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Given:

\(\begin{array}{l}n = 16\\\bar x = 0.52\\s = 0.26\end{array}\)

Let us assume that we want to calculate the confidence interval with\(95\% \)confidence (other confidence levels work similarly).

\(c = 95\% = 0.95\)Determine the t-value by looking in the row starting with degrees of freedom\(df = n - 1 = 16 - 1 = 15\)and in the column with\(1 - c/2 = \)\(0.025\)in the table of the Student's T distribution:

\({t_\alpha } = 2.131\)

The margin of error is then:

\(E = {t_{\alpha /2}} \times \frac{s}{{\sqrt n }} = 2.13 \times \frac{{0.26}}{{\sqrt {16} }} \approx 0.13845\)

The boundaries of the confidence interval then become: \(\begin{array}{l}\bar x - E = 0.52 - 0.13845 = 0.38155\\\bar x + E = 0.52 + 0.13845 = 0.65845\end{array}\)

\(\begin{array}{l}{{\bar x}_1} = 0.52\\{{\bar x}_2} = 0.35\\{n_1} = 16\\{n_2} = 8\\{s_1} = 0.26\\{s_2} = 0.15\end{array}\)

Let us assume that we want to calculate the confidence interval with\(95\% \)confidence (other confidence levels work sim

\(c = 95\% = 0.95\)

Determine the degrees of freedom (rounded down to the nearest integer): \(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{{{0.26}^2}}}{{16}} + \frac{{{{0.15}^2}}}{8}} \right)}^2}}}{{\frac{{{{\left( {{{0.26}^2}/16} \right)}^2}}}{{16 - 1}} + \frac{{{{\left( {{{0.15}^2}/8} \right)}^2}}}{{8 - 1}}}} \approx 21\)

Determine the t-value by looking in the row starting with degrees of freedom \(df = 21\) and in the column with \(1 - c/2 = 0.025\) in the Student's \(t\) distribution table in the appendix:

\({t_{\alpha /2}} = 2.080\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot \sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} = 2.080 \cdot \sqrt {\frac{{{{0.26}^2}}}{{16}} + \frac{{{{0.15}^2}}}{8}} \approx 0.1745\)

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

\(\begin{array}{l}\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E = (0.52 - 0.35) - 0.1745 = 0.17 - 0.1745 = - 0.0045\\\left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E = (0.52 - 0.35) + 0.1745 = 0.172 + 0.1745 = 0.3445\end{array}\)

The confidence interval suggest that there is no difference in the true average AN IAT compared to the true average control IAT, because the confidence interval contains \(0\)