Question

A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective.

a. Suppose a sample of 865 floppy disks is drawn. Of these disks, 112 were defective. Using the data, estimate the proportion of disks which are defective.
b. Suppose a sample of 865 floppy disks is drawn. Of these disks, 112 were defective. Using the data, construct the 95% confidence interval for the population proportion of disks which are defective.

Answer 1

a) 0.1295

b) The 95% confidence interval for the population proportion of disks which are defective is (0.1071, 0.1519).

Explanation:

Question a:

112 out of 865, so:

`\pi = (112)/(865) = 0.1295`

Question b:

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 z-score that has a p-value of
`1 - (\alpha)/(2)`.

For this problem, we have that:

`n = 865, \pi = 0.1295`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a p-value of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.1295 - 1.96\sqrt{(0.1295*0.8705)/(865)} = 0.1071`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.1295 + 1.96\sqrt{(0.1295*0.8705)/(865)} = 0.1519`

The 95% confidence interval for the population proportion of disks which are defective is (0.1071, 0.1519).