SOLUTION: Hinton Press hypothesizes that the average life of its largest web press is 14,500 hours. They know that the standard deviation of press life is 2,100 hours. From a sample of 25

Question 1171743: Hinton Press hypothesizes that the average life of its largest web press is 14,500 hours. They know that the standard deviation of press life is 2,100 hours. From a sample of 25 presses, the company finds a sample mean of 13,000 hours. At a 0.01 significance level, should the company conclude that the average life of the presses is less than the hypothesized 14,500 hours?
Answer by CPhill(2189) (Show Source): You can put this solution on YOUR website!
Here's how to solve this hypothesis testing problem:
**1. Define the Hypotheses:**
* **Null Hypothesis (H₀):** The average life of the web presses is 14,500 hours (μ = 14,500).
* **Alternative Hypothesis (H₁):** The average life of the web presses is less than 14,500 hours (μ < 14,500). This is a left-tailed test.
**2. Gather the Given Information:**
* Hypothesized mean (μ₀): 14,500 hours
* Population standard deviation (σ): 2,100 hours
* Sample mean (x̄): 13,000 hours
* Sample size (n): 25
* Significance level (α): 0.01
**3. Calculate the Test Statistic (z-score):**
Since we know the population standard deviation, we'll use a z-test.
z = (x̄ - μ₀) / (σ / √n)
z = (13,000 - 14,500) / (2,100 / √25)
z = -1,500 / (2,100 / 5)
z = -1,500 / 420
z ≈ -3.57
**4. Determine the P-value:**
* We have a left-tailed test, so we need to find the probability of getting a z-score of -3.57 or less.
* Using a z-table or calculator, the p-value is approximately 0.0002.
**5. Compare the P-value to the Significance Level:**
* p-value (0.0002) < α (0.01)
**6. Make a Decision:**
* Since the p-value is less than the significance level, we reject the null hypothesis.
**7. Conclusion:**
* There is sufficient evidence to conclude that the average life of the web presses is less than 14,500 hours at a 0.01 significance level.

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