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 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.

Answer 1

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

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 1536 floppy disks is drawn. Of these disks, 1383 were not defective.

1536 - 1383 = 153

This means that
`n = 1536, \pi = (153)/(1536) = 0.1`

98% confidence level

So
`\alpha = 0.02`, z is the value of Z that has a pvalue of
`1 - (0.02)/(2) = 0.99`, so
`Z = 2.327`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.1 - 2.327\sqrt{(0.1*0.9)/(1536)} = 0.082`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.1 + 2.327\sqrt{(0.1*0.9)/(1536)} = 0.118`

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