Math Problem Statement

A sample of 106 body temperatures has a mean of 98.20oF. Assuming that the sample is random and the population standard deviation is known to be 0.62 oF. Use a 5% significance level to test that the mean body temperature of healthy adults is not 98.60 oF.

Find the test p-value. Question 4 options:

0.00

0.03

3.10

-6.64

None of the above Question 5 (Mandatory) (4 points) A sample of 106 body temperatures has a mean of 98.20oF. Assuming that the sample is random and the population standard deviation is known to be 0.62 oF. Use a 5% significance level to test that the mean body temperature of healthy adults is not 98.60 oF.

State the conclusion. Question 5 options:

Reject H0.

Do not Reject H0 (the text refer as Accept H0)

Reject Ha.

Accept Ha

None of the above

Solution

Step 1: Formulating the Hypotheses

Given that we're testing whether the mean body temperature is different from 98.60°F, we need to formulate the null and alternative hypotheses:

  • Null Hypothesis (H₀): The mean body temperature of healthy adults is 98.60°F.
    H0:μ=98.60H_0: \mu = 98.60

  • Alternative Hypothesis (H₁): The mean body temperature of healthy adults is not 98.60°F (two-tailed test).
    H1:μ98.60H_1: \mu \neq 98.60

Step 2: Test Statistic Calculation

To test the hypothesis, we use the z-test formula because the population standard deviation (σ\sigma) is known:

z=xˉμ0σnz = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}

Where:

  • xˉ=98.20\bar{x} = 98.20 is the sample mean.
  • μ0=98.60\mu_0 = 98.60 is the population mean under the null hypothesis.
  • σ=0.62\sigma = 0.62 is the population standard deviation.
  • n=106n = 106 is the sample size.

Now, substitute the values into the formula:

z=98.2098.600.62106z = \frac{98.20 - 98.60}{\frac{0.62}{\sqrt{106}}}

Let's calculate the test statistic:

z=0.400.62106=0.400.06016.66z = \frac{-0.40}{\frac{0.62}{\sqrt{106}}} = \frac{-0.40}{0.0601} \approx -6.66

Step 3: Finding the P-value

The z-value we found is approximately 6.66-6.66. For a two-tailed test, the p-value corresponds to the area under the standard normal curve that is more extreme than the absolute value of the test statistic (since the alternative hypothesis is μ98.60\mu \neq 98.60).

Using the z-table or a statistical tool to find the p-value corresponding to z=6.66z = -6.66, we find that the p-value is extremely small, essentially 0.

Step 4: Decision Rule

At a significance level of α=0.05\alpha = 0.05, if the p-value is less than 0.05, we reject the null hypothesis.

Since the p-value is 0.00, which is less than 0.05, we reject the null hypothesis.

Conclusion

Question 4 (Test p-value):
The p-value is 0.00. Therefore, the correct answer is 0.00.

Question 5 (State the Conclusion):
Since the p-value is smaller than the significance level, we reject the null hypothesis. Therefore, the correct conclusion is:
Reject H₀.

Summary of Answers:

  • Question 4: 0.00
  • Question 5: Reject H₀

Would you like more details on the test calculation or any clarifications?

Here are a few related questions you might find interesting:

  1. How do you calculate a confidence interval for the population mean when the population standard deviation is known?
  2. What is the difference between a one-tailed and a two-tailed hypothesis test?
  3. How do you interpret a p-value of 0.03 in hypothesis testing?
  4. Why is it important to know whether the population standard deviation is known or not in hypothesis testing?
  5. What are the assumptions underlying the z-test for a population mean?

Tip: When performing hypothesis tests, always double-check if the test statistic corresponds to the correct distribution (z-test for known σ\sigma, t-test for unknown σ\sigma, etc.).

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Z-test
Two-Tailed Test
P-value

Formulas

z = (x̄ - μ₀) / (σ / √n)

Theorems

Central Limit Theorem
Z-test for population mean

Suitable Grade Level

Grades 11-12