SOLUTION: You receive a brochure from a large university. The brochure indicates that the mean class size for full-time faculty is fewer than 32students. You want to test this claim. You ran

Question 1168872: You receive a brochure from a large university. The brochure indicates that the mean class size for full-time faculty is fewer than 32students. You want to test this claim. You randomly select 18 classes taught by full time faculty and determine the class size of each. The results are listed below. At a=o.01, can you support the university's claim.
35 28 29 33 32 40 26 25 29
28 30 36 33 29 27 30 28 25
Research Question:
I. Hypotheses:
Ho: µ
Hi:µ

IV. Decision:

II. Criteria for Decision:
a =
Decision Rule:
Reject H 0 if p-value <
Ill.Test Statistics:

V. Summary:

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
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.

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