91Ó°ÊÓ

(a)

Given:

\(\begin{array}{l}{{\bar x}_1} = 80.95\\{s_1} = 9.59\\{{\bar x}_2} = 63.23\\{s_2} = 5.96\\{n_1} = {n_2} = 10\\c = 95\% = 0.95\end{array}\)

Determine the t-value by looking in the row starting with degrees of freedom $d\(f = n - 1 = 10 - 1 = 9\)and in the column with\(\alpha = 1 - c = 0.05\)in the table of the Student's T distribution:

\({t_\alpha } = 1.833\)

The margin of error is then:

\(E = {t_{\alpha /2}} \times \frac{s}{{\sqrt n }} = 1.833 \times \frac{{5.96}}{{\sqrt {10} }} \approx 3.4547\)

The boundaries of the confidence interval then become:

\(\bar x - E = 63.23 - 3.4547 = 59.7753\)

Thus the 95 % lower confidence bound is 59.7753.

(b)

Determine the t-value by looking in the row starting with degrees of freedom \(df = n - 1 = 10 - 1 = 9\) and in the column with \(\alpha = 1 - c = 0.05\) in the table of the Student's T distribution:

\({t_\alpha } = 1.833\)

The margin of error is then:

\(E = {t_{\alpha /2}} \times s\sqrt {1 + \frac{1}{n}} = 1.833 \times 5.96\sqrt {1 + \frac{1}{{10}}} \approx 11.4579\)

The boundaries of the prediction interval then become:

\(\bar x - E = 63.23 - 11.4579 = 51.7721\)

Thus the 95 % lower prediction bound is 51.7721.

(c)

Determine the t-value by looking in the row starting with degrees of freedom\(df = n - 1 = 10 - 1 = 9\)and in the column with\(\alpha = 1 - c/2 = 0.025\)in the table of the Student's T distribution:

\({t_\alpha } = 2.262\)

The margin of error is then:

\(E = {t_{\alpha /2}} \times \frac{s}{{\sqrt n }} = 2.262 \times \frac{{5.96}}{{\sqrt {10} }} \approx 4.2632\)

The boundaries of the confidence interval then become:

\(\begin{array}{l}\bar x - E = 80.95 - 4.2632 = 76.6868\\\\\bar x + E = 80.95 + 4.2632 = 85.2132\end{array}\)

(d)

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

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{9.5{9^2}}}{{10}} + \frac{{5.9{6^2}}}{{10}}} \right)}^2}}}{{\frac{{{{\left( {9.5{9^2}/10} \right)}^2}}}{{10 - 1}} + \frac{{{{\left( {5.9{6^2}/10} \right)}^2}}}{{10 - 1}}}} \approx 15\)

Determine the t-value by looking in the row starting with degrees of freedom df=15 and in the column with\(\alpha = 1 - c/2 = 0.025\) in the Student's t distribution table in the appendix:

\({t_{\alpha /2}} = 2.131\)

The margin of error is then:

\(E = {t_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} = 2.131 \times \sqrt {\frac{{9.5{9^2}}}{{10}} + \frac{{5.9{6^2}}}{{10}}} \approx 7.6089\)

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

\(\begin{array}{l}\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E = (80.95 - 63.23) - 7.6089 = 17.72 - 7.6089 = 10.1111\\\\\left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E = (80.95 - 63.23) + 7.6089 = 17.72 + 7.6089 = 25.3289\end{array}\)