Step 1

\(\displaystyle{A}{100}{\left({1}−\alpha\right)}\%\) confidence interval on the difference is

\(\displaystyle\overline{{x}}_{{1}}-\overline{{x}}_{{2}}-{t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}{s}{p}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}\le\mu_{{1}}-\mu_{{2}}\le\overline{{x}}_{{1}}-\overline{{x}}_{{2}},+{t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}{s}{p}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}\)

where \(\displaystyle{s}_{{p}}=\sqrt{{\frac{{{\left({n}_{{1}}-{1}\right)}{{s}_{{1}}^{{2}}}+{\left({n}_{{2}}-{1}\right)}{{s}_{{2}}^{{2}}}}}{{{n}_{{1}}+{n}_{{2}}-{1}}}}}\)

Step 2

Considering 95% confidence interval,

\(\displaystyle\alpha={0.05}\)

\(\displaystyle\overline{{x}}_{{1}}={4.8}\)

\(\displaystyle\overline{{x}}_{{2}}={5.6}\)

\(\displaystyle{{s}_{{1}}^{{2}}}={8.64}\)

\(\displaystyle{{s}_{{2}}^{{2}}}={7.88}\)

\(\displaystyle{n}_{{1}}={10}\)

\(\displaystyle{n}_{{2}}={10}\)

Therefore

\(\displaystyle{t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}={t}_{{\frac{{0.05}}{{2}},{10}+{10}-{1}}}\)

\(\displaystyle={t}_{{{0.025},{19}}}\)

=2.093

Step 3

To calculate \(\displaystyle{s}_{{p}}\)

\(\displaystyle{s}_{{p}}=\sqrt{{\frac{{{\left({n}_{{1}}-{1}\right)}{{s}_{{1}}^{{2}}}+{\left({n}_{{2}}-{1}\right)}{{s}_{{2}}^{{2}}}}}{{{n}_{{1}}+{n}_{{2}}-{1}}}}}\)

\(\displaystyle{s}_{{p}}=\frac{\sqrt{{{\left({10}-{1}\right)}{8.64}+{\left({10}-{1}\right)}{7.88}}}}{{{19}}}{)}\)

\(\displaystyle{s}_{{p}}={2.797}\)

Step 4

Now, by substituting all the values in the equation, we get

\(\displaystyle\overline{{x}}_{{1}}-\overline{{x}}_{{2}}-{t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}{s}{p}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}\le\mu_{{1}}-\mu_{{2}}\le\overline{{x}}_{{1}}-\overline{{x}}_{{2}}\)

\(\displaystyle={t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}{s}{p}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}\)

\(\displaystyle{4.8}-{5.6}-{2.093}\times{2.797}\times{0.447}\le\mu_{{1}}-\mu_{{2}}\le{4.8}-{5.6}+{2.093}\times{2.797}\times{0.447}-{3.424}\le\mu_{{1}}-\mu_{{2}}\le{1.82}\)

Hence 95% confidence interval in \(\displaystyle\mu_{{1}}-\mu_{{2}}{i}{s}-{3.424}\le\mu_{{1}}-\mu_{{2}}\le{1.82}\)

\(\displaystyle{A}{100}{\left({1}−\alpha\right)}\%\) confidence interval on the difference is

\(\displaystyle\overline{{x}}_{{1}}-\overline{{x}}_{{2}}-{t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}{s}{p}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}\le\mu_{{1}}-\mu_{{2}}\le\overline{{x}}_{{1}}-\overline{{x}}_{{2}},+{t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}{s}{p}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}\)

where \(\displaystyle{s}_{{p}}=\sqrt{{\frac{{{\left({n}_{{1}}-{1}\right)}{{s}_{{1}}^{{2}}}+{\left({n}_{{2}}-{1}\right)}{{s}_{{2}}^{{2}}}}}{{{n}_{{1}}+{n}_{{2}}-{1}}}}}\)

Step 2

Considering 95% confidence interval,

\(\displaystyle\alpha={0.05}\)

\(\displaystyle\overline{{x}}_{{1}}={4.8}\)

\(\displaystyle\overline{{x}}_{{2}}={5.6}\)

\(\displaystyle{{s}_{{1}}^{{2}}}={8.64}\)

\(\displaystyle{{s}_{{2}}^{{2}}}={7.88}\)

\(\displaystyle{n}_{{1}}={10}\)

\(\displaystyle{n}_{{2}}={10}\)

Therefore

\(\displaystyle{t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}={t}_{{\frac{{0.05}}{{2}},{10}+{10}-{1}}}\)

\(\displaystyle={t}_{{{0.025},{19}}}\)

=2.093

Step 3

To calculate \(\displaystyle{s}_{{p}}\)

\(\displaystyle{s}_{{p}}=\sqrt{{\frac{{{\left({n}_{{1}}-{1}\right)}{{s}_{{1}}^{{2}}}+{\left({n}_{{2}}-{1}\right)}{{s}_{{2}}^{{2}}}}}{{{n}_{{1}}+{n}_{{2}}-{1}}}}}\)

\(\displaystyle{s}_{{p}}=\frac{\sqrt{{{\left({10}-{1}\right)}{8.64}+{\left({10}-{1}\right)}{7.88}}}}{{{19}}}{)}\)

\(\displaystyle{s}_{{p}}={2.797}\)

Step 4

Now, by substituting all the values in the equation, we get

\(\displaystyle\overline{{x}}_{{1}}-\overline{{x}}_{{2}}-{t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}{s}{p}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}\le\mu_{{1}}-\mu_{{2}}\le\overline{{x}}_{{1}}-\overline{{x}}_{{2}}\)

\(\displaystyle={t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}{s}{p}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}\)

\(\displaystyle{4.8}-{5.6}-{2.093}\times{2.797}\times{0.447}\le\mu_{{1}}-\mu_{{2}}\le{4.8}-{5.6}+{2.093}\times{2.797}\times{0.447}-{3.424}\le\mu_{{1}}-\mu_{{2}}\le{1.82}\)

Hence 95% confidence interval in \(\displaystyle\mu_{{1}}-\mu_{{2}}{i}{s}-{3.424}\le\mu_{{1}}-\mu_{{2}}\le{1.82}\)