Question 1186603
Here's how to perform a hypothesis test for this situation:

**1. State the Hypotheses:**

*   **Null Hypothesis (H₀):** There is no significant difference between the sample mean and the population mean.  μ = 73
*   **Alternative Hypothesis (H₁ or Ha):** There is a significant difference between the sample mean and the population mean. μ ≠ 73 (This is a two-tailed test.)

**2. Determine the Level of Significance (alpha):**

*   α = 0.05 (This is given in the problem.)

**3. Calculate the Test Statistic:**

*   Since the sample size is large (n = 200), we can use a z-test.
*   **Sample Mean (x̄):** x̄ = 78
*   **Population Mean (μ):** μ = 73
*   **Population Standard Deviation (σ):** σ = 8
*   **Standard Error (SE):** SE = σ / sqrt(n) = 8 / sqrt(200) ≈ 0.566
*   **Z-score:** z = (x̄ - μ) / SE = (78 - 73) / 0.566 ≈ 8.83

**4. Determine the P-value:**

*   Since this is a two-tailed test, we need to find the probability of getting a z-score as extreme as 8.83 in *either* direction.
*   Given a z-score of 8.83, the p-value will be extremely small, essentially close to 0.  Most statistical tables won't even list such a high z-score.

**5. Make a Decision:**

*   **Compare the p-value to alpha:** The p-value (≈ 0) is *much less* than the level of significance (0.05).

*   **Decision:** Because the p-value is less than alpha, we reject the null hypothesis.

**6. Conclusion:**

There is very strong evidence at the 0.05 level of significance to conclude that there is a significant difference between the mean score of the school managers and the population mean.  The sample data suggests that the school managers have a significantly higher mean score on the Leadership Skills Test than the population used for standardization.