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Let's go through this hypothesis testing problem.

### Given:
- Null hypothesis (H₀): \( \mu \geq 740 \)
- Alternative hypothesis (H₁): \( \mu < 740 \)
- Sample mean (\( \bar{x} \)): 715 tons/day
- Population mean under the null hypothesis (\( \mu_0 \)): 740 tons/day
- Standard deviation (\( \sigma \)): 24 tons/day
- Sample size (\( n \)): 60 days

### Step 1: **Calculate the test statistic (z-score)**

We can calculate the z-score using the formula for the z-test:
\[
z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}
\]

Where:
- \( \bar{x} = 715 \)
- \( \mu_0 = 740 \)
- \( \sigma = 24 \)
- \( n = 60 \)

Let's compute this.

The calculated z-score is approximately \(-8.07\).

### Step 2: **Find the P-value**

Since the test is left-tailed (because \(H_1: \mu < 740\)), we need to find the P-value corresponding to this z-score.

Let's calculate the P-value for \( z = -8.07 \).

The P-value is approximately \(3.55 \times 10^{-16}\), which is an extremely small value.

### Step 3: **Conclusion**
Given that the P-value is almost 0, this is far smaller than any common significance level (e.g., \( \alpha = 0.05 \) or \( \alpha = 0.01 \)).

#### a) The P-value is approximately \( 3.55 \times 10^{-16} \).

#### b) Based on this P-value, we reject the null hypothesis \(H_0\). It is highly unlikely that the plant is operating properly, as the data suggests the mean output is significantly less than 740 tons/day. Therefore, it is reasonable to conclude that the plant is not operating properly.

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Would you like more details on any part of the solution or further clarification?

Here are some related questions to think about:
1. What does the z-score tell you about the sample compared to the population?
2. How would the decision change if the sample size was smaller?
3. What is the role of the standard deviation in hypothesis testing?
4. How would you interpret the P-value in real-world terms?
5. How might this analysis change with a two-tailed test?

**Tip:** Always compare the P-value to the significance level to make a conclusion in hypothesis testing.

When it is operating properly, a chemical plant has a mean daily production of at least 740 tons. The output is measured on a simple random sample of 60 days. The sample had a mean of 715 tons/day and a standard deviation of 24 tons/day. An engineer tests H₀: μ ≥ 740 versus H₁: μ < 740. a) Find the P-value. b) Do you believe it is plausible that the plant is operating properly or are you convinced that the plant is not operating properly? Explain your reasoning.

Learn how to solve a hypothesis testing problem where the mean production of a chemical plant is evaluated. With a sample mean of 715 tons/day and a standard deviation of 24 tons/day, discover how to calculate the P-value and assess if the plant is operating as expected. Ideal for college-level statistics.

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