Step 1

\(\begin{array}{|c|c|}\hline Test-1(X) & Test-2(Y) & (x-\bar{x}) & (x-\bar{x})^{2} & (y-\bar{y}) & (y-\bar{y})^{2} \\ \hline 1.2 & 1.4 & -0.31111 & 0.09679 & -0.24444 & 0.05975 \\ \hline 1.6 & 1.7 & 0.08889 & 0.00790 & 0.05556 & 0.00309 \\ \hline 1.5 & 1.5 & -0.01111 & 0.00012 & -0.14444 & 0.02086 \\ \hline 1.4 & 1.3 & -0.11111 & 0.01235 & -0.34444 & 0.11864 \\ \hline 1.8 & 2.0 & 0.28889 & 0.08346 & 0.35556 & 0.12642 \\ \hline 1.8 & 2.1 & 0.28889 & 0.08346 & 0.45556 & 0.20753 \\ \hline 1.4 & 1.7 & -0.11111 & 0.01235 & 0.05556 & 0.00309 \\ \hline 1.5 & 1.6 & -0.01111 & 0.00012 & -0.04444 & 0.00198\\ \hline 1.4 & 1.5 & -0.11111 & 0.01235 & -0.14444 & 0.02086\\ \hline \end{array}\)

Step 2

The mean is given by

\(\displaystyle\overline{{{x}}}={\frac{{{1}}}{{{n}}}}\sum{x}\)

Test 1

\(\displaystyle\overline{{{x}}}={\frac{{{13.6}}}{{{9}}}}={1.511}\)

Test 2

\(\displaystyle\overline{{{y}}}={\frac{{{14.8}}}{{{9}}}}={1.644}\)

The standard deviation is given by

\(\displaystyle\sigma=\sqrt{{{\frac{{{1}}}{{{n}-{1}}}}\sum{\left({x}-\overline{{{x}}}\right)}^{{{2}}}}}\)

Test 1

\(\displaystyle\sigma=\sqrt{{{\frac{{{0.30889}}}{{{8}}}}}}={0.1965}\)

Test 2

\(\displaystyle{s}_{{{2}}}=\sqrt{{{\frac{{{0.5622}}}{{{8}}}}}}={0.2651}\)

The \(\displaystyle{95}\%\) CI is given by

\(\displaystyle{C}.{I}.={\left(\overline{{{x}}}-\overline{{{y}}}\right)}\pm{t}_{{{0.05},\ {8}}}\sqrt{{{\frac{{{{s}_{{{1}}}^{{{2}}}}}}{{{n}_{{{1}}}}}}+{\frac{{{{s}_{{{2}}}^{{{2}}}}}}{{{n}_{{{2}}}}}}}}\)

\(\displaystyle{C}.{I}.={\left({1.511}-{1.644}\right)}\pm{2.306}\sqrt{{{\frac{{{\left({0.1965}\right)}^{{{2}}}}}{{{9}}}}+{\frac{{{\left({0.2651}\right)}^{{{2}}}}}{{{9}}}}}}\)

\(\displaystyle{C}.{I}.={\left[-{0.387},\ {0.120}\right]}\)

It shows that the true difference between the mean of test 1 and test 2 will lie within -0.387 and 0.120.