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 985 floppy disks is drawn. Of these disks, 917 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.

Suppose a sample of 985 floppy disks is drawn. Of these disks, 917 were not defective. Using the data, construct the 85% confidence interval for the population proportion of disks which are defective. Round your answers to three decimal places.

Answer 1

Proportion of disks which are defective = Number of disks which are defective / Total number of disks

Total number of disks = 985

Number of disks which are defective = 985 - 917 = 68

Proportion of disks which are defective = p = 68 / 985 = 0.069

Now, we find the confidence interval for the population proportion of disks which are defective using the formula :

$Upper\ limit : p + Z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}}$

$Lower\ limit : p - Z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}}$

We have p = 0.069

n = 985

$\alpha = 1-0.85 = 0.15$

Using Standard Normal Tables we get :

$Z_{alpha/2}=Z_{0.15/2} = Z_{0.075} =1.44$

Putting all the values in the above expression we get ,

$Upper\ limit = 0.069 + 1.44\sqrt{\frac{0.069(1-0.069)}{985}} =0.081$

$Lower\ limit = 0.069 - 1.44\sqrt{\frac{0.069(1-0.069)}{985}} =0.057$

So,  85% confidence interval for the population proportion of disks which are defective is : ( 0.057 , 0.081 )