Question

A survey of 120 randomly selected employees of a large insurance company shows that the sample proportion of employees which feel secure about their job is 0.35 and the standard error of this estimate is 0.044. Construct a 99% confidence interval for the proportion of all this company’s employees who feel secure about their jobs. Round your bounds to 2 decimal places.

Answer 1

p=proportion of employees which feel secure about their job out of selected sample (n)

ie P=0.35 n=120 SE=0.044

for 95% confidence level = (100-95)/100 = 0.01

/2 = 0.005

Z_/2 = Z_0.005= 2.58

#value of z is obtain from standard normal table  

99% confidence interval for the proportion of all the employees of dot com companies who feel secure about their jobs

(0.2365,0.4635)

#(0.24,0.46) is 99% confidence interval for the proportion