To calculate the p-value for this test, let's go through the given information:

• Sample size (n) = 57
• Sample mean ($\bar{x}$) = 16.42
• Standard deviation (s) = 6.55
• Test statistic (t) = 1.637 (rounded from 1.6368)
• Significance level ($\alpha$) = 0.05

The test is for whether the mean hours of outdoor activities is greater than 15. Since this is a one-tailed test, we need to determine the p-value for the given test statistic (1.637) under a t-distribution with degrees of freedom (df) = n - 1 = 57 - 1 = 56.

Step 1: Look up the t-distribution table

Using the t-distribution table for $df = 56$, we need to find the range of p-values for $t = 1.637$.

Since we may not have direct access to 56 degrees of freedom on most standard tables, you can often use 55 or 60 for approximation:

• For t = 1.637 with df = 56, the p-value lies between 0.05 and 0.10 in a one-tailed test, based on the closest values in a typical t-table.

Conclusion

So, we can conclude the following:

• Blank 1 (lower bound) = 0.05
• Blank 2 (upper bound) = 0.10

The p-value lies between 0.05 and 0.10.

Let me know if you have more questions or would like further explanation on any part of this!

Additional Related Questions:

1. How would the p-value change if we were conducting a two-tailed test instead?
2. What is the critical t-value for a one-tailed test with $\alpha = 0.05$ and 56 degrees of freedom?
3. How does the sample size affect the width of the confidence interval in a t-test?
4. Why is it important not to round intermediate calculations until the final step?
5. What assumptions are made in a t-test for a sample mean?

Tip: For higher accuracy, always refer to a detailed t-table or statistical software, especially when degrees of freedom don't match exactly with standard tables.