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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

Mileage tests are conducted for a particular model of automobile. If a \(98 \%\) confidence interval with a margin of error of 1 mile per gallon is desired, how many automobiles should be used in the test? Assume that preliminary mileage tests indicate the standard deviation is 2.6 miles per gallon.

A \(U S A\) Today/CNN/Gallup survey of 369 working parents found 200 who said they spend too little time with their children because of work commitments. a. What is the point estimate of the proportion of the population of working parents who feel they spend too little time with their children because of work commitments? b. At \(95 \%\) confidence, what is the margin of error? c. What is the \(95 \%\) confidence interval estimate of the population proportion of working parents who feel they spend too little time with their children because of work commitments?

A simple random sample of 40 items resulted in a sample mean of \(25 .\) The population standard deviation is \(\sigma=5\) a. What is the standard error of the mean, \(\sigma_{\bar{x}} ?\) b. At \(95 \%\) confidence, what is the margin of error?

During the first quarter of \(2003,\) the price/earnings (P/E) ratio for stocks listed on the New York Stock Exchange generally ranged from 5 to 60 (The Wall Street Journal, March 7 , 2003 ). Assume that we want to estimate the population mean P/E ratio for all stocks listed on the exchange. How many stocks should be included in the sample if we want a margin of error of \(3 ?\) Use \(95 \%\) confidence.

According to Thomson Financial, through January \(25,2006,\) the majority of companies reporting profits had beaten estimates (Business Week, February 6, 2006). A sample of 162 companies showed that 104 beat estimates, 29 matched estimates, and 29 fell short. a. What is the point estimate of the proportion that fell short of estimates? b. Determine the margin of error and provide a \(95 \%\) confidence interval for the proportion that beat estimates. c. How large a sample is needed if the desired margin of error is \(.05 ?\)
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