Question

A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective. Step 2 of 2 : Suppose a sample of 884 floppy disks is drawn. Of these disks, 831 were not defective. Using the data, construct the 98% confidence interval for the population proportion of disks which are defective. Round your answers to three decimal places.

Answer 1

Answer:

CI = ( 0, 042 ;  0,079 )

Step-by-step explanation:

If from 884 , 831  disks were OK  then  884  - 831  = 53 were defective

x  =  53

p = 53/884    p  = 0,0599     or p = 5,99 %   p ≈ 6 %

and q =  1  - 0,06       q = 0,94

p*n  and q*n   are big enough

For CI 98 %  significance level is α = 2 %   α/2 = 0,01

z(score) for α/2 = 0,01 is

z(c) ≈ 2,324

CI  =  p ±  z(c) *√p*q/n

√p*q/n   =  √ ( 0,94*0,06)/884

√p*q/n   =  √ 0,0564/884

√p*q/n   =  0,00799

CI  =  ( 0,06  ±  2,324*0,00799)

CI  =  ( 0,06  - 0,0185 ; 0,06 +  0,0185 )

CI = ( 0, 042 ;  0,079 )