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Obesity 2016 In \(2016,\) the Centers for Disease Control and Prevention reported that \(36.5 \%\) of adults in the United States are obese. A county health service planning a new awareness campaign polls a random sample of 750 adults living there. In this sample, 228 people were found to be obese based on their answers to a health questionnaire. Do these responses provide strong evidence that the \(36.5 \%\) figure is not accurate for this region? Correct the mistakes you find in a student's attempt to test an appropriate hypothesis. $$ \begin{array}{l} \mathrm{H}_{0}: \hat{p}=0.365 \\ \mathrm{H}_{\mathrm{A}}: \hat{p}<0.365 \\ \mathrm{SRS}, 750 \geq 10 \\ \frac{228}{750}=0.304 ; S D(\hat{p})=\sqrt{\frac{(0.304)(0.696)}{750}}=0.017 \end{array} $$ \(z=\frac{0.304-0.365}{0.017}=-3.588\) $$ \mathrm{P}=P(z>-3.588)=0.9998 $$ There is more than a \(99.98 \%\) chance that the stated percentage is correct for this region.

Step by step solution

01

Identifying the Null and Alternate Hypothesis

In this context, the null hypothesis \(H_{0}\) would be that the percentage of obese adults is the same as the national rate, i.e., \(H_{0}: \hat{p}=0.365\), which is correct. The alternate hypothesis \(H_{A}\), however, should represent that the percentage is not equal to the national rate, and not necessarily less than it. Hence, it should be \(H_{A}: \hat{p} \neq 0.365\), contrary to what the student suggested.

02

Calculating the Test Statistic

The formula for calculating the test statistic is correct: \(z=\frac{\hat{p}-p_{0}}{SD(\hat{p})}\). However, the standard deviation \(SD(\hat{p})\) should be computed under the null hypothesis and not using the sample proportion. Thus, the standard deviation should be \(\sqrt{\frac{(0.365)(0.635)}{750}} = 0.0176\), and the test statistic z becomes \(z=\frac{0.304-0.365}{0.0176}=-3.466\)

03

Calculating the P-value

The p-value corresponds to the probability of observing a result as extreme as our sample (or more) under the null hypothesis. Given that our alternate hypothesis is a two-sided test (i.e., \(H_{A}: \hat{p} \neq 0.365\), we should look at the probability in both tails. The student incorrectly calculated the p-value for a one-sided test, so the correct p-value is \(P(z > 3.466) + P(z < -3.466) = 2 * (1 - 0.99974) = 0.00052\), which is significantly smaller than the student's claim of 0.9998.

04

Conclusion

The extremely small p-value suggests there is strong evidence to reject the null hypothesis in favor of the alternative. Therefore, there is strong evidence that the true proportion of obese adults in this region is not the same as the national rate, contrary to the student's conclusion.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis

The null hypothesis is the default statement that there is no effect or difference of interest in a particular scenario. In hypothesis testing, it acts as the assumption to be tested and potentially rejected.
The null hypothesis for the given exercise is correctly stated as \( H_0: \hat{p}=0.365 \), indicating that the proportion of obese adults in the region is the same as the national rate of 36.5%. The purpose of including this concept is to demonstrate that you're looking for evidence to reject your initial assumption and not to prove it directly.

Alternate Hypothesis

The alternate hypothesis (also called the alternative hypothesis) is a statement that contrasts the null hypothesis. It represents the possibility that there is indeed an effect or difference that the test aims to detect.
For our problem, the alternate hypothesis needs correction and should be two-sided: \( H_A: \hat{p} eq 0.365 \), which means the actual proportion could be either less or more than the national average.
Introducing the alternate hypothesis is crucial as it defines the direction of the hypothesis test and what constitutes evidence against the null hypothesis.

P-value

The p-value is the probability that the test statistic will take on a value at least as extreme as the value observed in the sample data, assuming the null hypothesis is true. A small p-value (typically ≤ 0.05) indicates strong evidence against the null hypothesis, so it is rejected.
In the exercise, the p-value calculated by the student was one-sided and thus incorrect. The proper two-sided p-value is significantly smaller, which actually suggests strong evidence against the null hypothesis.
Understanding the p-value helps students determine the significance of their test results in the context of the chosen significance level.

Standard Deviation

Standard deviation measures the amount of variation or dispersion of a set of values. In hypothesis testing, it is used to assess how much the sample data deviate from the null hypothesis.
The student's mistake was using the sample proportion rather than the proportion under the null hypothesis to calculate the standard deviation. The corrected standard deviation should consider the expected variability under the null hypothesis, resulting in a slightly different value.
Grasping the role of standard deviation is essential to correctly calculate the test statistic and understand the spread of data.

Test Statistic

A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It's used to compare the observed data with what is expected under the null hypothesis.
For our obesity study, the test statistic is the z-value, which requires the correct standard deviation under the null hypothesis. Consequently, the z-score provides a measure of how many standard deviations the observed proportion is away from the null hypothesis.
Knowing how to compute the test statistic enables students to determine where the observed data falls in relation to the null hypothesis.

Z-score

A z-score is a statistical measurement describing the relation of a data point to the mean of a group of values, expressed in terms of standard deviations. In the context of hypothesis testing, the z-score represents the test statistic.
In this case, the student's z-score calculation was off due to the incorrect standard deviation. Recalculating with the proper standard deviation corrected the value. Understanding z-scores helps in interpreting the test statistic’s value in terms of the normal distribution, which is pivotal for finding p-values.

Confidence Interval

A confidence interval gives an estimated range of values which is likely to include an unknown population parameter, calculated from a given set of sample data. Though it was not explicitly covered in the incorrect solution, a confidence interval can offer additional insight.
In our scenario, a confidence interval around the sample proportion of obese individuals could help determine if the national average falls within that range. This approach provides a more intuitive understanding of the significance and potential error of our point estimate.
Confidence intervals are pivotal in conveying the precision and uncertainty associated with the sample estimate, complementing the p-value's information.