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 1536 floppy disks is drawn. Of these disks, 1383 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.

The 98% confidence interval for the population proportion of disks which are defective is (0.082, 0.118).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

Suppose a sample of 1536 floppy disks is drawn. Of these disks, 1383 were not defective.

1536 - 1383 = 153

This means that

98% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 98% confidence interval for the population proportion of disks which are defective is (0.082, 0.118).