A qualityconscious disk manufacturer wishes to know the frac

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.

Step 2 of 2: 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.

Solution

let p be the population proportion of disks which are defective.

step 1 of 2: we have a sample of size=n=985. out of these 917 were not defective.

so number of defective disks=985-917=68

so the sample proportion of disks which are defective=p\"=68/985

now the population proportion of defective disks p is estimated by the sample proportion of defective disks p\" as

p\" serves as an unbiased estimator of p

so the estimate of proportion of disks which are defective is 68/985 [answer]

step 2 of 2:   we have E[p\"]=p    and V[p\"]=p(1-p)/n

since here n=985 is very large hence by central limit theorem the distribution of p\" can be approximated by a Normal distribution with mean p and variance p(1-p)/n

so p\"~N(p,p(1-p)/n)

here the objective is to construct a 85% confidence interval for p.

so here alpha=0.15

from the distribution of p\"

(p\"-p)/sqrt[p(1-p)/n]~N(0,1)

so P[-tao_{alpha/2=0.075}<(p\"-p)/sqrt[p(1-p)/n]<tao_{alpha/2=0.075}]=1-alpha=1-0.15=0.85

where tao_alpha/2 is the upper alpha/2 point of a N(0,1) distribution.

or, P[p\"-tao_0.075*sqrt[p(1-p)/n]<p<p\"+tao_0.075*sqrt[p(1-p)/n]]=0.85

hence the 85% confidene interval for p is

[p\"-tao_0.075*sqrt[p(1-p)/n],p\"+tao_0.075*sqrt[p(1-p)/n]]

here p is unknown. so it is estimated by p\"

hence the confidence interval is [p\"-tao_0.075*sqrt[p\"(1-p\")/n],p\"+tao_0.075*sqrt[p\"(1-p\")/n]]

now p\"=68/985=0.069 n=985 tao_0.075=1.43953

hence the confiden interval for the population proportion of disks which are defective is

[0.069-1.43953*sqrt[0.069(1-0.069)/985],0.069+1.43953*sqrt[0.069(1-0.069)/985]]

=[0.057,0.081] [answer]