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A normal probability plot of the n=\({\rm{26}}\) observations on escape time shows a substantial linear pattern; the sample mean and sample standard deviation are \({\rm{370}}{\rm{.69 and 24}}{\rm{.36}}\), respectively.

a. Calculate an upper confidence bound for population mean escape time using a confidence level of \({\rm{95\% }}\)

b. Calculate an upper prediction bound for the escape time of a single additional worker using a prediction level of \({\rm{95\% }}\). How does this bound compare with the confidence bound of part (a)?

c. Suppose that two additional workers will be chosen to participate in the simulated escape exercise. Denote their escape times by \({\rm{X27 and X28}}\), and let X new denote the average of these two values. Modify the formula for a PI for a single x value to obtain a PI for X new, and calculate a 95% two-sided interval based on the given escape data.

Short Answer

Expert verified

a) The upper boundary of the confidence interval then becomes:

\({\rm{\bar x + E = 370}}{\rm{.69 + 8}}{\rm{.1598 = 378}}{\rm{.8498}}\)

b) The upper boundary of the prediction interval then becomes:

\({\rm{\bar x + E = 370}}{\rm{.69 + 42}}{\rm{.3995 = 413}}{\rm{.0895}}\)

Thus we note that the upper prediction bound is higher than then upper confidence bound.

c) The confidence interval becomes\((333.87,407.51)\)

Step by step solution

01

To Calculate an upper confidence bound

Given:

\(\begin{array}{l}{\rm{n = 26}}\\{\rm{\bar x = 370}}{\rm{.69}}\\{\rm{s = 24}}{\rm{.36}}\\{\rm{c = 95\% = 0}}{\rm{.95}}\end{array}\)

Determine the t-value by looking in the row starting with degrees of freedom \({\rm{df = n - 1 = 26 - 1 = 25}}\)and in the column with \({\rm{\alpha = 1 - c = 0}}{\rm{.05}}\) in the table of the critical values for t distributions in the appendix:

\({{\rm{t}}_{\rm{\alpha }}}{\rm{ = 1}}{\rm{.708}}\)

The margin of error is then:

\({\rm{E = }}{{\rm{t}}_{\rm{\alpha }}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{ = 1}}{\rm{.708 \times }}\frac{{{\rm{24}}{\rm{.36}}}}{{\sqrt {{\rm{26}}} }}{\rm{\gg 8}}{\rm{.1598}}\)

Hence The upper boundary of the confidence interval then becomes:

\({\rm{\bar x + E = 370}}{\rm{.69 + 8}}{\rm{.1598 = 378}}{\rm{.8498}}\)

02

To Calculate an upper prediction bound

Given:

\(\begin{array}{l}{\rm{n = 26}}\\{\rm{\bar x = 370}}{\rm{.69}}\\{\rm{s = 24}}{\rm{.36}}\\{\rm{c = 95\% = 0}}{\rm{.95}}\end{array}\)

Upper confidence bound found in part (a): \(378.8498\)

UPPER PREDICTION BOUND

Determine the t-value by looking in the row starting with degrees of freedom \({\rm{df = n - 1 = 26 - 1 = 25}}\) and in the column with \({\rm{\alpha = 1 - c = 0}}{\rm{.05}}\)in the table of the critical values for t distributions in the appendix:

\({{\rm{t}}_{{\rm{\alpha /2}}}}{\rm{ = 1}}{\rm{.708}}\)

The margin of error is then:

\({\rm{E = }}{{\rm{t}}_{{\rm{\alpha /2}}}}{\rm{ \times s}}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{n}}}} {\rm{ = 1}}{\rm{.708 \times 24}}{\rm{.36}}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{26}}}}} {\rm{\gg 42}}{\rm{.3995}}\)

Hence The upper boundary of the prediction interval then becomes:

\({\rm{\bar x + E = 370}}{\rm{.69 + 42}}{\rm{.3995 = 413}}{\rm{.0895}}\)

Thus we note that the upper prediction bound is higher than the upper confidence bound.

03

Step 3:To  calculate a \({\rm{95\% }}\)two-sided interval

(c):

The new random variable is

\({{\rm{\bar X}}_{{\rm{new }}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {{{{\rm{\bar X}}}_{{\rm{27}}}}{\rm{ + }}{{{\rm{\bar X}}}_{{\rm{28}}}}} \right)\)

In order to obtain PI for a single value, standard deviation of random variable

\({\rm{\bar X - }}{{\rm{\bar X}}_{{\rm{new }}}}\)

is needed. When the variance is computed, the t statistic can be obtained.

