Question 1185468
Here's how to conduct a Chi-Square Goodness-of-Fit test to compare the regional age distribution to the national distribution:

**1. State the Hypotheses:**

*   **Null Hypothesis (H₀):** The age distribution of participants in the region is the same as the national distribution.
*   **Alternative Hypothesis (H₁):** The age distribution of participants in the region is different from the national distribution.

**2. Calculate Expected Frequencies:**

Multiply each national proportion by the total number of regional participants (214):

*   5-year-olds: 0.09 * 214 = 19.26
*   4-year-olds: 0.42 * 214 = 89.88
*   3-year-olds: 0.20 * 214 = 42.8
*   Under 3: 0.29 * 214 = 62.06

**3. Calculate the Chi-Square Statistic:**

χ² = Σ [(Observed - Expected)² / Expected]

χ² = (21 - 19.26)² / 19.26 + (92 - 89.88)² / 89.88 + (45 - 42.8)² / 42.8 + (56 - 62.06)² / 62.06
χ² ≈ 0.163 + 0.051 + 0.157 + 0.592
χ² ≈ 0.963

**4. Determine Degrees of Freedom:**

df = Number of categories - 1 = 4 - 1 = 3

**5. Find the p-value:**

Consult a Chi-Square distribution table with df = 3.  A χ² value of 0.963 is quite low. The p-value will be much greater than 0.05

**6. Make a Decision:**

*   **Decision:** Since the p-value is greater than α = 0.05, we fail to reject the null hypothesis.

**7. Conclusion:**

There is not sufficient evidence at the α = 0.05 significance level to conclude that the age distribution of Head Start participants in this region differs from the national distribution.

**Answers:**

*   The test value (χ²) is approximately **0.963**.
*   The p-value is approximately **> 0.05** (A Chi-Square table will show a p-value > 0.05. Statistical software can give a more precise answer.)