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likoan [24]
2 years ago
13

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.
Mathematics
1 answer:
const2013 [10]2 years ago
7 0

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).

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