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 498 floppy disks is drawn. Of these disks, 34 were defective. Using the data, construct the 90% confidence interval for the population proportion of disks which are defective. Round your answers to three decimal places.

Answer 1

Solution :

Given that,

n = 498

x = 34

= x / n = 34 /498 = 0.068

1 - = 1 - 0.068= 0.932

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * (((0.068 * 0.932) / 498) = 0.019

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.068 - 0.019 < p < 0.068 + 0.019

0.049 < p < 0.087

The 90% confidence interval for the population proportion p is : ( 0.049 , 0.087)