To test whether the obesity rate in this county significantly differs from the national average, we can perform a hypothesis test. Let's examine each part carefully, correct any errors in the student's work, and provide a full solution.

Part 1: Identify the Hypotheses

For this hypothesis test:

• Null Hypothesis ($H_0$): The obesity rate in the county is the same as the national rate, so $p = 0.365$.
• Alternative Hypothesis ($H_A$): The obesity rate in the county differs from the national rate, so $p \neq 0.365$.

Thus, the hypotheses are: $H_0: p = 0.365$ $H_A: p \neq 0.365$

Part 2: Check the Assumptions and Conditions

To ensure valid conditions for the hypothesis test, we need to check the following:

1. Randomization Condition: The sample should be a simple random sample of adults in the county, which is assumed to be satisfied.

2. Independence Assumption and the 10% Condition: The sample size should be less than 10% of the population of the county. Since this condition isn't specified as violated, we assume it's met.

3. Success/Failure Condition: We need $np \geq 10$ and $nq \geq 10$ for the normal approximation to be valid.

   • Using $p = 0.365$: $np = 600 \times 0.365 = 219 \geq 10$ $nq = 600 \times (1 - 0.365) = 381 \geq 10$

   These values satisfy the Success/Failure Condition, so choice A is correct.

Part 3: Perform the Hypothesis Test

Now, let’s use the sample proportion to calculate the test statistic.

1. Sample Proportion ($\hat{p}$): $\hat{p} = \frac{186}{600} = 0.31$

2. Standard Deviation of $\hat{p}$ under $H_0$: $\text{SD}(\hat{p}) = \sqrt{\frac{p \cdot (1 - p)}{n}} = \sqrt{\frac{0.365 \cdot (1 - 0.365)}{600}} \approx 0.0194$

3. Test Statistic ($z$-score): $z = \frac{\hat{p} - p}{\text{SD}(\hat{p})} = \frac{0.31 - 0.365}{0.0194} \approx -2.83$

4. P-value: Since this is a two-tailed test (we are looking for any significant difference), we find the probability $P(z < -2.83 \text{ or } z > 2.83)$. Looking up $z = -2.83$ in a standard normal table gives: $P(z < -2.83) \approx 0.0023$

   For a two-tailed test, we double this value: $P = 2 \times 0.0023 = 0.0046$

Since $P = 0.0046$, which is less than the significance level of 0.05, we reject the null hypothesis.

Conclusion

There is strong evidence that the obesity rate in this county is different from the national average of 36.5%. The correct conclusion is that there is a significant difference, not a 99.8% chance that the national percentage is correct.

Summary of Corrections

1. Hypotheses: Use a two-tailed test, $H_0: p = 0.365$, $H_A: p \neq 0.365$.
2. Sample Proportion Calculation: $\hat{p} = 0.31$.
3. Standard Deviation of $\hat{p}$: Correct calculation gives $\approx 0.0194$, not 0.0189.
4. Test Statistic: Calculated as $z = -2.83$.
5. P-value and Conclusion: $P = 0.0046$, leading to rejection of $H_0$.

Would you like more details or have any questions about this solution?

Follow-up Questions

1. How would the test change if the sample size were smaller?
2. What impact does a different significance level have on the conclusion?
3. Can we use this result to estimate the obesity rate in other nearby counties?
4. How would you conduct this test if you only had access to sample data, not national rates?
5. What are other real-world factors that could affect the validity of this hypothesis test?

Tip

Always ensure your hypothesis test’s assumptions (random sampling, sample size, etc.) are met, as they are crucial for valid conclusions.