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 1475 floppy disks is drawn. Of these disks, 147 were defective. Using the data, construct the 80% confidence interval for the population proportion of disks which are defective. Round your answers to three decimal places.

Answer 1

Solution :  

Given that,

n = 1475

x = 147

Point estimate = sample proportion = = x / n = 147 / 1475 = 0.100

1 - = 1 - 0.100 = 0.9

At 80% confidence level the z is ,

= 1 - 80% = 1 - 0.8 = 0.2

/ 2 = 0.2 / 2 = 0.1

Z_/2 = Z_0.1 = 1.282

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

= 1.282 * (((0.100 * 0.9) / 1475)

= 0.010

A 80% confidence interval for population proportion p is ,

- E < p < + E

0.100 - 0.010 < p < 0.100 + 0.010

0.090 < p < 0.110

The 80% confidence interval for the population proportion p is : 0.090 , 0.110