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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: 2000 . 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
A statistically significant decline in the proportion has occurred.

Step by step solution

01

Define the Hypotheses

To determine if the proportion of workers not contributing has declined, we set up null and alternative hypotheses. - Null hypothesis \( H_0 \): The proportion \( p \) is 0.24 (same as before); \( H_0: p = 0.24 \).- Alternative hypothesis \( H_a \): The proportion \( p \) is less than 0.24 (has declined); \( H_a: p < 0.24 \).
02

Calculate the Point Estimate

To find the point estimate, calculate the sample proportion of workers who are not required to contribute.- Sample size \( n = 400 \).- Number of workers not contributing \( x = 81 \).- Point estimate \( \hat{p} = \frac{x}{n} = \frac{81}{400} = 0.2025 \).
03

Conduct the Hypothesis Test

Perform a hypothesis test to determine if there is a statistically significant decline.1. Calculate the standard error: \[ SE = \sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.24 \times 0.76}{400}} = 0.0214 \] 2. Compute the Z-statistic: \[ Z = \frac{\hat{p} - p_0}{SE} = \frac{0.2025 - 0.24}{0.0214} \approx -1.75 \]3. Determine the critical value from the Z-table for \( \alpha = 0.05 \) (left-tailed test): \( ext{Z}_{\alpha} \approx -1.645 \).4. Compare the Z-statistic with the critical value. Since \(-1.75 < -1.645\), we reject the null hypothesis.

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Key Concepts

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

Proportion Hypothesis
When performing hypothesis testing for proportions, it's crucial to start by establishing a clear hypothesis. The null hypothesis (\(H_0\)) usually reflects the idea that there has been no change or that the parameter of interest remains at a specific value. In this case, our null hypothesis is \(H_0: p = 0.24\), meaning that 24% of workers do not need to contribute to their health care premiums.
This hypothesis is essential to test against the alternative hypothesis (\(H_a\)), which posits that the proportion has declined, stated as \(H_a: p < 0.24\). Creating these hypotheses is the foundation of hypothesis testing. It sets the stage for calculating various statistical measures and, ultimately, making a decision based on data. Always ensure the hypotheses reflect the question you are interested in testing, in this case, whether there has been a decline.
Point Estimate
The point estimate is an essential tool that gives us a snapshot of the proportion we are testing. It's derived from the sample data and acts as an unbiased estimate of the population proportion. Here, the point estimate is calculated using the formula \( \hat{p} = \frac{x}{n} \), where \(x\) is the number of successes in our sample, and \(n\) is the total sample size.
In our exercise, 81 workers out of a sampled 400 do not contribute to the healthcare premium. Thus, the point estimate is \( \hat{p} = \frac{81}{400} = 0.2025 \).This means our best estimate, based on the sampled data, is that 20.25% of workers are not required to contribute. Understanding this estimate helps us gauge where the actual population proportion might lie.
Z-statistic
The Z-statistic is a vital component when determining the statistical significance of an observed difference from the expected proportion. It measures how many standard deviations our point estimate is from the null hypothesis proportion. To calculate the Z-statistic, use the formula:\[ Z = \frac{\hat{p} - p_0}{SE} \]where \( \hat{p} \) is the point estimate, \(p_0\) is the proportion under the null hypothesis, and \(SE\) is the standard error of the sample proportion. The standard error is given by:\[ SE = \sqrt{\frac{p_0 (1 - p_0)}{n}} \]In this exercise, our Z-statistic was calculated to be approximately -1.75.A Z-statistic helps in deciding how far and in what direction our sample proportion deviates from the null hypothesis proportion. It ultimately guides the decision to accept or reject the null hypothesis.
Statistical Significance
Statistical significance helps infer if an observed effect or difference in our study is genuine or due to random chance. It’s determined by comparing our calculated statistic to a critical value at a chosen significance level, \(\alpha\). In a hypothesis test, such as the one in this exercise, if the tested statistic (here, the Z-statistic) falls into the rejection region, it indicates statistical significance. For this test, we chose a significance level of \(\alpha = 0.05\), which implies a 5% risk of concluding that a difference exists when there is none. The critical Z-value for a left-tailed test at this level is approximately \(-1.645\). Since the calculated Z-statistic of \(-1.75\) is less than \(-1.645\), we determine there is a statistically significant decline in the proportion of workers not required to contribute. This means the evidence supports the conclusion that fewer workers aren't contributing compared to the past, beyond what could be expected by random variation alone.

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Most popular questions from this chapter

A study by the Centers for Disease Control (CDC) found that \(23.3 \%\) of adults are smokers and that roughly \(70 \%\) of those who do smoke indicate that they want to quit (Associated Press, July 26,2002 ). CDC reported that, of people who smoked at some point in their lives, \(50 \%\) have been able to kick the habit. Part of the study suggested that the success rate for quitting rose by education level. Assume that a sample of 100 college graduates who smoked at some point in their lives showed that 64 had been able to successfully stop smoking. a. State the hypotheses that can be used to determine whether the population of college graduates has a success rate higher than the overall population when it comes to breaking the smoking habit. b. Given the sample data, what is the proportion of college graduates who, having smoked at some point in their lives, were able to stop smoking? c. What is the \(p\) -value? At \(\alpha=.01\), what is your hypothesis testing conclusion?

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