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(a) Given:

\(T = \frac{{(\bar X - \bar Y) - \left( {{\mu _1} - {\mu _2}} \right)}}{{{S_p}\sqrt {\frac{1}{m} + \frac{1}{n}} }}\)

The t-value is the test statistic decreased by the mean, divided by the standard deviation:

\(\begin{array}{l}E(\bar X - \bar Y) = {\mu _1} - {\mu _2}\\V(\bar X - \bar Y) = {S_p}\sqrt {\frac{1}{m} + \frac{1}{n}} \end{array}\)

The boundaries of the confidence interval are then the test statistic decreased/increased by the product of the critical t-value and the variance of the test statistic:

\((\bar X - \bar Y) \pm {t_{\alpha /2}}{S_p}\sqrt {\frac{1}{m} + \frac{1}{n}} \)

The mean is the sum of all values divided by the number of values:

\(\begin{array}{l}{{\bar x}_1} = \frac{{14.0 + 14.3 + 12.2 + 15.1}}{4} \approx 13.9\\{{\bar x}_2} = \frac{{12.1 + 13.6 + 11.9 + 11.2}}{4} \approx 12.2\end{array}\)

The variance is the sum of squared deviations from the mean divided by\(n - 1\). The standard deviation is the square root of the variance: \(\begin{array}{l}{s_1} = \sqrt {\frac{{{{(14.0 - 13.9)}^2} + \ldots . + {{(14.0 - 13.9)}^2}}}{{4 - 1}}} \approx 1.2247\\{s_2} = \sqrt {\frac{{{{(12.1 - 12.2)}^2} + \ldots . + {{(11.2 - 12.2)}^2}}}{{4 - 1}}} \approx 1.0100\end{array}\)

Determine the pooled standard deviation:

\({s_p} = \sqrt {\frac{{\left( {{n_1} - 1} \right)s_1^2 + \left( {{n_2} - 1} \right)s_2^2}}{{{n_1} + {n_2} - 2}}} = \sqrt {\frac{{(4 - 1){{1.2247}^2} + (4 - 1){{1.0100}^2}}}{{4 + 4 - 2}}} \approx 1.1225\)

Determine the t-value by looking in the row starting with degrees of freedom \(df = {n_1} + {n_2} - 2 = 4 + 4 - 2 = 6\) and in the column with \(1 - \) \(c/2 = 0.025\) in the Student's t distribution table in the appendix:

\({t_{\alpha /2}} = 2.447\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot {s_p}\sqrt {\frac{1}{{{n_1}}} + \frac{1}{{{n_2}}}} = 2.447 \cdot 1.1225\sqrt {\frac{1}{4} + \frac{1}{4}} \approx 1.9423\)

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

\(\begin{array}{l}\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E = (13.9 - 12.2) - 1.9423 = 1.7 - 1.9423 = - 0.2423\\\left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E = (13.9 - 12.2) + 1.9423 = 1.7 + 1.9423 = 3.6423\end{array}\)

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

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{{{1.2247}^2}}}{4} + \frac{{{{1.0100}^2}}}{4}} \right)}^2}}}{{\frac{{{{\left( {{{1.2247}^2}/4} \right)}^2}}}{{4 - 1}} + \frac{{{{\left( {{{1.0100}^2}/4} \right)}^2}}}{{4 - 1}}}} \approx 5\)

Determine the t-value by looking in the row starting with degrees of freedom\(df = 5\)and in the column with\(1 - c/2 = 0.025\)in the Student's distribution table in the appendix:

\({t_{\alpha /2}} = 2.571\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot \sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} = 2.571 \cdot \sqrt {\frac{{{{1.2247}^2}}}{4} + \frac{{{{1.0100}^2}}}{4}} \approx 2.0407\)

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

\(\begin{array}{l}\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E = (13.9 - 12.2) - 2.0407 = 1.7 - 2.0407 = - 0.3407\\\left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E = (13.9 - 12.2) + 2.0407 = 1.7 + 2.0407 = 3.7407\end{array}\)

We note that this confidence interval is slightly wider than the confidence interval found in part (b).