p value?​(Round to three decimal places as​ needed.)

To find the p-value associated with a \( z \)-value of approximately 10.96, we look at the standard normal distribution. Since we are dealing with a one-tailed test (testing if \( p_1 \) is greater than \( p_2 \)), the p-value corresponds to the probability that \( Z \) is greater than 10.96.

Given that a \( z \)-value of 10.96 is far into the tail of the normal distribution, the p-value is extremely small. We can use statistical tables or a computational tool to find the p-value.

Using a standard normal distribution table or a computational tool, the p-value for \( z = 10.96 \) is essentially zero.

For practical purposes, we can say the p-value is:

\[ \text{p-value} \approx 0.000 \]

This p-value is rounded to three decimal places as requested.

Therefore, the p-value is essentially 0.000 when rounded to three decimal places.

Would you like more details or have any questions?

Here are five additional questions you might consider:

1. What is the critical value for a one-tailed test at a 0.05 significance level?
2. How do we interpret the results of this hypothesis test in a real-world context?
3. What are some potential sources of error in this study?
4. How could the study design be improved to provide more robust results?
5. Are there other statistical tests that could be used to compare the proportions?

**Tip:** When dealing with extremely small p-values, it's often useful to report them as being less than a certain threshold (e.g., p < 0.001) to indicate their significance clearly.

Learn how to calculate the p-value for a z-value of approximately 10.96 in a one-tailed hypothesis test comparing two proportions. This problem involves the standard normal distribution and statistical tables. Understand the significance of a p-value near zero in hypothesis testing.

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