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 815 floppy disks is drawn. Of these disks, 701 were not defective. Using the data, construct the 90% confidence interval for the population proportion of disks which are defective. Round your answers to three decimal places.

Answer 1

Answer: (0.120,0.160)

Step-by-step explanation:

Given : Sample size : $n=815$

Number of disks were not defective =701

Then , the number of disks which are defective = $815-701=114$

Now, the proportion of disks which are defective : $p=\dfrac{114}{815}\approx0.14$

Significance level : $\alpha: 1-0.9=0.1$

Critical value : $z_{\alpha/2}=1.645$

The confidence interval for population proportion is given by :-

$p\pm\ z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}\\\\=0.14\pm(1.645)\sqrt{\dfrac{(0.14)(1-0.14)}{815}}\\\\\approx0.14\pm0.020=(0.14-0.020,0.14+0.020)=(0.120,0.160)$

Hence, the 90% confidence interval for the population proportion of disks which are defective = (0.120,0.160)