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Chapter 10: Problem 13

In \(2001,23 \%\) of American university undergraduate students had at least one tattoo. A health practitioner obtains a random sample of 1026 university undergraduates and finds that 254 have at least one tattoo. Has the proportion of American university undergraduate students with at least one tattoo changed since \(2001 ?\) Use the \(\alpha=0.1\) level of significance.

Short Answer

Expert verified
Insufficient evidence at \(\alpha=0.1\) to conclude the proportion of undergraduates with tattoos has changed since 2001.

Step by step solution

01

- State the Hypotheses

Formulate the null and alternative hypotheses. Null Hypothesis (\textbf{H\textsubscript{0}}): The proportion of American university undergraduates with at least one tattoo has not changed since 2001, i.e., \ \[H_0: p = 0.23\]. Alternative Hypothesis (\textbf{H\textsubscript{1}}): The proportion has changed, i.e., \ \[H_1: p e 0.23\].
02

- Determine the Test Statistic

Identify and calculate the appropriate test statistic. Since we are dealing with proportions, use the z-test. The test statistic formula is: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \] Where \(\hat{p}\) is the sample proportion, \( p_0 \) is the hypothesized population proportion, and \( n \) is the sample size.
03

- Calculate the Sample Proportion

Calculate the sample proportion (\( \hat{p} \)). \[\hat{p} = \frac{\text{number of students with tattoos}}{\text{total number of students}} = \frac{254}{1026} \approx 0.2475\]
04

- Compute the Test Statistic

Substitute the known values into the formula to compute the test statistic. \[z = \frac{0.2475 - 0.23}{\sqrt{\frac{0.23(1-0.23)}{1026}}} \] Calculate the numerator and denominator separately for accuracy. Numerator: \(0.2475 - 0.23 = 0.0175\) Denominator: \[ \sqrt{\frac{0.23(0.77)}{1026}} = \sqrt{\frac{0.1771}{1026}} = \sqrt{0.0001726} \approx 0.0131 \] Therefore, \[ z = \frac{0.0175}{0.0131} \approx 1.34 \]
05

- Determine the Critical Value

For a two-tailed test at the \(\alpha=0.1\) significance level, the critical z-values are \( \pm 1.645 \).
06

- Make a Decision

Compare the computed test statistic with the critical values: The computed \(z\)-value is approximately \(1.34\). Since \(1.34\) is less than the critical value \(1.645\), we fail to reject the null hypothesis \ \(H_0\). This means there is insufficient evidence to suggest the proportion of American university undergraduates with at least one tattoo has changed since 2001 at the \(\alpha=0.1\) significance level.

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Key Concepts

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

proportion testing
When analyzing a population by examining a sample, we often want to know if the proportion of observed characteristics has changed over time. Proportion testing is used for this purpose. For example, we might be interested in knowing if the proportion of university students with tattoos has changed. This involves comparing the sample proportion to the known proportion from an earlier time.

Using proportion testing can help make informed decisions based on sample data rather than surveying the entire population. By calculating sample proportion, we get insight into the characteristics of a larger group.
z-test
The z-test is a statistical tool used to determine if there is a significant difference between the sample proportion and the hypothesized population proportion. In our case, to check whether the proportion of university students with tattoos has changed since 2001, we use the z-test.

The test statistic for the z-test is calculated using the formula:

\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \]
Where \( \hat{p} \) is the observed sample proportion, \( p_0 \) is the hypothesized population proportion, and \( n \) is the sample size.

This formula helps to standardize the difference between the sample proportion and the population proportion relative to the variability expected in a random sample.
significance level
In hypothesis testing, the significance level (denoted by alpha: \( \alpha \)) represents the probability of rejecting the null hypothesis when it is actually true. A smaller significance level means stricter criteria for rejecting the null hypothesis.

For our example, we use the \( \alpha = 0.1 \) level of significance. This means we have a 10% risk of concluding that the proportion of university students with tattoos has changed when, in fact, it has not. Choosing the appropriate significance level is crucial as it balances the risk of errors in our conclusions.
critical value
The critical value is a point on the test distribution used to decide whether to reject the null hypothesis. For a two-tailed test, we calculate two critical values, one on the lower end and one on the higher end of the distribution.

In our hypothesis test, for a 0.1 significance level, the critical z-values are \( \pm 1.645 \). If our computed test statistic falls outside this range, we reject the null hypothesis. If it falls within, we fail to reject it.

In our test, the computed z-value was approximately 1.34, which lies within the range of \( -1.645 \) to \( 1.645 \), leading us to fail to reject the null hypothesis.
null hypothesis
The null hypothesis (\( H_0 \)) is a statement that there is no effect or no difference. It serves as the starting point for hypothesis testing.

In our example, the null hypothesis is that the proportion of American university undergraduates with at least one tattoo has not changed since 2001, i.e., \( H_0: p = 0.23 \). The alternative hypothesis (\( H_1 \)) suggests there has been a change, i.e., \( H_1: p e 0.23 \).

We use a sample to test these hypotheses. Based on our z-test and the significance level, we make a decision about whether there is enough evidence to reject the null hypothesis. If our data do not show enough evidence, we do not reject \( H_0 \), indicating no change in the proportion.

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Most popular questions from this chapter

According to the Insurance Information Institute, the mean expenditure for auto insurance in the United States was \(\$ 774\) for \(2002 .\) An insurance salesman obtains a random sample of 35 auto insurance policies and determines the mean expenditure to be \(\$ 735\) with a standard deviation of \(\$ 48.31 .\) Is there enough evidence to conclude that the mean expenditure for auto insurance is different from the 2002 amount at the \(\alpha=0.01\) level of significance?

Simulation Simulate drawing 40 simple random samples of size \(n=35\) from a population that is exponentially distributed with mean 8 and standard deviation \(\sqrt{8}\). (a) Test the null hypothesis \(H_{0}: \mu=8\) versus the alternative hypothesis \(H_{1}: \mu \neq 8\) (b) Suppose we were testing this hypothesis at the \(\alpha=0.05\) level of significance. How many of the 40 samples would you expect to result in a Type I error? (c) Count the number of samples that lead to a rejection of the null hypothesis. Is it close to the expected value determined in part (b)? (d) Describe how we know that a rejection of the null hypothesis results in making a Type I error in this situation.

To test \(H_{0}: \mu=50\) versus \(H_{1}: \mu<50,\) a random sample of size \(n=24\) is obtained from a population that is known to be normally distributed with \(\sigma=12\) (a) If the sample mean is determined to be \(\bar{x}=47.1\) compute the test statistic. (b) If the researcher decides to test this hypothesis at the \(\alpha=0.05\) level of significance, determine the critical value. (c) Draw a normal curve that depicts the critical region. (d) Will the researcher reject the null hypothesis? Why?

To test \(H_{0}: \mu=20\) versus \(H_{1}: \mu<20,\) a simple random sample of size \(n=18\) is obtained from a population that is known to be normally distributed. (a) If \(\bar{x}=18.3\) and \(s=4.3,\) compute the test statistic. (b) Draw a \(t\) -distribution with the area that represents the \(P\) -value shaded. (c) Approximate and interpret the \(P\) -value. (d) If the researcher decides to test this hypothesis at the \(\alpha=0.05\) level of significance, will the researcher reject the null hypothesis? Why?

Test the hypothesis, using (a) the classical approach and then (b) the P-value approach. Be sure to verify the requirements of the test. $$\begin{aligned}&H_{0}: p=0.25 \text { versus } H_{1}: p<0.25\\\&n=400 ; x=96 ; \alpha=0.1\end{aligned}$$
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