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.
| Number | Observed Frequency |
|---|---|
| 1 | 15 |
| 2 | 24 |
| 3 | 19 |
| 4 | 22 |
| 5 | 20 |
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.