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 65% 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 633 companies. Compute the margin of error and a 90% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage in 2009. The margin of error: 0417 (to 4 decimals) The 90% confidence interval: 5283 , 6117 ) (to 4 decimals)

Answer 1

Margin of error E = Z$\alpha$/2 * sqrt [ $\hat{p}$ ( 1- $\hat{p}$ ) / n ]

= 1.6449 * sqrt [ 0.65 ( 1 - 0.65) / 633 ]

= 0.0312

90% confidence interval is

$\hat{p}$ - E < p < $\hat{p}$ + E

0.65 - 0.0312 < p < 0.65 + 0.0312

0.6188 < p < 0.6812

90% CI is ( 0.6188 , 0.6812 )