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A 2008 AARP survey reported that \(85 \%\) of U.S. workers aged 50 years and older with at least one 4-year college degree had taken employer-based training within the previous 2 years, compared to only \(50 \%\) of workers aged 50 years and older with a high school degree or less. In a current survey of \(640 \mathrm{U.S}\). workers aged 50 years and older with a high school degree or less, 341 had taken employer-based training within the previous 2 years. a. Using the critical-value approach and \(\alpha=.05\), test whether the current percentage of all U.S. workers aged 50 years and older with a high school degree or less who have taken employerbased training within the previous 2 years is different from \(50 \%\). b. How do you explain the Type I error in part a? What is the probability of making this error in part a? c. Calculate the \(p\) -value for the test of part a. What is your conclusion if \(\alpha=.05 ?\)

Step by step solution

01

State the Hypotheses

For part a, we need to formulate the null hypothesis (H0) and the alternative hypothesis (H1). \n\( H0: p = 0.5 \) \n\( H1: p \neq 0.5 \), where \( p \) is the current percentage of all U.S. workers aged 50 years or older with a high school degree or less who have taken employer-based training within the previous 2 years.

02

Compute Test Statistic

The test statistic for the proportion is given by \( Z = \frac{\hat{p} - p_0}{\sqrt{\ \frac{p_0(1 - p_0)}{n}}}\), where \( \hat{p} \) is the sample proportion, \( p_0 \) is the hypothesized proportion, and \( n \) is the sample size. Given that \( n = 640 \), \( p_0 = 0.5 \), and \( \hat{p} = \frac{341}{640} = 0.5328125 \), then we can substitute these values in to get the value of \( Z \).

03

Compute Critical Values and Make Decision

Using the significance level \( \alpha = 0.05 \) and because this is a two-tailed test, we find the critical values from a standard normal distribution table to be \( ±1.96 \). If the computed test statistic is beyond these values, we reject the null hypothesis.

04

Explain Type I Error

For part b, a Type I error occurs when we reject a true null hypothesis. In this context, it means that we would mistakenly conclude that the proportion of U.S. workers aged 50 years and older with a high school degree or less who have taken employer based training is not 50% when in fact it is. The probability of making this error, represented by \( \alpha \), is 0.05.

05

Compute the P-value

For part c, the p-value is the smallest level of significance at which we could reject the null hypothesis given the observed sample data. To find it, we use the test statistic computed in Step 2 and find its corresponding probability under the standard normal distribution, then multiply by 2 since this is a two-tailed test.

06

Make Conclusion

If the p-value is less than \( \alpha = 0.05 \), then we reject \( H0 \). If not, we do not reject \( H0 \). This determines whether the current percentage of all U.S. workers aged 50 years and older with a high school degree or less who have taken employer-based training within the previous 2 years is different from 50%

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Key Concepts

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

Null Hypothesis

In hypothesis testing, the null hypothesis, often denoted as \( H_0 \), serves as the starting assumption for any statistical test. It's the hypothesis that there is no effect or no difference, and it often represents the status quo or a baseline which is assumed to be true until evidence suggests otherwise.
For example, in the context of proportion tests, the null hypothesis might assert that a particular proportion is equal to a specified value. In the given exercise, the null hypothesis \( H_0: p = 0.5 \) states that the proportion of all U.S. workers aged 50 years and older with a high school degree or less who have taken employer-based training is 50\%.
This hypothesis is essential because it forms the foundation against which the alternative hypothesis is compared. The goal of hypothesis testing is to make a decision about the validity of this null hypothesis based upon sample data.

Type I Error

A Type I error occurs when we incorrectly reject a null hypothesis that is actually true. In other words, we mistakenly see an effect or difference that doesn't exist. This type of error is also known as a false positive. In many cases, researchers strive to control the probability of this error occurring, which is represented by the significance level \( \alpha \).
In the exercise, making a Type I error means concluding that the proportion of workers who took training is different from 50\%, when in fact it is still 50\%. The significance level \( \alpha = 0.05 \) indicates a 5\% probability of making such an error.

• Type I errors can lead to incorrect beliefs about the difference or effect and potentially erroneous policy decisions or strategies.
• The probability of making a Type I error is chosen by the researcher and can be adjusted based on the context and the consequences of making such an error.

By understanding what a Type I error is, researchers can better interpret their findings and communicate the results of a hypothesis test more effectively.

P-value

The p-value is a fundamental concept in hypothesis testing, often misinterpreted. It assists in decision-making by quantifying the evidence against the null hypothesis. Specifically, it's the probability of observing data as extreme as, or more extreme than, the observed data, assuming the null hypothesis is true.
In our exercise, the p-value helps indicate how consistent the sample data is with the null hypothesis. A smaller p-value suggests more substantial evidence against \( H_0 \).
To find the p-value in the exercise, we calculate it from the test statistic. If this p-value is less than the significance level \( \alpha = 0.05 \), it provides enough evidence to reject the null hypothesis.

• A p-value less than 0.05 typically means the results are statistically significant, suggesting strong evidence against the null hypothesis.
• Conversely, a high p-value implies that the observed data is consistent with the null hypothesis.

The p-value allows researchers to determine the strength of their evidence without solely relying on a fixed threshold of decision-making.

Proportion Test

A proportion test is used when you're interested in comparing a sample proportion to a known proportion or determining whether two proportions are different. It's especially useful in studies involving large sample sizes when the parameter of interest is a proportion.
In our exercise, the test is constructed to see if the proportion of workers aged 50 and older, who took employer-based training, is different from a specified proportion (50\%).
The test statistic for a proportion 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 sample proportion
• \( p_0 \) is the hypothesized proportion
• \( n \) is the sample size

The result of this calculation helps in determining whether the observed proportion differs significantly from the hypothesized proportion. With this understanding, a researcher can effectively evaluate claims about proportions and draw inferences with precision.