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According to the federal government, \(24 \%\) of workers covered by their company's health care plan were not required to contribute to the premium (Statistical Abstract of the United States: 2006 . A recent study found that 81 out of 400 workers sampled were not required to contribute to their company's health care plan. a. Develop hypotheses that can be used to test whether the percent of workers not required to contribute to their company's health care plan has declined. b. What is a point estimate of the proportion receiving free company-sponsored health care insurance? c. Has a statistically significant decline occurred in the proportion of workers receiving free company-sponsored health care insurance? Use \(\alpha=.05\).

Short Answer

Expert verified
The proportion of non-contributing workers has declined significantly.

Step by step solution

01

Define the Hypotheses

To determine if there is a decline in the proportion of workers not required to contribute to their company's health care plan, we first need to establish the null hypothesis (H0) and the alternative hypothesis (H1):- Null Hypothesis (H0): The proportion of workers not required to contribute is equal to 24%, i.e., \( p = 0.24 \).- Alternative Hypothesis (H1): The proportion of workers not required to contribute is less than 24%, i.e., \( p < 0.24 \). This signifies a decrease in the proportion.
02

Find the Point Estimate

The point estimate for the proportion of workers not required to contribute to their company's health care plan is calculated using the data from the sample. The formula for the sample proportion \( \hat{p} \) is:\[ \hat{p} = \frac{x}{n} \]where \( x = 81 \) is the number of workers not required to contribute and \( n = 400 \) is the sample size.Calculating this gives:\[ \hat{p} = \frac{81}{400} = 0.2025 \].Thus, the point estimate is 0.2025.
03

Conduct the Hypothesis Test

To test the hypothesis, we use a one-sample Z-test for proportions. The test statistic \( Z \) is calculated as follows:\[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]where \( p_0 = 0.24 \) (the population proportion), and \( n = 400 \).Substitute the values:\[ Z = \frac{0.2025 - 0.24}{\sqrt{\frac{0.24(0.76)}{400}}} = \frac{0.2025 - 0.24}{0.0218} \approx -1.71 \].
04

Determine the Critical Value and Conclusion

For a left-tailed test with \( \alpha = 0.05 \), the critical value from the Z-table is approximately \( -1.645 \).Compare the calculated Z-value to the critical value:- If \( Z < -1.645 \), we reject the null hypothesis.In this case, \( -1.71 < -1.645 \), so we reject the null hypothesis. This means that there is significant statistical evidence to suggest that the proportion of workers not required to contribute to their company's health care plan has declined.

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Key Concepts

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

Proportion Test
A proportion test is a statistical method used to compare a sample proportion to a known population proportion. It helps determine whether there is a significant difference between the two. In the context of the exercise, we want to find out if the current proportion of workers who do not have to contribute to their health care plan has decreased from the previous 24%. This is done using a one-sample Z-test for proportions.

The Z-test calculates how many standard deviations our sample proportion is from the population proportion. If this difference is too large, it implies that the current proportion is likely different from the historical one.

To conduct this test, you'll need:
  • The sample proportion, which is the ratio of workers in the study not required to contribute to their health care plan.
  • The known population proportion, which in this example is 24% or 0.24.
  • The sample size, which is the total number of workers in the study.
Performing this test provides a statistical foundation to determine if any observed changes are genuine or just due to random sampling variation.
Null and Alternative Hypotheses
In hypothesis testing, formulating null and alternative hypotheses is the initial step. The null hypothesis ( H_0) is a statement of no effect or no difference. It's what you assume to be true before analyzing the data. For this problem, the null hypothesis is that the proportion of workers not required to contribute remains at 24%.

Conversely, the alternative hypothesis ( H_1) represents what you seek to prove. It's a claim about the population parameter that suggests a change or effect. In this scenario, the alternative hypothesis claims that the proportion has decreased from 24%, stated as H_1: p < 0.24.

