Question

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 1067 floppy disks is drawn. Of these disks, 74 were 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.

Answer 1

The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

Suppose a sample of 1067 floppy disks is drawn. Of these disks, 74 were defective.

This means that
`n = 1067, \pi = (74)/(1067) = 0.069`

80% confidence level

So
`\alpha = 0.2`, z is the value of Z that has a pvalue of
`1 - (0.2)/(2) = 0.9`, so
`Z = 1.28`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.069 - 1.28\sqrt{(0.069*0.931)/(1067)} = 0.059`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.069 + 1.28\sqrt{(0.069*0.931)/(1067)} = 0.079`

The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).