The variance is

\(\begin{array}{l}{\rm{V}}\left( {{\rm{\bar X - }}{{{\rm{\bar X}}}_{{\rm{new }}}}} \right){\rm{ }}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{V(\bar X) + ( - 1}}{{\rm{)}}^{\rm{2}}}{\rm{V}}\left( {{{{\rm{\bar X}}}_{{\rm{new }}}}} \right){\rm{ = V(\bar X) + V}}\left( {\frac{{\rm{1}}}{{\rm{2}}}\left( {{{{\rm{\bar X}}}_{{\rm{27}}}}{\rm{ + }}{{{\rm{\bar X}}}_{{\rm{28}}}}} \right)} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{V(\bar X) + }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{V}}\left( {{{{\rm{\bar X}}}_{{\rm{27}}}}} \right){\rm{ + }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{V}}\left( {{{{\rm{\bar X}}}_{{\rm{28}}}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(3)}}} \frac{{\rm{1}}}{{\rm{n}}}{{\rm{\sigma }}^{\rm{2}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{4}}}{{\rm{\sigma }}^{\rm{2}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{4}}}{{\rm{\sigma }}^{\rm{2}}}{\rm{ = }}{{\rm{\sigma }}^{\rm{2}}}\left( {\frac{{\rm{1}}}{{\rm{n}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}} \right)\end{array}\)

(1): random variables are independent,

(2): random variables are independent and \({\rm{V(cX) = }}{{\rm{c}}^{\rm{2}}}{\rm{V(X)}}\),

(3) : random variables are from the same distribution with variance \({{\rm{\sigma }}^{\rm{2}}}\)and the result for variance of \({\rm{\bar X}}\)is known.

This indicates that the random variable

\({\rm{T = }}\frac{{{\rm{\bar X - }}{{{\rm{\bar X}}}_{{\rm{new }}}}}}{{\sqrt {{{\rm{s}}^{\rm{2}}}\left( {\frac{{\rm{1}}}{{\rm{n}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}} \right)} }}\)

has student's t distribution with \({\rm{n - 1}}\)degrees of freedom, where the sample standard deviation is an estimate of the standard deviation.

Therefore, the two sided prediction interval is

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

The values are known from the exercise are

\(\begin{array}{l}{\rm{x = 370}}{\rm{.69}}\\{\rm{s = 24}}{\rm{.36,n = 26}}\end{array}\)

The value of \({{\rm{t}}_{{\rm{\alpha /2,n - 1}}}}{\rm{ = }}{{\rm{t}}_{{\rm{0}}{\rm{.025,25}}}}{\rm{ = 2}}{\rm{.06}}\)can be found in the appendix of the book or computed by PC, where \({\rm{\alpha /2 = 0}}{\rm{.025}}\), because the \({\rm{95\% }}\)two-sided interval needs to be computed and

\(\begin{array}{l}{\rm{100(1 - \alpha ) = 95 }}\\{\rm{\alpha = 0}}{\rm{.05}}\end{array}\)

and \({\rm{n - 1 = 26 - 1 = 25}}\)degrees of freedom.

The confidence interval becomes

\(\begin{array}{l}\left( {{\rm{\bar x - }}{{\rm{t}}_{{\rm{\alpha /2,n - 1}}}}{\rm{ \times s \times }}\sqrt {\frac{{\rm{1}}}{{\rm{n}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}} {\rm{,\bar x + }}{{\rm{t}}_{{\rm{\alpha /2,n - 1}}}}{\rm{ \times s \times }}\sqrt {\frac{{\rm{1}}}{{\rm{n}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}} } \right)\\{\rm{ = }}\left( {{\rm{370}}{\rm{.69 - 2}}{\rm{.06 \times 24}}{\rm{.36 \times }}\sqrt {\frac{{\rm{1}}}{{{\rm{26}}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}} {\rm{,370}}{\rm{.69 + 2}}{\rm{.06 \times 24}}{\rm{.36 \times }}\sqrt {\frac{{\rm{1}}}{{{\rm{26}}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}} } \right)\\{\rm{ = (370}}{\rm{.69 - 36}}{\rm{.82,370}}{\rm{.69 + 36}}{\rm{.82)}}\\{\rm{ = (333}}{\rm{.87,407}}{\rm{.51)}}\end{array}\)

Hence The confidence interval becomes\((333.87,407.51)\)

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