91影视

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

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

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

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