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Anxiety Disorders According to Time magazine (December 5,2011 ), \(18 \%\) of adult Americans suffer from an anxiety disorder. Suppose a test of 300 random college students showed that 50 suffered from an anxiety disorder a. How many out of 300 would you expect to have an anxiety disorder if the \(18 \%\) is correct? b. Suppose you are testing the hypothesis that the population proportion of college students suffering from an anxiety disorder is not \(0.18\) at the \(0.05\) significance level. Choose the correct figure and interpret the p-value.

Short Answer

Expert verified
a. We would expect 54 out of 300 students to suffer from an anxiety disorder if the 18% figure is correct. b. The p-value, obtained from the z-value calculated using the sample proportion and standard error, gives the probability of observing a proportion as extreme as our sample. Depending on its value relative to the 0.05 significance level, we either have evidence to reject or fail to reject the null hypothesis.

Step by step solution

01

Calculate Expected Number of Students with Anxiety Disorder

To find the expected number of college students suffering from an anxiety disorder out of 300, we should use the given percentage of adult Americans suffering from an anxiety disorder, which is 18%. You multiply this percentage by the total number of students tested: \(300 * 0.18 = 54\).
02

Formulate Null and Alternative Hypothesis

We are testing if the population proportion of college students having an anxiety disorder is not 0.18. Our null hypothesis (Ho) would be 'the proportion is 0.18'. Our alternative hypothesis (Ha) would then be 'the proportion is not 0.18'.
03

Perform Hypothesis Testing

We use the sample proportion to carry out the test. The sample proportion is 50 out of 300, or 0.167. We mustfind the standard error of the sample proportion: \(\sqrt{(0.18*(1-0.18)/300)}\). Next, calculate the z-value: \((0.167-0.18) / SE\). Using this z-value, we can find our p-value from the z-table.
04

Interpret the P-value

The p-value is the probability of obtaining test results at least as extreme as the results actually observed. If the p-value is greater than 0.05, we fail to reject Ho. This means that the percentage of college students having an anxiety disorder could indeed be 18%. If the p-value is less than 0.05, we reject Ho. This means that there's weak evidence that the percentage of college students having an anxiety disorder is 18%.

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Key Concepts

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

Anxiety Disorders
Anxiety disorders are a common mental health condition that affect a significant portion of the population. These disorders can manifest in various forms such as generalized anxiety disorder, panic disorder, and social anxiety disorder, among others. Symptoms may include excessive worry, restlessness, and trouble concentrating, which can significantly impact daily life. According to Time magazine, about 18% of adult Americans deal with anxiety disorders.

Understanding the prevalence of anxiety disorders is crucial for prioritizing mental health resources and creating supportive environments. While adult statistics are helpful, it's also important to assess specific groups, like college students, due to factors such as academic stress and lifestyle changes that may affect their mental health. Testing the hypothesis that college students may exhibit different rates than the general population helps in identifying and addressing the unique needs of this group.
Significance Level
In hypothesis testing, the significance level, denoted as alpha (\(\alpha\)), plays a critical role. It represents the probability of rejecting the null hypothesis when it is true. Essentially, it's a threshold for determining statistical significance.

In this exercise, the significance level is set at 0.05. This means there is a 5% risk of concluding that the proportion of college students with anxiety disorders is different from 0.18, when in fact it is not. A lower significance level suggests more stringent criteria for rejecting the null hypothesis, minimizing false positives.

Choosing an appropriate significance level depends on the context and consequences of errors in decision-making. In medical or psychological research, a conventional 0.05 level balances being cautious about errors with practicality.
Population Proportion
Population proportion refers to the fraction of individuals within a population that exhibit a particular characteristic. In our exercise, the population proportion under investigation is the 18% of adults thought to suffer from anxiety disorders.

For the college students sample, we hypothesize that their proportion of anxiety disorders may differ from the general adult population figure. The null hypothesis assumes the college population proportion is 0.18, aligning with adult statistics while the alternative suggests otherwise.

Sample surveys and tests, like the one conducted on 300 college students, help estimate population proportions. The observed number (50 with anxiety disorders) offers a sample proportion, which can be compared against the claimed population proportion using a test statistic, such as the z-value, to draw conclusions.
P-value Interpretation
Interpreting the p-value is a vital step in hypothesis testing. The p-value indicates the probability of obtaining a sample proportion as extreme as, or more extreme than, the observed proportion, assuming the null hypothesis is true.

If the p-value obtained is less than the significance level (0.05 in this case), we reject the null hypothesis. This implies the sample provides strong evidence that the proportion of college students with anxiety disorders differs from 0.18. Conversely, a p-value greater than 0.05 means the data does not provide sufficient evidence to reject the null hypothesis, and we might conclude the college students' proportion could be similar to the reported adult rate.

Careful p-value interpretation guides researchers on whether their results reflect actual population differences or if they occurred by random chance.

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Most popular questions from this chapter

Suppose a sample of 600 surgeries with the new scrub shows 18 infections. Find the value of the test statistic, \(z\), and explain its meaning in context. The old infection rate was \(4 \%\).

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A statistician studying ESP tests 500 students. Each student is asked to predict the outcome of a large number of dice rolls. For each student, a hypothesis test using a \(10 \%\) significance level is performed. If the p-value for the student is less than or equal to \(0.10\), the researcher concludes that the student has ESP. Out of 500 students who do not have ESP, about how many could you expect the statistician to declare do have ESP?

A biased lotto draws even numbers faster than odd numbers. A token is drawn 50 times, and 35 even numbers come up. a. Pick the correct null hypothesis: i. \(\hat{p}=0.50\) ii. \(\hat{p}=0.70\) iii. \(p=0.50\) iv. \(p=0.70\) b. Pick the correct alternative hypothesis: i. \(\hat{p}=0.50\) ii. \(\hat{p}=0.70\) iii. \(p>0.50\) iv. \(p>0.70\)

A teacher giving a true/false test wants to make sure her students do better than they would if they were simply guessing, so she forms a hypothesis to test this. Her null hypothesis is that a student will get \(50 \%\) of the questions on the exam correct. The alternative hypothesis is that the student is not guessing and should get more than \(50 \%\) in the long run. $$ \begin{aligned} &\mathrm{H}_{0}=p=0.50 \\ &\mathrm{H}_{\mathrm{a}}: p>0.50 \end{aligned} $$ A student gets 30 out of 50 questions, or \(60 \%\), correct. The p-value is \(0.079 .\) Explain the meaning of the p-value in the context of this question.

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