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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
a. \(H_0: p = 0.125\), \(H_a: p > 0.125\); b. \(p\)-value is 0.3803; c. Do not reject \(H_0\).

Step by step solution

01

Formulate Hypotheses

We want to test if the proportion of workers in unions has increased from 12.5%. Define \( p \) as the true proportion of workers in unions in 2006. We set up the null and alternative hypotheses as follows:- Null Hypothesis: \( H_0: p = 0.125 \)- Alternative Hypothesis: \( H_a: p > 0.125 \).This is a right-tailed test since we are testing for an increase.
02

Find Sample Proportion

Calculate the sample proportion of workers in unions in 2006. The sample size is 400, and 52 workers belonged to unions.\[ p_{\text{sample}} = \frac{52}{400} = 0.13 \]
03

Standard Error Calculation

Calculate the standard error of the sample proportion using the formula:\[ SE = \sqrt{\frac{p_0 (1 - p_0)}{n}} \]where \( p_0 = 0.125 \) is the hypothesized proportion and \( n = 400 \) is the sample size.\[ SE = \sqrt{\frac{0.125 \times 0.875}{400}} \approx 0.0164 \]
04

Calculate Z-Score

Compute the Z-score to compare the sample proportion to the hypothesized proportion:\[ Z = \frac{p_{\text{sample}} - p_0}{SE} = \frac{0.13 - 0.125}{0.0164} \approx 0.305 \]
05

Find P-Value

Determine the p-value for the Z-score calculated. Since this is a right-tailed test, we find the probability that Z is greater than 0.305. Using standard normal distribution tables, the p-value associated with a Z-value of 0.305 is approximately 0.3803.
06

Make a Conclusion

Compare the p-value with the significance level \(\alpha = 0.05\). Since the p-value (0.3803) is greater than 0.05, we fail to reject the null hypothesis.Therefore, there is not sufficient evidence to conclude that the union membership among workers increased in 2006.

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Key Concepts

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

Z-Score Calculation
The Z-score is a statistical measure that helps determine how far a single data point is from the mean in terms of standard deviations. In hypothesis testing, it is used to decide whether to reject the null hypothesis. To compute the Z-score, we take the difference between the sample proportion (\( p_{\text{sample}} \) which was 0.13 in our example) and the hypothesized proportion (\( p_0 \) which was 0.125). Then, we divide that difference by the standard error (SE). In our exercise:\[ Z = \frac{0.13 - 0.125}{0.0164} \approx 0.305 \]It tells us how many standard errors the sample proportion is away from the hypothesized proportion. A Z-score close to zero indicates that the sample proportion is very close to the hypothesized value, while a Z-score farther from zero suggests a substantial difference.
Standard Error
The Standard Error (SE) quantifies the amount of variation or "spread" that can be expected from the sample proportion when comparing to the true proportion. It acts as an important component in deriving the Z-score, giving us a reference for the variability within the sample.The formula to calculate the Standard Error of the sample proportion is:\[ SE = \sqrt{\frac{p_0 (1 - p_0)}{n}} \]where:- \( p_0 \) is the hypothesized proportion (0.125 in our problem),- \( n \) is the sample size (400 for our example).Using these values, we calculated:\[ SE = \sqrt{\frac{0.125 \times 0.875}{400}} \approx 0.0164 \]This SE serves as the denominator in our Z-score formula and helps determine how significant the results of the hypothesis test are.
P-Value Interpretation
The p-value is a crucial element in hypothesis testing. It measures the probability of observing a sample statistic as extreme or more extreme than the one observed, assuming the null hypothesis is true. In this context, a smaller p-value indicates stronger evidence against the null hypothesis. Our exercise sought a right-tailed test, aiming to prove an increase in union membership. Thus, we found the p-value corresponding to the Z-score of 0.305:Approximately 0.3803.Understanding this value:- A p-value greater than the significance level \( \alpha = 0.05 \) suggests insufficient evidence to reject the null hypothesis.- Conversely, a p-value less than \( \alpha \) would indicate strong evidence to reject it.For our scenario, a p-value of 0.3803 was not significant enough to assert an increase in union membership.
Null and Alternative Hypotheses
Formulating hypotheses is the first and most critical step in hypothesis testing. It sets the stage for the analysis and what the test aims to prove or disprove.- **Null Hypothesis (\( H_0 \))**: This posits that there is no effect or change in the parameter being tested. In our problem, it was stated as \( H_0: p = 0.125 \). This means that, statistically, the proportion of workers in unions hasn't increased.- **Alternative Hypothesis (\( H_a \))**: This asserts the opposite of the null hypothesis and what we wish to prove. Here, it was \( H_a: p > 0.125 \), indicating the belief that union membership has increased.This framework is crucial because the aim of hypothesis testing is to determine whether there is enough statistical evidence to reject the null hypothesis in favor of the alternative.

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Most popular questions from this chapter

Nielsen reported that young men in the United States watch 56.2 minutes of prime-time TV daily (The Wall Street Journal Europe, November 18,2003 ). A researcher believes that young men in Germany spend more time watching prime- time TV. A sample of German young men will be selected by the researcher and the time they spend watching TV in one day will be recorded. The sample results will be used to test the following null and alternative hypotheses. \\[ \begin{array}{l} H_{0}: \mu \leq 56.2 \\ H_{\mathrm{a}}: \mu>56.2 \end{array} \\] a. What is the Type I error in this situation? What are the consequences of making this error? b. What is the Type II error in this situation? What are the consequences of making this error?

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The manager of an automobile dealership is considering a new bonus plan designed to increase sales volume. Currently, the mean sales volume is 14 automobiles per month. The manager wants to conduct a research study to see whether the new bonus plan increases sales volume. To collect data on the plan, a sample of sales personnel will be allowed to sell under the new bonus plan for a one-month period. a. Develop the null and alternative hypotheses most appropriate for this research situation. b. Comment on the conclusion when \(H_{0}\) cannot be rejected. c. Comment on the conclusion when \(H_{0}\) can be rejected.

A production line operates with a mean filling weight of 16 ounces per container. Overfilling or underfilling presents a serious problem and when detected requires the operator to shut down the production line to readjust the filling mechanism. From past data, a population standard deviation \(\sigma=.8\) ounces is assumed. A quality control inspector selects a sample of 30 items every hour and at that time makes the decision of whether to shut down the line for readjustment. The level of significance is \(\alpha=.05\) a. State the hypothesis test for this quality control application. b. If a sample mean of \(\bar{x}=16.32\) ounces were found, what is the \(p\) -value? What action would you recommend? c. If a sample mean of \(\bar{x}=15.82\) ounces were found, what is the \(p\) -value? What action would you recommend? d. Use the critical value approach. What is the rejection rule for the preceding hypothesis testing procedure? Repeat parts (b) and (c). Do you reach the same conclusion?

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