91Ó°ÊÓ

\(\begin{array}{l}{{\bar x}_1} = 33.4\\{{\bar x}_2} = 42.8\\{n_1} = 26\\{n_2} = 26\\{s_1} = 2.2\\{s_2} = 4.3\\c = 99\% = 0.99\end{array}\)

Determine the degrees of freedom (rounded down to the nearest integer):

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{{{2.2}^2}}}{{26}} + \frac{{{{4.3}^2}}}{{26}}} \right)}^2}}}{{\frac{{{{\left( {{{2.2}^2}/26} \right)}^2}}}{{26 - 1}} + \frac{{{{\left( {{{4.3}^2}/26} \right)}^2}}}{{26 - 1}}}} \approx 37 > 36\)

Determine the\(t\)-value by looking in the row starting with degrees of freedom\(df = 36\)and in the column with\(1 - c/2 = 0.005\)in the Student's\(t\)distribution table in the appendix:

\({t_{\alpha /2}} = 2.719\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot \sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} = 2.719 \cdot \sqrt {\frac{{{{2.2}^2}}}{{26}} + \frac{{{{4.3}^2}}}{{26}}} \approx 2.5756\)

The endpoints of the confidence interval for\({\mu _1} - {\mu _2}\)are:

\(\begin{array}{l}\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E = (33.4 - 42.8) - 2.5756 = - 9.4 - 2.5756 = - 11.9756\\\left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E = (33.4 - 42.8) + 2.5756 = - 9.4 + 2.5756 = - 6.8244\end{array}\)

We are \(99\% \)confident that the difference between the true average load for the fiberglass beams and that for the carbon beams is between \( - 11.9756\) and \( - 6.8244\)

The upper confidence bound from part (a) is a \(99.5\% \) confidence bound instead of a \(99\% \) confidence bound.

\(\begin{array}{l}{{\bar x}_1} = 33.4\\{{\bar x}_2} = 42.8\\{n_1} = 26\\{n_2} = 26\\{s_1} = 2.2\\{s_2} = 4.3\\c = 99\% = 0.99\end{array}\)

Determine the degrees of freedom (rounded down to the nearest integer):

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{{{2.2}^2}}}{{26}} + \frac{{{{4.3}^2}}}{{26}}} \right)}^2}}}{{\frac{{{{\left( {{{2.2}^2}/26} \right)}^2}}}{{26 - 1}} + \frac{{{{\left( {{{4.3}^2}/26} \right)}^2}}}{{26 - 1}}}} \approx 37 > 36\)

Determine the t-value by looking in the row starting with degrees of freedom \(df = 36\) and in the column with \(1 - c = 0.01\)in the Student's \(t\) distribution table in the appendix: \({t_{\alpha /2}} = 2.434\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot \sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} = 2.434 \cdot \sqrt {\frac{{{{2.2}^2}}}{{26}} + \frac{{{{4.3}^2}}}{{26}}} \approx 2.3056\)

The upper confidence bound of the confidence interval for \({\mu _1} - {\mu _2}\) is then:

\(\left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E = (33.4 - 42.8) + 2.3056 = - 9.4 + 2.3056 = - 7.0944\)

There is sufficient evidence that the true average load for the carbon beams is more than that for the fiberglass beams, because the upper confidence bound is smaller than \(0.\)