https://file.aiquickdraw.com/mathbot/file/172686007011777ruc76x.jpg

The image describes a hypothesis test where:

- The null hypothesis \( H_0 \) is that the mean body temperature \( \mu \leq 98.6^\circ F \).
- The alternative hypothesis \( H_1 \) is that \( \mu > 98.6^\circ F \) (a one-tailed test).
- The sample size is \( n = 5 \).
- The test statistic is \( t = 1.567 \).
- The significance level is \( \alpha = 0.05 \).

To find the P-value for this one-tailed t-test, we use the degrees of freedom \( \text{df} = n - 1 = 5 - 1 = 4 \).

Next, we can calculate the P-value using the t-distribution with 4 degrees of freedom, by looking up the t-score of 1.567 or using a calculator.

Let me compute the P-value for you.The P-value for the given hypothesis test is approximately 0.0961.

Since the P-value (0.0961) is greater than the significance level \( \alpha = 0.05 \), we fail to reject the null hypothesis. This means that the data does not provide strong enough evidence to support the claim that the mean body temperature is greater than \( 98.6^\circ F \).

Do you have any questions, or would you like further details?

Here are five related questions for you:
1. What is the importance of the P-value in hypothesis testing?
2. How do we interpret the result if the P-value is less than the significance level?
3. What are the consequences of rejecting the null hypothesis when it is true (Type I error)?
4. Why is a one-tailed test used in this scenario?
5. How does the sample size affect the degrees of freedom and the test outcome?

**Tip**: Always ensure to check the assumptions of the t-test, such as normality, especially with small sample sizes.

The claim is that for 12 AM body temperatures, the mean is μ > 98.6°F. The sample size is 5 and the test statistic is t = 1.567. Use technology to find the P-value for the hypothesis test described.

Perform a t-test for body temperatures where the mean is hypothesized to be greater than 98.6°F, using a sample size of 5. Learn how to calculate the P-value with a test statistic of t = 1.567 and a significance level of α = 0.05. Suitable for college-level statistics learners.

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