Fashion Industries randomly tests its employees throughout t

Fashion Industries randomly tests its employees throughout the year. Last year in the 440 random tests conducted, 22 employees failed the test. (Use Student\'s z Distribution Table.)

Develop a 98% confidence interval for the proportion of employees that fail the test. (Round your answers to 3 decimal places.)

Would it be reasonable to conclude that less than 5% of the employees are not able to pass the random drug test?

Solution

1)
Z FOR 98% CONFIDNCE INTERVAL = 2.33
p = 22/440 = 0.05

CI : (p- z* sqrt(p(1-p)/n) , p+ z* sqrt(p(1-p)/n) )
= (0.05- 2.33* sqrt(0.05(1-0.05)/440),0.05+ 2.33* sqrt(0.05(1-0.05)/440))
=(0.026 , 0.074 ) Answer

2)

No , because upper limit is greater than 0.05