Suppose a sample of 1475 floppy disks is drawn. Of these disks, 1328 were not defective. Using the data, construct the 80%confidence interval for the population proportion of disks which are defective. Round your answers to three decimal places. Lower:
Upper:

The 80% confidence interval estimate of the true population proportion of disks which are defective is 0.100 +/- 0.010

= (0.090, 0.110)

Lower: 0.090

Upper: 0.110

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

p+/-z√(p(1-p)/n)

Given that;

Proportion of disks which are not defective p' = 1328/1475 = 0.900

Proportion of disks which are defective p = 1 - p' = 1 - 0.900 = 0.100

p = 0.100

Number of samples n = 1475

Confidence interval = 80%

z value(at 80% confidence) = 1.28

Substituting the values we have;

0.100 +/- 1.28√(0.100(1-0.100)/1475)

0.100 +/- 1.28(0.007811334658)

0.100 +/- 0.009998508363

0.100 +/- 0.010

= (0.090, 0.110)

The 80% confidence interval estimate of the true population proportion of disks which are defective is 0.100 +/- 0.010

= (0.090, 0.110)

Lower: 0.090

Upper: 0.110

step-by-step explanation:

a

step-by-step explanation: