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The image shows a hypothesis test with a two-tailed alternative hypothesis \( H_1: \mu \neq 6 \). 

For part (a), the test statistic is \( z = 1.21 \), and the calculated \( P \)-value is 0.2263. 

Now, in part (b), the test statistic is \( z = -0.99 \). For a two-tailed test, we need to find the \( P \)-value corresponding to \( |z| = 0.99 \).

1. **Calculate the P-value for \( z = -0.99 \)**:
   - Using standard normal distribution tables or a calculator:
     - For \( z = 0.99 \), the one-tailed area is approximately 0.1611.
     - For a two-tailed test, the \( P \)-value is \( 2 \times 0.1611 = 0.3222 \).

So, the \( P \)-value is approximately **0.3222**.

Would you like a detailed explanation of the interpretation of the \( P \)-value, or have any other questions?

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Here are five questions to deepen understanding:

1. Why is the \( P \)-value multiplied by 2 in a two-tailed test?
2. How do we interpret a \( P \)-value of 0.3222 in the context of hypothesis testing?
3. What does the \( z \)-score represent in this type of hypothesis test?
4. How would the \( P \)-value change if this were a one-tailed test?
5. When is it appropriate to use a two-tailed vs. a one-tailed test?

**Tip:** In hypothesis testing, a smaller \( P \)-value (e.g., less than 0.05) suggests stronger evidence against the null hypothesis.

A test is made of: H₀ : μ = 6 versus H₁ : μ ≠ 6. (b) The test statistic is z = -0.99. Find and interpret the P-value. Round your answer to at least 4 decimal places.

Learn how to calculate and interpret the P-value for a two-tailed hypothesis test with z = -0.99. This solution demonstrates the use of the standard normal distribution and provides step-by-step guidance for rounding the P-value to four decimal places.

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