An assembly line does a quality check by sampling 50 of its products. It finds t

Question

An assembly line does a quality check by sampling 50 of its products. It finds that 16% of the parts are defective. a. Create a 95% confidence interval for the percent of defective parts for the company and 4. interpret this interval b. If we decreased the confidence level to 90% what would happen to: i. the critical value? ii. the margin of error? iii. the confidence interval? If the sample size were increased to 200, the same sample proportion were found, and we did a 95% confidence interval, what would happen to: c. i. the critical value? ii. the margin of error? ii. the confidence interval?

Explanation / Answer

Ans:

Given that

sample size,n=50

sample proportion=0.16

a)critical z value=1.96

95% confidence interval for proportion if defectives

=0.16+/-1.96*sqrt(0.16*(1-0.16)/50)

=0.16+/-0.102

=(0.058,0.262)

b)i)if confidence level is 90%,crititcal z value will decrease to 1.645

ii)margin of error=1.645*sqrt(0.16*(1-0.16)/50)=0.085 (decreased)

iii)confidence interval=(0.16-0.085, 0.16+0.085)=(0.075, 0.245), narrower

c)Now,if sample size,n=200

i)critical value remains same,as confidence level is same.

ii)margin of error=1.96*sqrt(0.16*(1-0.16)/400)=0.051 (reduced to half)

iii)confidence interval=(0.16-0.051, 0.16+0.051)=(0.109, 0.211)

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