Suppose the survey was based on a sample of 700 companies. Find the 80% confidence interval for the proportion of all companies likely to require higher employee contributions for health care coverage. Find the LEFT endpoint and RIGHT endpoint.

The 80% confidence interval would be given by (0.586;0.634)

Step-by-step explanation:

A recent survey showed that 61% 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. Find the 80% confidence interval for the proportion of all companies likely to require higher employee contributions for health care coverage.

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 80% of confidence, our significance level would be given by and . And the critical value would be given by:

The confidence interval for the mean is given by the following formula:  

If we replace the values obtained we got:

The 80% confidence interval would be given by (0.586;0.634)

step-by-step explanation:

given perimeter: 28cm

length: 9cm

atp,

2(l+w)=p of the reactangle

2(9+x)=28

18+2x=28

2x=28-18

x=10/2

x=5

therefore width is 5 cm

central angle of a triangle has the same measure as its intercepted arc.

step-by-step explanation: