Fashion Industries randomly tests its employees throughout the year. Last year i

Question

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?

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.)

Explanation / Answer

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

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