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91Ó°ÊÓ

A study found that, in \(2005,12.5 \%\) of U.S. workers belonged to unions (The Wall Street Journal, January 21,2006 . Suppose a sample of 400 U.S. workers is collected in 2006 to determine whether union efforts to organize have increased union membership. a. Formulate the hypotheses that can be used to determine whether union membership increased in 2006 b. If the sample results show that 52 of the workers belonged to unions, what is the \(p\) -value for your hypothesis test? c. \(\quad\) At \(\alpha=.05,\) what is your conclusion?

Short Answer

Expert verified
No increase in union membership is detected (p-value = 0.377).

Step by step solution

01

Formulating Hypotheses

To determine whether union membership has increased, we need to define our null hypothesis and alternative hypothesis. The null hypothesis (\(H_0\)) states that the proportion of workers in unions has not increased, which means it is less than or equal to 12.5%. The alternative hypothesis (\(H_a\)) states that the proportion has increased beyond 12.5%.\[ H_0: p \leq 0.125 \ H_a: p > 0.125\]
02

Identifying Sample Proportion and Sample Size

Given a sample size of 400 workers, where 52 belong to unions, we calculate the sample proportion (\(\hat{p}\)). This is done by dividing the number of union members by the total sample size:\[ \hat{p} = \frac{52}{400} = 0.13 \]
03

Calculating the Test Statistic

To find the test statistic, we use the formula for the test statistic of a proportion:\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]Substituting the values:\[ z = \frac{0.13 - 0.125}{\sqrt{\frac{0.125 \times (1-0.125)}{400}}} \approx \frac{0.005}{0.016} \approx 0.3125 \]
04

Finding the p-value

Using the standard normal distribution, we find the \( p \)-value for our calculated \( z \)-value of 0.3125. This is a right-tailed test, so we find the probability \( P(Z > 0.3125) \). By looking up a standard normal distribution table or using a software tool, we find:\[ p \text{-value} \approx 0.377 \]
05

Making the Conclusion

With a \( p \)-value of approximately 0.377 and a significance level (\( \alpha \)) of 0.05, we compare the \( p \)-value with \( \alpha \). Since 0.377 > 0.05, we do not reject the null hypothesis.Thus, there is not enough statistical evidence to conclude that union membership increased.

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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, often denoted as \( H_0 \), is a fundamental concept in hypothesis testing. It represents a statement of no effect or no difference, which in our exercise translates to the condition that union membership has not increased.

The null hypothesis is essential because it provides a baseline that statistical tests can work against. In the case of union membership, the null hypothesis is that the unionization rate is still 12.5% or less. Formally, this is represented as:
\( H_0: p \leq 0.125 \)
  • "\( H_0 \)" acts as a default or starting assumption that the study aims to challenge.
  • It's important as this hypothesis will either be rejected or not rejected based on statistical evidence.
Rejection of the null hypothesis would imply that we have enough evidence to support that union membership has increased. However, in cases where statistical tests do not provide significant evidence, the null hypothesis remains unchallenged.
Alternative Hypothesis
In hypothesis testing, the Alternative Hypothesis, denoted as \( H_a \), represents the statement that researchers want to prove. It is directly opposed to the null hypothesis and suggests that there is a significant effect or difference.

In our scenario with union membership, the alternative hypothesis claims that union efforts have been successful and the rate has risen above 12.5%. Thus, the formal expression of the alternative hypothesis is:
\( H_a: p > 0.125 \)
  • This hypothesis is the research assumption, usually involving a positive change or a novel finding.
  • Testing aims to collect enough evidence to support \( H_a \) by rejecting \( H_0 \).
If the data supports the alternative hypothesis through statistical evidence, it would suggest that union membership has indeed increased beyond the previous percentage.
P-value
The P-value, or probability value, is a critical component in determining the outcome of a hypothesis test. It measures the probability of obtaining test results as extreme as the observed results, assuming that the null hypothesis is true.

