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 700 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

The 95% confidence interval estimate of the true population proportion of U.S. employers that were likely to require higher employee contributions for health care coverage is 0.52 +/- 0.0370

= (0.4830, 0.5570)

The margin of error M.E = 0.0370

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

p+/-z√(p(1-p)/n)

p+/-M.E

Given that;

M.E = margin of error

Proportion p = 52% = 0.52

Number of samples n = 700

Confidence interval = 95%

z value (at 95% confidence) = 1.96

Substituting the values we have;

0.52 +/- 1.96√(0.52(1-0.52)/700)

0.52 +/- 1.96(0.0189)

0.52 +/- 0.0370

( 0.4830, 0.5570)

The 95% confidence interval estimate of the true population proportion of U.S. employers that were likely to require higher employee contributions for health care coverage is 0.52 +/- 0.0370

= (0.4830, 0.5570)

The margin of error M.E = 0.0370

i assume that by 5 you mean 5 gallons.

(20)(0.25) = 15

removing 15 gallons reduces the tank from 20 to 5.

exactly you go it

step-by-step explanation: