Question

A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective.

Step 1 of 2:

Suppose a sample of 884 floppy disks is drawn. Of these disks, 796 were not defective. Using the data, estimate the proportion of disks which are defective. Enter your answer as a fraction or a decimal number rounded to three decimal places.

Step 2 of 2:

Suppose a sample of 884 floppy disks is drawn. Of these disks, 796 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 = 884

x = 796

Point estimate = sample proportion = = x / n = 796/884=0.900

1 - = 1-0.900=0.1

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

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

= 2.326 (((0.900*0.1) /884 )

E = 0.023

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.900 -0.023 < p < 0.900+0.023

0.877< p < 0.923

The 98% confidence interval for the population proportion p is : 0.877, 0.923