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 \(\displaystyle{\left({t}_{{{c}}}\right)}={2.132}\)

\(\begin{array}{|c|c|} \hline & X & (X−12.54)^{2} \\ \hline & 12.6 & 0.0036 \\ \hline & 11.9 & 0.4096 \\ \hline & 13 & 0.2116 \\ \hline & 12.7 & 0.0256 \\ \hline & 12.5 & 0.0016 \\ \hline Total & 62.7 & 0.652 \\ \hline \end{array}\)

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 )