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With double-digit annual percentage increases in the cost of health insurance, more andmore workers are likely to lack health insurance coverage $(U S A$ Today, January $23,2004) .$The following sample data provide a comparison of workers with and without health in-surance coverage for small, medium, and large companies. For the purposes of this study.small companies are companies that have fewer than 100 employees. Medium companieshave 100 to 999 employees, and large companies have 1000 or more employees. Sampledata are reported for 50 employees of small companies, 75 employees of medium companies, and 100 employees of large companies.a. Conduct a test of independence to determine whether employee health insurance coverage is independent of the size of the company. Use $\alpha=.05 .$ What is the $p$ -value, and what is your conclusion?b. The USA Todayarticle indicated employees of small companies are more likely to lack health insurance coverage. Use percentages based on the preceding data to support this conclusion.

a. There is sufficient evidence to support the claim that the variables arenot independent.b. $28 \%, 13.33 \%, 12 \%$

Intro Stats / AP Statistics

Chapter 11

Comparisons Involving Proportions and a Test of Independence

Descriptive Statistics

Confidence Intervals

The Chi-Square Distribution

University of North Carolina at Chapel Hill

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in question 59. We're told that we have a population with a proportion of 0.24 and we're also given a more recent sample from the population in which the proportion is 81 out of 400 apart A. Whereas to set up a hypothesis test to determine whether the proportion has declined. The alternative hypothesis is that portion is less than 0.24 and the null hypothesis is that it is greater than or equal to 0.2 for for Party for Part B were asked, What is the point estimate of from portion. So our best point estimate of proportion is a sample proportion, which is 81 out of 400 and that equals 0.20 to 5. And then, for part C were asked Teoh test whether there has been a statistically significant decline in the proportion at a significance level of 0.5 So just blood these values into her test statistic that comes out to one minus 1.756 and so we can go to a standard normal table then and look up the area to the left of his escort my minus 1.756 This also happens to correspond to RP Valley because this is a lower tail test. So the P value is the area to the left of the test statistic. And so from the table, we can determine that the P value is equal to 0.392 and then to conclude, we can say that the P value is creator than Alfa and therefore we can fail to reject the null hypothesis.

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