A recent survey showed that 63 of US employers were likely t

A recent survey showed that 63% 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. Find the 99% confidence interval for the proportion of all companies likely to require higher employee contributions for health care coverage. (round to 4 decimal places)

Left Endpoint=

Right Endpoint=

Solution

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.63          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.019710403          
              
Now, for the critical z,              
alpha/2 =   0.005          
Thus, z(alpha/2) =    2.33          
Thus,              
              
left endpoint = p^ - z(alpha/2) * sp =   0.58407476          
right endpoint = p^ + z(alpha/2) * sp =    0.67592524   [ANSWERS]