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

CI = ( 0, 042 ; 0,079 )

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 )