In our exercise example, the P-value helps us understand how likely it is to observe a sample proportion of unionized workers (or more extreme) if the true proportion were still 12.5%. After calculations, a P-value of approximately 0.377 was obtained.
  • A low P-value (typically less than the significance level) indicates strong evidence against the null hypothesis.
  • A high P-value suggests that the data is consistent with the null hypothesis.
In summary, since our P-value (0.377) is much greater than the common significance level of 0.05, we lack sufficient evidence to reject the null hypothesis. Therefore, it points to no significant increase in union membership.
Significance Level
The Significance Level, denoted as \( \alpha \), is the criterion used in hypothesis testing to decide whether to reject the null hypothesis. It's the threshold for determining the statistical significance of the test results.

Typically set at 0.05, the significance level indicates the probability of rejecting the null hypothesis when it is actually true (also known as Type I error). In our case, \( \alpha = 0.05 \).
  • If the P-value is less than or equal to \( \alpha \), we reject the null hypothesis, signifying statistical significance.
  • If the P-value is greater than \( \alpha \), we fail to reject the null hypothesis.
In the union membership problem, since the P-value (0.377) exceeds the significance level of 0.05, it leads us to conclude that there is not enough evidence to claim an increase in union membership based on our sample.

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Most popular questions from this chapter

On Friday, Wall Street traders were anxiously awaiting the federal government's release of numbers on the January increase in nonfarm payrolls. The early consensus estimate among economists was for a growth of 250,000 new jobs (CNBC, February 3, 2006). However, a sample of 20 economists taken Thursday afternoon provided a sample mean of 266,000 with a sample standard deviation of 24,000 . Financial analysts often call such a sample mean, based on late-breaking news, the whisper number. Treat the "consensus estimate" as the population mean, Conduct a hypothesis test to determine whether the whisper number justifies a conclusion of a statistically significant increase in the consensus estimate of economists, Use \(\alpha=.01\) as the level of significance.

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \geq 80 \\ H_{\mathrm{a}}: \mu<80 \end{array} \\] A sample of 100 is used and the population standard deviation is \(12 .\) Compute the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.01\) a. \(\quad \bar{x}=78.5\) b. \(\bar{x}=77\) c. \(\bar{x}=75.5\) d. \(\bar{x}=81\)

The manager of the Danvers-Hilton Resort Hotel stated that the mean guest bill for a weekend is \(\$ 600\) or less. A member of the hotel's accounting staff noticed that the total charges for guest bills have been increasing in recent months. The accountant will use a sample of future weekend guest bills to test the manager's claim. a. Which form of the hypotheses should be used to test the manager's claim? Explain. \\[\begin{array}{lll}H_{0}: \mu \geq 600 & H_{0}: \mu \leq 600 & H_{0}: \mu=600 \\ H_{\mathrm{a}}: \mu<600 & H_{\mathrm{a}}: \mu>600 & H_{\mathrm{a}}: \mu \neq 600\end{array}\\] b. What conclusion is appropriate when \(H_{0}\) cannot be rejected? c. What conclusion is appropriate when \(H_{0}\) can be rejected?

A recent article concerning bullish and bearish sentiment about the stock market reported that \(41 \%\) of investors responding to an American Institute of Individual Investors (AAII) poll were bullish on the market and \(26 \%\) were bearish (USA Today, January 11,2010 ). The article also reported that the long-term average measure of bullishness is .39 or \(39 \%\) Suppose the AAII poll used a sample size of \(450 .\) Using .39 (the long-term average) as the population proportion of investors who are bullish, conduct a hypothesis test to determine if the current proportion of investors who are bullish is significantly greater than the longterm average proportion. a. State the appropriate hypotheses for your significance test. b. Use the sample results to compute the test statistic and the \(p\) -value. c. Using \(\alpha=.10,\) what is your conclusion?

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \geq 45 \\ H_{\mathrm{a}}: \mu<45 \end{array} \\] A sample of 36 is used. Identify the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.01\) a. \(\bar{x}=44\) and \(s=5.2\) b. \(\bar{x}=43\) and \(s=4.6\) c. \(\bar{x}=46\) and \(s=5.0\)

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