Question 1168872
Let's break down this hypothesis test step-by-step.

**I. Hypotheses:**

* **Null Hypothesis (H₀):** The mean class size for full-time faculty is greater than or equal to 32 students.
    * H₀: µ ≥ 32
* **Alternative Hypothesis (H₁):** The mean class size for full-time faculty is less than 32 students.
    * H₁: µ < 32

**II. Criteria for Decision:**

* **a (alpha):** 0.01 (given in the problem)
* **Decision Rule:** Since the alternative hypothesis is a "less than" test (left-tailed), we will reject H₀ if the p-value is less than 0.01.

**III. Test Statistics:**

1.  **Calculate the Sample Mean (x̄):**
    * Sum of class sizes: 35 + 28 + 29 + 33 + 32 + 40 + 26 + 25 + 29 + 28 + 30 + 36 + 33 + 29 + 27 + 30 + 28 + 25 = 523
    * Sample size (n): 18
    * x̄ = 523 / 18 ≈ 29.0556

2.  **Calculate the Sample Standard Deviation (s):**
    * Using a calculator or statistical software, we find the sample standard deviation (s) ≈ 4.195

3.  **Calculate the Test Statistic (t):**
    * Since the population standard deviation is unknown and the sample size is small (n < 30), we will use a t-test.
    * t = (x̄ - µ) / (s / √n)
    * t = (29.0556 - 32) / (4.195 / √18)
    * t ≈ -2.94

4.  **Degrees of Freedom (df):**
    * df = n - 1 = 18 - 1 = 17

5.  **Calculate the p-value:**
    * Using a t-distribution table or statistical software, we find the p-value for a t-statistic of -2.94 with 17 degrees of freedom (left-tailed test).
    * p-value ≈ 0.0046

**IV. Decision:**

* **Compare the p-value to alpha:** 0.0046 < 0.01
* **Decision:** Reject the null hypothesis (H₀).

**V. Summary:**

* The calculated t-statistic is approximately -2.94.
* The p-value is approximately 0.0046.
* Since the p-value (0.0046) is less than the significance level (0.01), we reject the null hypothesis.
* **Conclusion:** There is sufficient evidence at the α = 0.01 level to support the university's claim that the mean class size for full-time faculty is fewer than 32 students.