91Ó°ÊÓ

(a):

Normal probability plot suggests that the population distribution is normal. This stands because the data is not far from normal probability line.

Hence it is plausible that the population distribution from this sample was selected is normal

The sample mean, which has \({\rm{n = 13}}\)observations is

\({\rm{\bar x = }}\frac{{\rm{1}}}{{{\rm{13}}}}{\rm{(23 + 39 + \ldots + 72) = }}\frac{{\rm{1}}}{{{\rm{13}}}}{\rm{ \times 679 = 52}}{\rm{.231}}\)

The Sample Variance $s^{2}$ is

\({{\rm{s}}^{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{n - 1}}}}{\rm{ \times }}{{\rm{S}}_{{\rm{xx}}}}\)

where

\({{\rm{S}}_{{\rm{xx}}}}{\rm{ = }}\sum {{{\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - \bar x}}} \right)}^{\rm{2}}}} {\rm{ = }}\sum {{\rm{x}}_{\rm{i}}^{\rm{2}}} {\rm{ - }}\frac{{\rm{1}}}{{\rm{n}}}{\rm{ \times }}{\left( {\sum {{{\rm{x}}_{\rm{i}}}} } \right)^{\rm{2}}}\)

The Sample Standard Deviation $s$ is

First compute \({{\rm{S}}_{{\rm{xx}}}}\) using

\(\sum {{\rm{x}}_{\rm{i}}^{\rm{2}}} {\rm{ - }}\frac{{\rm{1}}}{{\rm{n}}}{\rm{ \times }}{\left( {\sum {{{\rm{x}}_{\rm{i}}}} } \right)^{\rm{2}}}\)

where

\(\sum {{{\rm{x}}_{\rm{i}}}} {\rm{ = 679}}\)

which implies that

\(\frac{{\rm{1}}}{{\rm{n}}}{\left( {\sum {{{\rm{x}}_{\rm{i}}}} } \right)^{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{13}}}}{\rm{ \times 67}}{{\rm{9}}^{\rm{2}}}{\rm{ = 35464}}{\rm{.69}}\)

From the squared observations \(\left( {{\rm{x}}_{\rm{i}}^{\rm{2}}} \right)\), calculated below

\({\rm{529,1521,1600,1681,1849,2209,2601,3364,3969,4356,4489,4761,5184, }}\)

the following is true

\(\sum {{\rm{x}}_{\rm{i}}^{\rm{2}}} {\rm{ = 529 + 1521 + \ldots + 5184 = 38113}}\)

Now \({{\rm{S}}_{{\rm{xx}}}}\)becomes

\({{\rm{S}}_{{\rm{xx}}}}{\rm{ = }}\sum {{\rm{x}}_{\rm{i}}^{\rm{2}}} {\rm{ - }}\frac{{\rm{1}}}{{\rm{n}}}{\rm{ \times }}{\left( {\sum {{{\rm{x}}_{\rm{i}}}} } \right)^{\rm{2}}}{\rm{ = 38113 - 35464}}{\rm{.69 = 2648}}{\rm{.31}}\)

Finally, The Sample Variance is

\({{\rm{s}}^{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{n - 1}}}}{{\rm{S}}_{{\rm{zx}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{12}}}}{\rm{ \times 2648}}{\rm{.31 = 220}}{\rm{.69}}\)

And The Sample Standard Deviation is

\({\rm{s = }}\sqrt {{{\rm{s}}^{\rm{2}}}} {\rm{ = }}\sqrt {{\rm{220}}{\rm{.69}}} {\rm{ = 14}}{\rm{.856}}\)

The tolerance interval is

\({\rm{\bar x \pm ( tolererance critical value) \times s}}\)

by substituting the values, the interval becomes

\({\rm{(52}}{\rm{.231 - 3}}{\rm{.081 \times 14}}{\rm{.856,52}}{\rm{.231 + 3}}{\rm{.081 \times 14}}{\rm{.856) = (6}}{\rm{.46,98)}}\)

Hence the interval becomes \(\left( {6.46,98} \right)\)

(c):

The prediction interval can be computed using formula

\({\rm{\bar x \pm }}{{\rm{t}}_{{\rm{\alpha /2,n - 1}}}}{\rm{ \times s}}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{n}}}} \)

The only value missing is

\({{\rm{t}}_{{\rm{\alpha /2,n - 1}}}}{\rm{ = }}{{\rm{t}}_{{\rm{0}}{\rm{.025,12}}}}{\rm{ = 2}}{\rm{.179}}\)

which can be found in the appendix of the book. Value $\alpha=0.05$ is obtained from

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

because a \(95\% \)prediction interval is needed.

Therefore, the prediction interval is

\(\left( {{\rm{52}}{\rm{.231 - 2}}{\rm{.179 \times 14}}{\rm{.856 \times }}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{13}}}}} {\rm{,52}}{\rm{.231 + 2}}{\rm{.179 \times 14}}{\rm{.856 \times }}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{13}}}}} } \right){\rm{ = (18}}{\rm{.638,85}}{\rm{.824)}}\)The width of the prediction interval is smaller than the width of the tolerance interval.