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To determine the P-value for this hypothesis test, where the test statistic \( z = 1.31 \), we proceed as follows:

1. Since this is a two-tailed test (as indicated by \( H_1: \mu \neq 6 \)), the P-value is calculated as:
   \[
   P = 2 \times P(Z > |z|)
   \]
   where \( Z \) is the standard normal variable.

2. Using a standard normal distribution table or a calculator:
   - Find the probability \( P(Z > 1.31) \).
   - For \( z = 1.31 \), \( P(Z > 1.31) \approx 0.0951 \).

3. Therefore, the P-value is:
   \[
   P = 2 \times 0.0951 = 0.1902
   \]

### Interpretation:
The P-value of 0.1902 suggests that if the null hypothesis \( H_0: \mu = 6 \) is true, there is about a 19.02% probability of observing a test statistic as extreme as \( z = 1.31 \) (or more extreme) purely due to random chance. 

Since the P-value is typically compared against a significance level (like \( \alpha = 0.05 \)), if \( \alpha = 0.05 \), this P-value would be too large to reject the null hypothesis. Therefore, we do not have enough evidence to reject \( H_0 \).

Would you like more details or have any questions?

Here are some related questions:
1. What does a two-tailed test imply in hypothesis testing?
2. How does one determine the significance level in hypothesis testing?
3. What does it mean if the P-value is less than the significance level?
4. How would the interpretation change if this were a one-tailed test?
5. What are the implications of not rejecting the null hypothesis?

**Tip:** Always check if a test is one-tailed or two-tailed before calculating the P-value, as this affects the interpretation.

A test is made of: H0: μ = 6 versus H1: μ ≠ 6. Part (a) The test statistic is z = 1.31. Find and interpret the P-value. Round your answer to at least 4 decimal places.

Learn how to find and interpret the P-value for a hypothesis test where H0: μ = 6 versus H1: μ ≠ 6 with a test statistic z = 1.31. This solution includes detailed steps, key formulas, and a statistical interpretation, suitable for undergraduate statistics students.

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