Question

A certain industry wants to know the proportion of defective parts it manufactures. For this, a...

A certain industry wants to know the proportion of defective parts it manufactures. For this, a random sample of 50 pieces was observed, in which 9 pieces showed defect.

a) Determine the point estimate for the proportion of defective parts in that industry.
b) Obtain the 82% confidence interval for the proportion of defective parts. Use a conservative approach. Interpret the interval found.

c) Was it necessary to make any assumptions to obtain the above confidence interval? Explain.
d) What sample size is necessary to ensure with an 82% confidence that the estimated error of this ratio is at most 0.06?

Homework Answers

Answer #1

a. Point estimate for the proportion is

b. z value for 82% CI is 1.341 as P(-1.341<z<1.341)=0.82

So Margin of Error is

Hence CI is

So population proportion of defective parts will lie in the range of 0.107 to 0.253

c. We need to make a assumption that population is normal

d. Here E=0.06 and z value is 1.341

So n from E is

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