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Compute a \(99\% \) \(CI\) for \({\mu _1} - {\mu _2}\)using the data in Exercise \(12\).

Short Answer

Expert verified

Therefore,confidence interval is

\(( - 0.79,0.73)\)

Step by step solution

01

Given.

Given:

Let

\({d_{ij}} = {x_i} - {y_i}\)

where\({x_1},{x_2}, \ldots ,{x_m}\)and\({y_1},{y_2}, \ldots ,{y_n}\)are observed value of continuous distributions that differ only in location but not in shape.

The general form of a\(100(1 - \alpha )\)confidence interval for\({\mu _1} - {\mu _2}\)is

\(\left( {{d_{ij(mn - c + 1)}},{d_{ij(c)}}} \right),\)where\({d_{ij(1)}},{d_{ij(2)}} \ldots ,{d_{ij(mn)}}\) are the ordered differences.

Use Appendix Table A.l6 to determine values of\(c\).
For\(m = 6,n = 8\)and the Table A.16, for\(99.2\% \)confidence interval,\(c = 44\), the confidence interval is in the form of

\(\left( {{d_{ij(mn - c + 1)}},{d_{ij(c)}}} \right) = \left( {{d_{ij(6 \cdot 8 - 44 + 1)}},{d_{ij(44)}}} \right) = \left( {{d_{ij(5)}},{d_{ij(21)}}} \right)\)

Compute all differences, find\({5^{{\rm{th }}}}\)and\({44^{{\rm{th }}}}\)ordered difference values.

The five smallest ordered differences are

\( - 1.04, - 0.99, - 0.83, - 0.82, - 0.79\)

and the five largest ordered differences are

\(1.04,0.92,0.85,0.85,0.73.\)

The\({5^{{\rm{th }}}}\)and\({44^{{\rm{th }}}}\)ordered difference are\( - 0.79\)and\(0.73\), respectively; thus, a\(99.2\% \)confidence interval is

\(\left( {{d_{ij(5)}},{d_{ij(44)}}} \right) = ( - 0.79,0.73)\)

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