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 52% of U.S. employers were likely to require higher employee contributions for health care coverage. Suppose the survey was based on a sample of 600 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. If required, round your answer to four decimal places. Round intermediate calculations to four decimal places.

Margin of Error:
Confidence Interval:	to

Answer 1

Solution :

Given that,

n = 600

Point estimate = sample proportion = =0.52

1 -   = 1- 0.52 =0.48

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E = Z/2   * ((( * (1 - )) / n)

= 1.96 (((0.52*0.48) /600 )

E = 0.0400

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.52-0.0400 < p <0.52+ 0.0400

0.4800< p < 0.5600

The 95% confidence interval for the population proportion p is :0.4800, 0.5600