Answer:
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).