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.