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 1536floppy disks is drawn. Of these disks, 1383 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

Solution :

Given that,

n = 1563

x = 1383

= x / n = 1383 /1563 = 0.900

1 - = 1 - 0. 900= 0.100

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

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

= 2.326 * (((0.900 * 0.100) / 1536) = 0.018

A 98 % confidence interval for population proportion p is ,

- E < P < + E

0.900 - 0.018 < p < 0.900 + 0.018

0.882 < p < 0.918

The 98% confidence interval for the population proportion p is : ( 0.882 , 0.918)