Question

A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective. Step 1 of 2: Suppose a sample of 7002 floppy disks is drawn. Of these disks, 6232 were not defective. Using the data, estimate the proportion of disks which are defective. Enter your answer as a fraction or a decimal number rounded to three decimal places. Step 2 of 2: Suppose a sample of 7002 floppy disks is drawn. Of these disks, 6232 were not defective. Using the data, construct the 99% confidence interval for the population proportion of disks which are defective. Round your answers to three decimal places.

Answer 1

Step 1 :

Sample Size : n=7002

Number of non-defective disks =6232

Number of defective disks : x = 7002-6232=770

Sample proportion $\widehat{p}$ = x/n = 770/ 7002 = 0.110

Estimated proportion of disks which are defective : Sample propotion : $\widehat{p}$ = 0.110

Step 2 :

Confidence interval for population proportion :

$=\widehat{p}\pm z_{\alpha /2}\sqrt{(\widehat{p}(1-\widehat{p}))/n}$

$\alpha$ = (100-99)/100 = 0.01; $\alpha$/2 = 0.005

99% Confidence interval =

$=\widehat{p}\pm z_{0.005}\sqrt{(\widehat{p}(1-\widehat{p}))/n}$

from standard normal tables z_0.005 = 2.58

99% Confidence interval for population proportion of disks which ar defective=

$=\widehat{p}\pm z_{0.005}\sqrt{(\widehat{p}(1-\widehat{p}))/n}$

$=0.11\pm 2.58\sqrt{(0.11(1-0.11))/7002}$

$=0.11\pm 2.58\sqrt{(0.110\times 0.89)/7002}=0.11\pm 2.58\sqrt{\frac{0.0979}{7002}}$

$=0.11\pm 2.58\times 0.0037 = 0.11\pm 0.0096 =(0.1004, 0.1196)$

99% Confidence interval for population proportion of disks which ar defective= (0.1004, 0.1196)