Confidence intervals of population mean when is known Past

(Confidence intervals of population mean ? when ? is known.) Past experience has indecated that the breaking strength of yarn used in manufacturing drapery material is normally distributed and that ? = 2 psi. A random sample of nine specimens is tested, and the average breaking strength is found to be 92.514 psi.

(a) Construct 90%, 95%, and 99% two-sided confidence intervals on the true mean breaking strength ?. Compare the length of these intervals, tell me your conclusion.

(b) Give an interpretation of the 95% confidence interval on the true mean breaking strength ?.

(c) Construct a 95% lower one-sided confidence interval on the true mean breaking strength ? and

give an interpretation.

(d) Construct a 95% upper one-sided confidence interval on the true mean breaking strength ? and give an interpretation.

(e) If the scientists want the confidence interval to be no wider than 0.5 psi, how many observations should they take? (Note that, the width of the confidence interval is two times the margin error E, so E = 0.5/2).

Solution

(a) Given a=1-0.9=0.1, Z(0.05) = 1.645 (from standard normal table)

So the lower bound is

xbar - Z*s/vn =92.514 -1.645*2/sqrt(9) =91.41733

So the upper bound is

xbar + Z*s/vn=92.514 +1.645*2/sqrt(9)=93.61067

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Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So the lower bound is

xbar - Z*s/vn =92.514 -1.96*2/sqrt(9) =91.20733

So the upper bound is

xbar + Z*s/vn=92.514 +1.96*2/sqrt(9)=93.82067

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Given a=1-0.99=0.01, Z(0.005) = 2.58 (from standard normal table)

So the lower bound is

xbar - Z*s/vn =92.514 -2.58*2/sqrt(9) =90.794

So the upper bound is

xbar + Z*s/vn=92.514 +2.58*2/sqrt(9)=94.234

99% two-sided confidence intervals is the largest.

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(b) We have 95% confident that the population mean will be within this interval.

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(c) Given a=1-0.95 =0.05, Z(0.05) = 1.645 (from standard normal table)

So the lower bound is

xbar - Z*s/vn =92.514 -1.645*2/sqrt(9) =91.41733

We have 95% confident that the population mean will greater than this value.

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(d)Given a=1-0.95 =0.05, Z(0.05) = 1.645 (from standard normal table)

So the upper bound is

xbar + Z*s/vn =92.514 +1.645*2/sqrt(9) =93.61067

We have 95% confident that the population mean will less than this value.

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(e) E = 0.5/2 =0.25

n=(Z*s/E)^2

=(1.96*2/0.25)^2

=245.8624

Take n=246