = n \cdot P^*_i \]
This leads to the following extended table:
| Category | Observed Count \(O_i\) | Probability \(P^*_i\) | Expected Count \(E_i = n \cdot P^*_i\) |
|---|---|---|---|
| 1 | \(O_1\) | \(P^*_1\) | \(E_1\) |
| 2 | \(O_2\) | \(P^*_2\) | \(E_2\) |
| 3 | \(O_3\) | \(P^*_3\) | \(E_3\) |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
| \(K\) | \(O_K\) | \(P^*_K\) | \(E_K\) |
| Total | \(n\) | \(1\) | \(n\) |
We use the chi-square test statistic defined by: \[ \chi^2 = \sum_{i=1}^{K} \frac{(O_i - E_i)^2}{E_i} \] where:
- \(O_i\) is the observed frequency for category \(i\)
- \(E_i\) is the expected frequency for category \(i\) under \(H_0\)
This test statistic follows approximately a chi-square distribution with \((K - 1)\) degrees of freedom, assuming that all expected counts \(E_i\) are sufficiently large (usually at least 5).
A large value of \(\chi^2\) indicates a large discrepancy between observed and expected counts, which provides evidence against the null hypothesis. We compare the test statistic to a critical value from the \(\chi^2\) distribution table or compute a \(p\)-value.
If the \(p\)-value is less than the significance level (e.g., 0.05), we reject \(H_0\) and conclude that the observed distribution does not fit the expected distribution.
Example 24.1: Chi-square Goodness-of-Fit Test for Color Preference
Let’s conduct a chi-square goodness-of-fit test to examine whether a product is equally preferred in four different colors. We ask 80 randomly selected potential customers about their preferred color, and the responses are summarized as follows:
| Color | Observed Count (\(O_i\)) |
|---|---|
| 1 | 12 |
| 2 | 40 |
| 3 | 8 |
| 4 | 20 |
| Total | 80 |
Step 1: Hypotheses
We want to test if all colors are equally preferred.
\(H_0\): The colors are equally preferred, i.e., \(P_1 = P_2 = P_3 = P_4 = 0.25\)
\(H_1\): Not all colors are equally preferred.
Step 2: Significance Level
We choose a significance level of
\[\alpha = 0.05\]
Step 3: Test Statistic
We use the chi-square test statistic:
\[ \chi^2 = \sum_{i=1}^{k} \frac{(O_i - E_i)^2}{E_i} \] where:
- \(O_i\) = observed frequency
- \(E_i\) = expected frequency under \(H_0\)
- \(k = 4\) categories
Degrees of freedom:
\[
\text{df} = k - 1 = 3
\]
Step 4: Decision Rule
From the chi-square distribution table (Appendix C), the critical value at
\[ \chi^2_{3, 0.05} = 7.815 \]
Reject \(H_0\) if:
\[ \chi^2_{\text{obs}} > 7.815 \]
Step 5: Observation
We compute the expected frequency for each category:
\[ E_i = n \cdot P_i = 80 \cdot 0.25 = 20 \]
| Color | \(O_i\) | \(E_i\) | \((O_i - E_i)^2 / E_i\) |
|---|---|---|---|
| 1 | 12 | 20 | 3.20 |
| 2 | 40 | 20 | 20.00 |
| 3 | 8 | 20 | 7.20 |
| 4 | 20 | 20 | 0.00 |
| Total | 30.40 |
Now we can compute the test statistic:
\[
\chi^2_{\text{obs}} = 3.20 + 20.00 + 7.20 + 0.00 = 30.40
\]
Step 6: Conclusion
Since
\[
\chi^2_{\text{obs}} = 30.40 > 7.815 = \chi^2_{3, 0.05}
\] We reject the null hypothesis at the 5% significance level. There is a statistically significant difference in color preferences among consumers.
Example 23.2: Chi-square Goodness-of-Fit
We want to test whether the market shares of three competing products (A, B, and C) have changed after a recent modification to product C.
