Question

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 1222 floppy disks is drawn. Of these disks, 1076 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.

Find Upper and Lower Endpoints

Answer 1

Solution :

Given that,

n = 1222

x = 1076

Point estimate = sample proportion = = x / n = 1076 / 1222 = 0.881

1 - = 1 - 0.881 = 0.119

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 * (((0.881*0.119) / 1222)

= 0.022

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.881 - 0.022 < p < 0.881 + 0.022

0.859 < p < 0.903

Upper Endpoint = 0.903

Lower Endpoint = 0.859