Question

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 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 companies.Compute the margin of error and a confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage in 2009.

Answer 1

A. Margin of error=0.0346

B. Confidence interval=0.4854 to 0.5546

Explanation:

A. Computation the margin of error

First step is to calculate the Confidence level using this formula

Confidence level= 1 -α

Where,

α=0.95

Let plug in the formula

Confidence level= 1-0.95

Confidence level= 0.05

Second step is to find Z using this formula

Zα/2

Let plug in the formula

Z= 0.05/2

Z=0.025

Third step is to find the Z score of Z=0.025

Zα/2=1.96

Now let calculate the margin of error using this formula

Margin of error=Zα/2√p(1-p)/n

Where,

Zα/2=1.96

p=0.52

n=800

Let plug in the formula

Margin of error=1.96√0.52(1-0.52)/800

Margin of error=1.96√0.52(0.48)/800

Margin of error=1.96√0.2496/800

Margin of error=1.96√0.000312

Margin of error=0.0346

Therefore Margin of error will be 0.0346

B. Computation for the 95% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage in 2009

Computation for the boundaries of confidence interval

Using this formula

p- Zα/2√p(1-p)/n

Where,

Zα/2=1.96

p=0.52

n=800

Let plug in the formula

Confidence interval=0.52-1.96√0.52(1-0.52)/800

Confidence interval=0.52-1.96√0.52(0.48)/800

Confidence interval=0.52-1.96√0.2496/800

Confidence interval=-0.52-1.96√0.000312

Confidence interval=0.4854

Computation for the boundaries of confidence interval

Using this formula

p+ Zα/2√p(1-p)/n

Where,

Zα/2=1.96

p=0.52

n=800

Let plug in the formula

Confidence interval=0.52+1.96√0.52(1-0.52)/800

Confidence interval=0.52+1.96√0.52(0.48)/800

Confidence interval=0.52+1.96√0.2496/800

Confidence interval=-0.52+1.96√0.000312

Confidence interval=0.5546

Therefore the 95% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage in 2009 is 0.4854 to 0.5546