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 62% 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 943 companies.

Compute the margin of error and a 99% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage in 2009.

The margin of error:  (to 4 decimals)

The 99% confidence interval: ( ,  ) (to 4 decimals)

Answer 1

solution :-

Given :-

p = 0.62

Sample Size ( n ) = 943

At 99 % Confidence level ' z ' is,

by using Standard normal table,

Z / 2 = 2.58

At 99 %, Margin of error is,

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

Margin of Error ( E ) = 2.58 * 0.62 ( 1 - 0.62 ) / 943

Margin of Error ( E ) = 0.0407

At 99 % CI for proportion is,

p Margin of Error

0.62   0.0407

( 0.5793 , 0.6607 )

At 99 % CI for proportion is,( 0.5793 , 0.6607 )