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Chapter 7: Q34E (page 302)

A sample of 14 joint specimens of a particular type gave a sample mean proportional limit stress of \({\rm{8}}{\rm{.48}}\)MPa and a sample standard deviation of . \({\rm{79}}\)MPa a. Calculate and interpret a \({\rm{95\% }}\)lower confidence bound for the true average proportional limit stress of all such joints. What, if any, assumptions did you make about the distribution of proportional limit stress?

b. Calculate and interpret a \({\rm{95\% }}\)lower prediction bound for the proportional limit stress of a single joint of this type.

Short Answer

Expert verified

a) The \({\rm{100(1 - \alpha )\% }}\)confidence interval is

\({\rm{(8}}{\rm{.11, + \currency)}}\)

b) The \({\rm{100(1 - \alpha )\% }}\)confidence interval is

\({\rm{(7}}{\rm{.03, + \currency)}}\)

Step by step solution

01

To Calculate and interpret a \({\rm{95\% }}\)lower confidence bound for the true average

The following is given in the exercise

\(\begin{array}{*{20}{l}}{{\rm{n = 16}}}&{{\rm{ joint specimens, }}}\\{{\rm{\bar x = 8}}{\rm{.48}}}&{{\rm{ sample mean, }}}\\{{\rm{s = 0}}{\rm{.79}}}&{{\rm{ sample standard deviation}}{\rm{. }}}\end{array}\)

(a):

Upper confidence bound for \({\rm{\mu }}\)and lower confidence bound for \({\rm{\mu }}\)are given by

\(\begin{array}{l}{\rm{\bar x + }}{{\rm{t}}_{{\rm{\alpha ,n - 1}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{ upper bound, }}\\{\rm{\bar x - }}{{\rm{t}}_{{\rm{\alpha ,n - 1}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{ lower bound, }}\end{array}\)

with confidence level of \({\rm{100(1 - \alpha )\% }}\). The distribution the random sample is taken from is normal.

A $95 \%$ lower confidence bound can be computed using formula given above, but with the assumption about the sample distribution assume that the sample is taken from a normally distributed population

Therefore, the lower confidence bound is

\({\rm{\bar x - }}{{\rm{t}}_{{\rm{\alpha ,n - 1}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{ = 8}}{\rm{.48 - 1}}{\rm{.771 \times }}\frac{{{\rm{0}}{\rm{.79}}}}{{\sqrt {{\rm{14}}} }}{\rm{ = 8}}{\rm{.48 - 0}}{\rm{.37 = 8}}{\rm{.11}}{\rm{.}}\)

The \({{\rm{t}}_{{\rm{\alpha ,n - 1}}}}\)is obtained from the table in appendix for \({\rm{n - 1 = 14 - 1 = 13}}\)and $\alpha=0.05$, where \({\rm{\alpha }}\)is computed from

\({\rm{100(1 - \alpha ) = 95}}{\rm{.}}\)

Hence The \({\rm{100(1 - \alpha )\% }}\)confidence interval is

\({\rm{(8}}{\rm{.11, + \currency)}}\)

02

To Calculate and interpret a \({\rm{95\% }}\)lower prediction bound

b)

For the single observation, the formula for lower confidence bound is

\({\rm{\bar x - }}{{\rm{t}}_{{\rm{\alpha ,n - 1}}}}{\rm{ \times s \times }}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{n}}}} \)

Therefore, the lower bound for the proportional limit stress of a single joint of this type is

\(\begin{array}{l}{\rm{\bar x - }}{{\rm{t}}_{{\rm{\alpha ,n - 1}}}}{\rm{ \times s \times }}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{n}}}} {\rm{ \& = 8}}{\rm{.48 - 1}}{\rm{.771 \times 0}}{\rm{.79 \times }}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{14}}}}} \\{\rm{ = 8}}{\rm{.48 - 1}}{\rm{.45 = 7}}{\rm{.03}}{\rm{.}}\end{array}\)

The \({{\rm{t}}_{{\rm{\alpha ,n - 1}}}}\)is obtained from the table in appendix for $n-1=14-1=13$ and $\alpha=0.05$, where \({\rm{\alpha }}\)is computed from

\({\rm{100(1 - \alpha ) = 95}}\)

Hence the \({\rm{100(1 - \alpha )\% }}\)confidence interval is

\({\rm{(7}}{\rm{.03, + \currency)}}\)

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Most popular questions from this chapter

A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (229.764, 233.504). You decide that a confidence level of 99% is more appropriate than the 95% level used. What are the limits of the 99% interval? (Hint: Use the center of the interval and its width to determine \(\overline x \) and s.)

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