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

Answer:

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

Step-by-step 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.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

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

This means that $n = 1067, \pi = \frac{74}{1067} = 0.069$

80% confidence level

So $\alpha = 0.2$, z is the value of Z that has a pvalue of $1 - \frac{0.2}{2} = 0.9$, so $Z = 1.28$.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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