Suppose that Crown Bottling Company decides to use a level of significance of α = 0.01, and suppose a random sample of 32 bottle fills is obtained from a test run of the filler. For each of the following four sample means—  x⎯⎯
  = 16.06,  x⎯⎯
  = 15.95,  x⎯⎯
  = 16.01, and  x⎯⎯
  = 15.97 — determine whether the filler’s initial setup should be readjusted. In each case, use a critical value, a p-value, and a confidence interval. Assume that σ equals .1. (Round your z to 2 decimal places and p-value to 4 decimal places and CI to 3 decimal places.)

 

To determine whether the filler’s initial setup should be readjusted based on the given sample means, we need to perform hypothesis testing for each sample mean. We'll use the following information:
- Level of significance (α) = 0.01
- Sample size (n) = 32
- Population standard deviation (σ) = 0.1
- Hypothesized population mean (µ₀) = 16.0 (assuming this is the target fill amount)

We'll test the hypothesis:
- Null hypothesis (H₀): µ = 16.0
- Alternative hypothesis (H₁): µ ≠ 16.0

We will calculate the z-score, p-value, and 99% confidence interval for each sample mean. Let's proceed step by step for each sample mean:

### 1. For \( \bar{x} = 16.06 \)

**Z-score Calculation:**
\[ z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{16.06 - 16.0}{0.1 / \sqrt{32}} = \frac{0.06}{0.01768} \approx 3.39 \]

**P-value Calculation:**
Since this is a two-tailed test:
\[ \text{p-value} = 2 \times P(Z > 3.39) \approx 2 \times 0.0003 = 0.0006 \]

**99% Confidence Interval:**
\[ \bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} \]
For 99% confidence, \( z_{\alpha/2} \approx 2.576 \):
\[ 16.06 \pm 2.576 \times 0.01768 \approx 16.06 \pm 0.046 \]
\[ \text{CI} \approx (16.014, 16.106) \]

### 2. For \( \bar{x} = 15.95 \)

**Z-score Calculation:**
\[ z = \frac{15.95 - 16.0}{0.1 / \sqrt{32}} = \frac{-0.05}{0.01768} \approx -2.83 \]

**P-value Calculation:**
Since this is a two-tailed test:
\[ \text{p-value} = 2 \times P(Z < -2.83) \approx 2 \times 0.0023 = 0.0046 \]

**99% Confidence Interval:**
\[ 15.95 \pm 2.576 \times 0.01768 \approx 15.95 \pm 0.046 \]
\[ \text{CI} \approx (15.904, 15.996) \]

### 3. For \( \bar{x} = 16.01 \)

**Z-score Calculation:**
\[ z = \frac{16.01 - 16.0}{0.1 / \sqrt{32}} = \frac{0.01}{0.01768} \approx 0.57 \]

**P-value Calculation:**
Since this is a two-tailed test:
\[ \text{p-value} = 2 \times P(Z > 0.57) \approx 2 \times 0.2843 = 0.5686 \]

**99% Confidence Interval:**
\[ 16.01 \pm 2.576 \times 0.01768 \approx 16.01 \pm 0.046 \]
\[ \text{CI} \approx (15.964, 16.056) \]

### 4. For \( \bar{x} = 15.97 \)

**Z-score Calculation:**
\[ z = \frac{15.97 - 16.0}{0.1 / \sqrt{32}} = \frac{-0.03}{0.01768} \approx -1.70 \]

**P-value Calculation:**
Since this is a two-tailed test:
\[ \text{p-value} = 2 \times P(Z < -1.70) \approx 2 \times 0.0446 = 0.0892 \]

**99% Confidence Interval:**
\[ 15.97 \pm 2.576 \times 0.01768 \approx 15.97 \pm 0.046 \]
\[ \text{CI} \approx (15.924, 16.016) \]

### Conclusion:
- For \( \bar{x} = 16.06 \): Reject H₀ (since p-value < 0.01 and the CI does not contain 16.0)
- For \( \bar{x} = 15.95 \): Reject H₀ (since p-value < 0.01 and the CI does not contain 16.0)
- For \( \bar{x} = 16.01 \): Do not reject H₀ (since p-value > 0.01 and the CI contains 16.0)
- For \( \bar{x} = 15.97 \): Do not reject H₀ (since p-value > 0.01 and the CI contains 16.0)

Would you like any further details or explanations on these results? Here are five questions you might ask:

1. Can you explain more about the hypothesis testing process?
2. What is the significance of the p-value in this context?
3. How is the confidence interval interpreted in this scenario?
4. What steps should be taken if the null hypothesis is rejected?
5. Can you provide more examples of similar hypothesis tests?

Tip: Always check if the sample size is sufficiently large when performing hypothesis tests involving the normal distribution.

Explore how to determine if the filler setup at Crown Bottling Company needs adjustment through hypothesis testing. Analyze four sample means (16.06, 15.95, 16.01, 15.97) using a significance level of 0.01, critical values, p-values, and 99% confidence intervals.

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