Question

Suppose a sample of 382 floppy disks is drawn. Of these disks, 363 were not 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 :

363 were not defective .

So, 382 - 363 = 19 are defective .

Given that,

n = 382

x = 19

= x / n = 0.050

1 - = 1 - 0.050 = 0.95

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 * (((0.050 * 0.95) / 382)

= 0.019

A 90% confidence interval for population proportion p is ,

- E < P < + E

0.050 - 0.019 < p < 0.050 + 0.019

0.031 < p < 0.068

(0.031 , 0.068)