A quality-conscious disk manufacturer wishes to know the fraction of disks his c

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.

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.

Explanation / Answer

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[-taoalpha/2=0.075<(p"-p)/sqrt[p(1-p)/n]<taoalpha/2=0.075]=1-alpha=1-0.15=0.85

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

or, P[p"-tao0.075*sqrt[p(1-p)/n]<p<p"+tao0.075*sqrt[p(1-p)/n]]=0.85

hence the 85% confidene interval for p is

[p"-tao0.075*sqrt[p(1-p)/n],p"+tao0.075*sqrt[p(1-p)/n]]

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

hence the confidence interval is [p"-tao0.075*sqrt[p"(1-p")/n],p"+tao0.075*sqrt[p"(1-p")/n]]

now p"=68/985=0.069 n=985 tao0.075=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]

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