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 1996 floppy disks is drawn. Of these disks, 319 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 = 1996

x = 319

= x / n = 319 /1996 = 160

1 - = 1 - 0.160= 0.0.840

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 * (((0.160 * 0.840) / 1996) = 0.021

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.160 - 0.021 < p < 0.160 + 0.021

0.139 < p < 0.181

The 99% confidence interval for the population proportion p is : ( 0.139 , 0.181)