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, 1677 were not 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 = 1677

Point estimate = sample proportion = = x / n = 1677 / 1996 = 0.840

1 - = 1 - 0.840 = 0.16

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 * (((0.840 * 0.16) / 1996)

= 0.021

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.840 - 0.021 < p < 0.840 + 0.021

0.819 < p < 0.861

The 99% confidence interval for the population proportion p is : 0.819 , 0.861