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

Answer:

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

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

1536 - 1383 = 153

This means that $n = 1536, \pi = \frac{153}{1536} = 0.1$

98% confidence level

So $\alpha = 0.02$, z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$, so $Z = 2.327$.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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