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

Answer:

a) 0.1295

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

Step-by-step explanation:

Question a:

112 out of 865, so:

$\pi = \frac{112}{865} = 0.1295$

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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

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 $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.  

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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