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 1358

floppy disks is drawn. Of these disks, 108

were defective. Using the data, construct the 99%

confidence interval for the population proportion of disks which are defective. Round your answers to three decimal places.

Answer 1

Solution :

Given that,

n = 1358

x = 108

= x / n =108 / 1358 = 0.080

1 - = 1 - 0.080 = 0.920

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 * (((0.080 * 0.920) / 1358) = 0.019

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.080 - 0.019 < p < 0.080 + 0.019

0.061 < p < 0.099

The 99% confidence interval for the population proportion p is : (0.061 , 0.099)