Question

This problem has been solved!See the answerFor 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. Suppose the survey was based on a sample of 800 companies. 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.

Answer 1

Answer with explanation:

Given : The proportion of U.S. employers were likely to require higher employee contributions for health care coverage in 2009 : p=0.52

Sample size : n= 800

Significance level : $\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

Margin of error : $E=z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$

$\\\\\Rightarrow\ E=(1.96)\sqrt{\dfrac{0.52(1-0.52)}{800}}=0.03462050259\approx0.03$

The 95% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage in 2009 will be :_

$p\pm E\\\\=0.52\pm0.03=(0.52-0.03,0.52+0.03)=(0.49,\ 0.55)$

Answer 2

The 95% confidence interval for the proportion of companies likely to require higher employee contributions is; CI = (0.485, 0.555)

What is the confidence interval?

We are given;

Sample proportion; p = 52% = 0.52

Sample size; n = 800

Confidence level = 95%

Formula for margin of error is;

M = z√(p(1 - p)/n)

z at 95% confidence level = 1.96

Thus;

M = 1.96√(0.52(1 - 0.52)/800)

M = 0.035

Confidence interval is;

CI = p ± M

CI = 0.52 ± 0.035

CI = (0.485, 0.555)

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