Step 1
Data values are given population standard deviation is not known hence we use T Interval.
Sample size \(\displaystyle{\left({n}\right)}={5}\)
Degrees of freedom(df):
\(\displaystyle{d}{f}={n}-{1}\)

\(\displaystyle={5}-{1}\)

\(\displaystyle={4}\) Confidence level =0.90 Use T table or excel command =T.INV.2T \(\displaystyle{\left({1}-{0.90},{4}\right)}\) to get critical value Critical value \(\displaystyle{\left({t}_{{{c}}}\right)}={2.132}\) \(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}&{X}&{\left({X}−{12.54}\right)}^{{{2}}}\backslash{h}{l}\in{e}&{12.6}&{0.0036}\backslash{h}{l}\in{e}&{11.9}&{0.4096}\backslash{h}{l}\in{e}&{13}&{0.2116}\backslash{h}{l}\in{e}&{12.7}&{0.0256}\backslash{h}{l}\in{e}&{12.5}&{0.0016}\backslash{h}{l}\in{e}{T}{o}{t}{a}{l}&{62.7}&{0.652}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\) Sample mean: \(\displaystyle\overline{{{x}}}={\frac{{{\sum_{{{i}={1}}}^{{5}}}\xi}}{{{n}}}}\)

\(\displaystyle={\frac{{{62.7}}}{{{5}}}}\)

\(\displaystyle{12.54}\) Sample standard deviation: \(\displaystyle{s}=\sqrt{{{\frac{{{\sum_{{{i}={1}}}^{{5}}}{\left({x}_{{{i}}}-\overline{{{x}}}\right)}^{{{2}}}}}{{{n}-{1}}}}}}\)

\(\displaystyle=\sqrt{{{\frac{{{\sum_{{{i}={1}}}^{{5}}}{\left({x}_{{{i}}}-{12.54}\right)}^{{{2}}}}}{{{5}-{1}}}}}}\)

\(\displaystyle=\sqrt{{{\frac{{{0.652}}}{{{4}}}}}}\)

\(\displaystyle=\sqrt{{{0.163}}}\)

\(\displaystyle={0.403733}\) Step 3 \(\displaystyle=\overline{{{x}}}\pm{t}_{{{c}}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

\(\displaystyle={12.54}\pm{2.132}\times{\frac{{{0.403733}}}{{\sqrt{{{5}}}}}}\)

\(\displaystyle={12.54}\pm{2.132}\times{0.180555}\)

\(\displaystyle={12.54}\pm{0.3849}\)

\(\displaystyle={\left({12.54}-{0.3849},{12.54}+{0.3849}\right)}\)

\(\displaystyle={\left({12.1551},{12.9249}\right)}\) Step 4 Answer: 90% confidence interval is ( 12.16, 12.92 )

\(\displaystyle={5}-{1}\)

\(\displaystyle={4}\) Confidence level =0.90 Use T table or excel command =T.INV.2T \(\displaystyle{\left({1}-{0.90},{4}\right)}\) to get critical value Critical value \(\displaystyle{\left({t}_{{{c}}}\right)}={2.132}\) \(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}&{X}&{\left({X}−{12.54}\right)}^{{{2}}}\backslash{h}{l}\in{e}&{12.6}&{0.0036}\backslash{h}{l}\in{e}&{11.9}&{0.4096}\backslash{h}{l}\in{e}&{13}&{0.2116}\backslash{h}{l}\in{e}&{12.7}&{0.0256}\backslash{h}{l}\in{e}&{12.5}&{0.0016}\backslash{h}{l}\in{e}{T}{o}{t}{a}{l}&{62.7}&{0.652}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\) Sample mean: \(\displaystyle\overline{{{x}}}={\frac{{{\sum_{{{i}={1}}}^{{5}}}\xi}}{{{n}}}}\)

\(\displaystyle={\frac{{{62.7}}}{{{5}}}}\)

\(\displaystyle{12.54}\) Sample standard deviation: \(\displaystyle{s}=\sqrt{{{\frac{{{\sum_{{{i}={1}}}^{{5}}}{\left({x}_{{{i}}}-\overline{{{x}}}\right)}^{{{2}}}}}{{{n}-{1}}}}}}\)

\(\displaystyle=\sqrt{{{\frac{{{\sum_{{{i}={1}}}^{{5}}}{\left({x}_{{{i}}}-{12.54}\right)}^{{{2}}}}}{{{5}-{1}}}}}}\)

\(\displaystyle=\sqrt{{{\frac{{{0.652}}}{{{4}}}}}}\)

\(\displaystyle=\sqrt{{{0.163}}}\)

\(\displaystyle={0.403733}\) Step 3 \(\displaystyle=\overline{{{x}}}\pm{t}_{{{c}}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

\(\displaystyle={12.54}\pm{2.132}\times{\frac{{{0.403733}}}{{\sqrt{{{5}}}}}}\)

\(\displaystyle={12.54}\pm{2.132}\times{0.180555}\)

\(\displaystyle={12.54}\pm{0.3849}\)

\(\displaystyle={\left({12.54}-{0.3849},{12.54}+{0.3849}\right)}\)

\(\displaystyle={\left({12.1551},{12.9249}\right)}\) Step 4 Answer: 90% confidence interval is ( 12.16, 12.92 )