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Chapter 8: Problem 40

For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to less inflation in health care prices and employees paying for a larger portion of health care benefits. A recent Mercer survey showed that \(52 \%\) of U.S. employers were likely to require higher employee contributions for health care coverage in 2009 (Business Week, February 16,2009 ). Suppose the survey was based on a sample of \(800 \mathrm{com}\) panies. Compute the margin of error and a \(95 \%\) confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage in 2009

Short Answer

Expert verified
The margin of error is approximately 0.0347 and the 95% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage in 2009 is (0.4853, 0.5547).

Step by step solution

01

Identify the given values

We are given that the sample size (n) is 800 and the sample proportion (p̂) is 0.52.
02

Calculate the standard error

The standard error (SE) is calculated using the formula: SE = \(\sqrt{\frac{p̂(1-p̂)}{n}}\). Plugging in our given values: SE = \(\sqrt{\frac{0.52(1-0.52)}{800}} = \sqrt{\frac{0.52(0.48)}{800}} = \sqrt{0.000312} ≈ 0.0177\)
03

Calculate the margin of error

The margin of error (ME) is calculated using the following formula: ME = Z * SE. For a 95% confidence interval, the Z-score (Z) is 1.96. ME = 1.96 * 0.0177 ≈ 0.0347
04

Calculate the confidence interval

To calculate the confidence interval, we need to find the lower and upper limits of the interval. To do this, we will subtract the margin of error from the sample proportion for the lower limit and add the margin of error to the sample proportion for the upper limit: Lower Limit: p̂ - ME = 0.52 - 0.0347 ≈ 0.4853 Upper Limit: p̂ + ME = 0.52 + 0.0347 ≈ 0.5547 The 95% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage in 2009 is (0.4853, 0.5547).

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Most popular questions from this chapter

How large a sample should be selected to provide a \(95 \%\) confidence interval with a margin of error of \(10 ?\) Assume that the population standard deviation is 40

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