Understanding these hypotheses is essential as the entire procedure of testing and conclusions revolves around them.
  • H_0: p = 0.24 (No change in proportion).
  • H_1: p < 0.24 (Proportion has decreased).
Testing these hypotheses through a suitable statistical test helps in making data-driven decisions.
Point Estimate
The point estimate is a single value that provides the best estimate of the population parameter based on the sample data. In this exercise, it estimates the proportion of workers who did not pay for their health care plan. The point estimate is calculated through the formula: \(\hat{p} = \frac{x}{n}\)

Here, \(x\) is the number of workers who did not have to contribute, in this case, 81, and \(n\) is the total number of workers surveyed, which is 400. Plugging these numbers into the formula gives the point estimate:\[\hat{p} = \frac{81}{400} = 0.2025\]

This means that, according to the sample data, about 20.25% of workers were not required to contribute to their healthcare plan. This point estimate is used for further testing to see if there is statistical evidence suggesting a change from the original 24% figure.
Significance Level
The significance level, denoted by \(\alpha\), is a crucial part of hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true, commonly known as the Type I error. In other words, it's the risk you're willing to take when making a claim about the population based on your sample data.

In this exercise, the significance level is set at \(\alpha = 0.05\). This implies that you are accepting a 5% chance of concluding that there is a decline in the proportion when there isn't one.

Choosing the right significance level is important. A higher \(\alpha\) increases the probability of a Type I error, while a lower \(\alpha\) might make the test too conservative. Always consider the context when setting your significance level. For example, more critical decisions may require a lower \(\alpha\), while others may tolerate higher ones.

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Most popular questions from this chapter

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?

The Internal Revenue Service (IRS) provides a toll-free help line for taxpayers to call in and get answers to questions as they prepare their tax returns. In recent years, the IRS has been inundated with taxpayer calls and has redesigned its phone service as well as posted answers to frequently asked questions on its website (The Cincinnati Enquirer, January 7,2010 ). According to a report by a taxpayer advocate, callers using the new system can expect to wait on hold for an unreasonably long time of 12 minutes before being able to talk to an IRS employee. Suppose you select a sample of 50 callers after the new phone service has been implemented; the sample results show a mean waiting time of 10 minutes before an IRS employee comes on the line. Based upon data from past years, you decide it is reasonable to assume that the standard deviation of waiting time is 8 minutes. Using your sample results, can you conclude that the actual mean waiting time turned out to be significantly less than the 12 -minute claim made by the taxpayer advocate? Use \(\alpha=.05\).

Wall Street securities firms paid out record year-end bonuses of \(\$ 125,500\) per employee for 2005 (Fortune, February 6,2006 ). Suppose we would like to take a sample of employees at the Jones \(\&\) Ryan securities firm to see whether the mean year-end bonus is different from the reported mean of \(\$ 125,500\) for the population. a. State the null and alternative hypotheses you would use to test whether the year-end bonuses paid by Jones \& Ryan were different from the population mean. b. Suppose a sample of 40 Jones \(\&\) Ryan employees showed a sample mean year- end bonus of \(\$ 118,000\), Assume a population standard deviation of \(\sigma=\$ 30,000\) and compute the \(p\) -value. c. With \(a=.05\) as the level of significance, what is your conclusion? d. Repeat the preceding hypothesis test using the critical value approach.

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?

AOL Time Warner Inc.'s CNN has been the longtime ratings leader of cable television news. Nielsen Media Research indicated that the mean CNN viewing audience was 600,000 viewers per day during 2002 (The Wall Street Journal, March 10,2003 ). Assume that for a sample of 40 days during the first half of \(2003,\) the daily audience was 612,000 viewers with a sample standard deviation of 65,000 viewers. a. What are the hypotheses if CNN management would like information on any change in the CNN viewing audience? b. What is the \(p\) -value? c. Select your own level of significance. What is your conclusion? d. What recommendation would you make to CNN management in this application?

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