Step 1: Hypotheses
We assume that the historical market shares were:
- Product A: 30%
- Product B: 50%
- Product C: 20%
We test:
\(H_0: p_A = 0.30,\quad p_B = 0.50,\quad p_C = 0.20\)
\(H_1:\) The proportions are not equal to those specified under \(H_0\)
Step 2: Significance level
We use a significance level of:
\[ \alpha = 0.05 \]
Step 3: Test statistic
We use the chi-square goodness-of-fit test statistic: \[
\chi^2 = \sum_{i=1}^k \frac{(O_i - E_i)^2}{E_i}
\] Degrees of freedom: \[
df = k - 1 = 3 - 1 = 2
\]
Step 4: Decision rule
We reject the null hypothesis if: \[
\chi^2_{\text{obs}} > \chi^2_{2, 0.05} = 5.991
\]
Step 5: Observation
From a survey of 200 customers:
| Product | Observed (\(O_i\)) | Expected proportion (\(p^*_i\)) | Expected count (\(E_i = 200 \cdot p^*_i\)) |
|---|---|---|---|
| A | 48 | 0.30 | 60 |
| B | 98 | 0.50 | 100 |
| C | 54 | 0.20 | 40 |
Now compute the observed value of the test statistic: \[ \chi^2_{\text{obs}} = \frac{(48 - 60)^2}{60} + \frac{(98 - 100)^2}{100} + \frac{(54 - 40)^2}{40} = 7.34 \]
Step 6: Conclusion
Since: \[
\chi^2_{\text{obs}} = 7.34 > \chi^2_{2, 0.05} = 5.991,
\] we reject the null hypothesis at the 5% significance level. There is sufficient evidence to suggest that the market shares for products A, B, and C have changed.
Exercises
- A factory produces light bulbs that are supposed to have three brightness levels in the following proportions:
Low: 20%
Medium: 50%
High: 30%
In a random sample of 200 light bulbs, the observed distribution is:
Low: 38
Medium: 92
High: 70
Test whether the observed distribution fits the expected proportions at the 5% significance level.
Step 1: Hypotheses
- \(H_0\): The population distribution follows the claimed proportions (20%, 50%, 30%).
- \(H_1\): The distribution does not follow these proportions.
Step 2: Significance level
- Use \(\alpha = 0.05\)
Step 3: Test statistic
We use the chi-square goodness-of-fit test statistic: \[
\chi^2 = \sum_{i=1}^k \frac{(O_i - E_i)^2}{E_i}
\]
Step 4: Observation
Total sample size: \(n = 200\)
Claimed proportions:
- Low: \(p_1 = 0.20\)
- Medium: \(p_2 = 0.50\)
- High: \(p_3 = 0.30\)
Expected counts:
\[ \begin{split} & E_1 = 200 \times 0.20 = 40 \\ & E_2 = 200 \times 0.50 = 100 \\ & E_3 = 200 \times 0.30 = 60 \end{split} \]
Expected frequencies:
| Category | Observed (\(O_i\)) | Expected (\(E_i\)) |
|---|---|---|
| Low | 38 | 40 |
| Medium | 92 | 100 |
| High | 70 | 60 |
Step 4: Decision rule
Degrees of freedom: \(k - 1 = 3 - 1 = 2\)
Critical value: \(\chi^2_{0.95, 2} = 5.991\)
Reject \(H_0\) if observed value > 5.991.
Step 5: Observation
Our observed value is given by: \[
\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}
\]
\[ \chi^2 = \frac{(38 - 40)^2}{40} + \frac{(92 - 100)^2}{100} + \frac{(70 - 60)^2}{60} \\ = \frac{4}{40} + \frac{64}{100} + \frac{100}{60} \\ = 0.10 + 0.64 + 1.67 = 2.41 \]
Step 6: Conclusion
Since \(2.41 < 5.991\), we fail to reject \(H_0\). There is no significant difference between the observed and expected brightness levels. The data is consistent with the claimed distribution at the 5% level.
- Suppose a state lottery randomly draws one number from the integers 1 through 5 each week. Over 100 weeks, the frequency of each number is recorded. Test whether these numbers are drawn uniformly, i.e., whether each number has an equal chance of being drawn.
Step 1: Hypotheses
\(H_0\): The numbers are uniformly distributed (random).
\(H_1\): The numbers are not uniformly distributed.
Step 2: Significance Level
Let \(\alpha = 0.05\)
Step 3: Decision Rule
Degrees of freedom = \(k - 1 = 5 - 1 = 4\)
From table; \(\chi^2_{0.05, 4} = 9.488\) which we reject if our observed value is greater than.
Step 4: Test Statistic
The Chi-square statistic is:
\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]
Step 5: Observation
If the distribution is uniform, each number is expected to occur:
\[ E = \frac{100}{5} = 20 \text{ times} \]
So, expected counts for all five categories = 20.
Calculating: \[ \begin{aligned} \chi^2 &= \frac{(15 - 20)^2}{20} + \frac{(24 - 20)^2}{20} + \frac{(19 - 20)^2}{20} + \frac{(22 - 20)^2}{20} + \frac{(20 - 20)^2}{20} \\ &= \frac{25}{20} + \frac{16}{20} + \frac{1}{20} + \frac{4}{20} + \frac{0}{20} \\ &= 1.25 + 0.8 + 0.05 + 0.2 + 0 \\ &= 2.3 \end{aligned} \]
Step 6: Conclusion
Since \(2.3 < 9.488\), we fail to reject \(H_0\). There is no evidence to suggest the lottery numbers are not drawn uniformly. The observed differences could be due to chance.